CBSE Term II Mathematics X (Standard) 93 Example 12. From the top of a 7 m building, the angle Given, speed of plane is 720 km/h and time of flight is 10 s. of elevation of the top of a cable tower is 60° and Also, given ∠AOC = 60° and ∠BOD = 30° the angle of depression of its foot is 45°. Determine In right angled ΔOCA, the height of the tower. (use 3 = 1.73). cot 60° = OC AC [CBSE 2020 (Standard)] Sol. Let AB = 7 m be the height of the building and DE = h m ⇒ 1 = OC 3h be the height of cable tower. E ⇒ OC = h m 3 B 60º C In right angled ΔODB, 45º hm cot 30° = OD BD 7m ⇒ 3 = OD h ⇒ OD = h 3 Now, CD = OD − OC A 45º D =h 3− h 2h m xm = 33 Then, ∠CBE = 60° (alternate angle) Thus, distance covered by aeroplane in 10 s is 2h m. and ∠CBD = 45° ⇒ ∠ADB = 45° … (i) 3 Let distance between two towers be Q Speed of aeroplane = Distance Time AD = BC = x m In right angled ΔBCE, tan 60° = CE 2h BC ∴ 720 × 5 = 3 ⎢⎡⎣Q 1 km = 5 m / s⎤⎦⎥ ⇒ 3 = CE ⇒ x = CE 18 10 18 x3 ⇒ 40 × 5 × 10 3 = 2h and in right angled ΔADB, ⇒ h = 1000 3 m Hence, height at which the aeroplane is flying is 1000 3 m. tan 45° = AB ⇒1 = 7 ⇒ x = 7 m AD x Put x = 7 in Eq. (i), we get Example 14. A straight highway leads to the foot of a 7 = CE ⇒ CE = 7 3 m 3 tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, Now, height of cable tower, which is approaching the foot of the tower with a uniform speed. After covering a distance of 50 m, h = DC + CE the angle of depression of the car becomes 60°. = 7 + 7 3 = 7(1 + 3) = 7(1 + 1.73) = 7 × 2 .73 = 19.11 m Find the height of the tower. (use 3 = 1.73). Example 13. The angle of elevation of an aeroplane [CBSE 2020 (Standard)] from point O on the ground is 60°. After a flight of Sol. Let AB = h m be the height of the tower. Let C be the initial 10 s, on the same height, the angle of elevation position of the car and D be the final position of the car, when from point O becomes 30°. If the aeroplane is flying it covers a distance, CD = 50 m . at the speed of 720 km/h, find the constant height at which the aeroplane is flying. [CBSE 2020 (Standard)] EB 30° 60° Sol. Let OX be the horizontal ground, A and B be the two h positions of the plane and O be the points of observation. 30° 60° A Let height of an aeroplane from A to the ground is C AC = BD = h m D 50 m x AB h Given, the angle of depressions from point B to the points C h and D are 60º 30º DX ∠EBC = 60° and ∠EBD = 30° OC ⇒ ∠BCA = 60° and ∠BDA = 30° [alternate angles]
94 CBSE Term II Mathematics X (Standard) In right angled ΔCAB, A F tan 60° = AB 45º 60º AC 60 Ö3 m Bx 45º E ⇒ 3=h x …(i) ⇒ x= h m 3 And in right angled ΔDAB, hm tan 30° = AB AD 60º C xD 1h ⇒= Given, the angle of depressions from point A are 3 50 + x ⇒ 50 + x = h 3 ∠FAE = 45° and ∠FAD = 60° ⇒ x = h 3 − 50 …(ii) ⇒ ∠BEA = 45° and ∠CDA = 60°[alternate angles] From Eqs. (i) and (ii), we have In right angled ΔAEB, h = h 3 − 50 3 tan 45° = AB ⇒ 1 = AB BE x ⇒ h = 3h − 50 3 ⇒ x = AB …(i) ⇒ 2h = 50 3 and in right angled ΔADC, tan 60° = AC ⇒ 3 = 60 3 ⇒ x= 60 3 CD x 3 ⇒ h = 25 3 ⇒ h = 25 × 1 .73 ⇒ x = 60 m …(ii) ⇒ h = 43.25 m Hence, height of the tower is 43.25 m. ∴ From Eqs. (i) and (ii), we get AB = x = 60 m Example 15. The angles of depression of the top and Now height of cliff, bottom of a tower as seen from the top of a 60 3 m h = AC − AB high cliff are 45° and 60°, respectively. Find the height of the tower. (use 3 = 1.73) = 60 3 − 60 = 60( 3 − 1) Sol. Let AC = 60 3 m be the height of the cliff, DE = h m be the = 60(1.73 − 1) = 60 × 0.73 = 43.8 m height of the tower and distance between tower and cliff be Hence, height of the cliff is 43.8 m. CD = BE = x m.
CBSE Term II Mathematics X (Standard) 95 Chapter Practice PART 1 7. A ladder, leaning against a wall, makes an angle of Objective Questions 60° with the horizontal. If the foot of the ladder is 9.5 m away from the wall. The length of the ladder is [NCERT Exemplar] (a) 10 m (b) 16 m G Multiple Choice Questions (c) 18 m (d) 19 m 1. The angle of elevation of the Sun when the shadow 8. A ramp for disabled people in a hospital have slope of a pole h m high is 3 h m long is 30°. If the height of the ramp be 1 m, then the (a) 0° (b) 30° (c) 45° (d) 60° length of ramp is (a) 2 m (b) 0.5 m 2. If a pole 6 m high casts a shadow 2 3 m long on the (c) 2 3 m (d) 1 m ground, then the Sun’s elevation is 9. A kite is flying at a height of 80 m above the (a) 60° (b) 45° ground. The string attached to the kite is (c) 30° (d) 90° temporarily tied to a point on the ground. The 3. If 300 3 m high tower makes an angle of elevation inclination of the string with ground is 60°, then the at a point on ground which is 300 m away from its length of the string is foot, then the angle of elevation is (a) 62.37 m (b) 92.37 m (a) 0° (b) 30° (c) 52.57 m (d) 72.57 m (c) 45° (d) 60° 10. The length of a string between a kite and a point on 4. From the top of a 60 m high tower, the angle of the ground is 85 m. If the string makes an angle θ depression of a point on the ground is 30°. The with the ground level such that tan θ = 15, then the 8 distance of the point from the foot of tower is (a) 180 m (b) 60 3 m height of kite is (c) 150 m (d) 30 3 m (a) 75 m (b) 78.05 m 5. The figure shows the observation of point C from (c) 226 m (d) None of these point A. The angle of depression from A is 11. A tower stands near an airport. The angle of [CBSE 2013] elevation θ of the tower from a point on the ground DA is such that its tangent is 5/12. The height of the tower, if the distance of the observer from the tower 4m is 120 m is [CBSE 2015] (a) 40 m (b) 50 m (c) 60 m (d) 70 m C B 12. The top of two poles of height 20 m and 4√3 m (a) 30° (b) 45° (c) 60° (d) 90° 14 m are connected by a wire. If the wire makes an 6. A circus artist is climbing a 20 m long rope, which angle of 30° with the horizontal, then the length of is tightly stretched and tied from the top of a the wire is vertical pole to the ground, then the height of pole, (a) 12 m (b) 10 m (c) 8 m (d) 6 m if the angle made by the rope with the ground level 13. An observer, 1.5 m tall is 20.5 m away from a tower is 30°, is 22 m high, then the angle of elevation of the top of (a) 5 m (b) 10 m the tower from the eye of the observer is (c) 15 m (d) 20 m (a) 30° (b) 45° (c) 60° (d) 90°
96 CBSE Term II Mathematics X (Standard) 14. The angle of elevation of the top of the tower from a (ii) Value of DF is equal to point, which is 40 m away from the base of the (a) h m (b) h 3 m (c) h m (d) h m 3 2 tower in the horizontal level, is 45°. Find the height of the tower. (iii) Value of h is (a) 2 (a) 70 m (b) 60 m (c) 4 (b) 3( 3 + 1) (d) 3( 3 − 1) (c) 40 m (d) 30 m 15. The angle of elevation of the top of a building (iv) Height of the Parachute from the ground is 150 m high, from a point on the ground is 45°. The (a) 4 m (b) 3(4− 3) distance of the point from foot of the building is (c) 8 m (d) 3(4+ 3) (a) 120 m (b) 130 m (c) 140 m (d) 150 m (v) If the Parachute is moving towards the building, 16. The angle of depression of the car parked on the then both angles of elevation will road from the top of a 150 m high tower is 30°. The (a) remain same (b) increases distance of the car from the tower is [CBSE 2014] (c) decreases (d) Can’t be determined (a) 150 m (b) 75 m (c) 150 3 m (d) 150 m 20. A cyclist is climbing through a 20 m long rope 3 which is highly stretched and tied from the top of a vertical pole to the ground as shown below 17. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively, then the height of the tower is (a) 14.64 m (b) 28.64 m (c) 38.64 m (d) 19.64 m 18. A bridge on a river makes an angle of 45° with its edge. If the length along the bridge from one edge Based on the above information, answer the following questions. to the other is 150 m, then the width of the river is (a) 107.75 m (b) 105 m (i) The height of the pole, if angle made by rope with the ground level is 60°, is (c) 75 m (d) 106.05 m G Case Based MCQs (a) 15 m (b) 10 3 m (c) 10 m (d) 15 m 19. There are two balcony in a house. First balcony is at a height of 3 m above the ground and other 3 2 balcony is 6 m vertically above the lower balcony. Ankit and Radha are sitting inside the two balcony (ii) If the angle made by the rope with the ground at points G and F, respectively. At any instant, the angles of elevation of a Parachute from these level is 60°, then the distance between artist and balcony are observed to be 60° and 45° as shown below pole at ground level is E (a) 10 m (b) 10 2 m 2 (c) 10 m (d) 10 3 m (iii) If the angle made by the rope with the ground level is 45°. The height of the pole is F 45° hm (a) 2.5 m (b) 10 m D (c) 7.5 m (d) 10 2 m 6m G 60° C (iv) If the angle made by the rope with the ground 3m B level is 45° and 3 m rope is broken, then the height A of the pole is Based on the above information, answer the (a) 17 m (b) 7 m following questions. 2 (i) Who is more closer to the Parachute. (c) 14 m (d) 7 2 m (a) Ankit (b) Radha (v) Which mathematical concept is used here? (c) Both are at equal distance (d) Can’t be determined (a) Similar triangles (b) Pythagoras theorem (c) Application of trigonometry (d) None of the above
CBSE Term II Mathematics X (Standard) 97 21. A group of students of class X visited India Gate on respectively. If the distance between the peaks of an educational trip. The teacher and students had two mountains is 1937 km, and the satellite is interest in history as well. The teacher narrated that vertically above the mid-point of the distance India Gate, official name Delhi Memorial, between the two mountains. (use 3 = 1.73) originally called All-India War Memorial, monumental sandstone arch in New Delhi, F dedicated to the troops of British India who died in A wars fought between 1914 and 1919.The teacher also said that India Gate, which is located at the G eastern end of the Rajpath (formerly called the H P Kingsway), is about 138 feet (42 m) in height. QR B D CI Nanda devi S Mullayanagiri (i) The distance of the satellite from the top of Nanda Devi is (a) 1139.4 km (b) 1119.65 km (c) 1937 km (d) 1025.36 km (ii) The distance of the satellite from the top of Mullayanagiri is (a) 1139.4 km (b) 577.52 km (i) What is the angle of elevation if they are standing (c) 1937 km (d) 1025.36 km at a distance of 42 m away from the monument? (iii) The distance of the satellite from the ground is (a) 30° (b) 45° (a) 1139.4 km (b) 567.64 km (c) 60° (d) 0° (c) 1937 km (d) 1025.36 km (ii) They want to see the tower at an angle of 60°. So, (iv) What is the angle of elevation, if a man is standing they want to know the distance where they should at a distance of 7816 m from Nanda Devi? stand and hence find the distance. (a) 30° (b) 45° (a) 25.24 m (b) 20.12 m (c) 60° (d) 0° (c) 42 m (d) 24.24 m (v) If a mile stone very far away, makes 45° to the top (iii) If the altitude of the Sun is at 60°, then the height of Mullayanagiri mountain. Hence, find the distance of this mile stone from the mountain. of the vertical tower that will cast a shadow of (a) 1118.327 m (b) 566.976 m length 20 m is (c) 1930 m (d) 1025.36 m (b) 20 m (a) 20 3 m 3 (c) 15 m (d) 15 3 m PART 2 3 Subjective Questions (iv) The ratio of the length of a rod and its shadow is 1 : 1. The angle of elevation of the Sun is (a) 30° (b) 45° G Short Answer Type Questions (c) 60° (d) 90° 1. If the height of a tower and the distance of the point of observation from its foot, both are (v) The angle formed by the line of sight with the increased by 10%, then the angle of elevation of its horizontal when the object viewed is below the top remains unchanged. Explain. horizontal level is 2. A straight tree is broken due to thunderstorm. The (a) corresponding angle broken part is bent in such a way that the peak of (b) angle of elevation the tree touches the ground at an angle of 60° at a (c) angle of depression distance of 2 3 m. Find the whole height of the tree. (d) complete angle 3. Determine the height of a mountain, if the 22. A Satellite flying at height h is watching the top of elevation of its top at an unknown distance from the the two tallest mountains in Uttarakhand and base is 30° and at a distance 10 km farther off from Karnataka ,they being Nanda Devi (height 7,816m) the mountain, along the same line, the angle of and Mullayanagiri (height 1,930 m). The angles of elevation is 15°. (take tan 15° = 0.27) depression from the satellite to the top of Nanda Devi and Mullayanagiri are 30° and 60°,
98 CBSE Term II Mathematics X (Standard) 4. There is a flag staff on a tower of height 20 m. At a 15. The angle of elevation of an aeroplane from a point point on the ground, the angles of elevation of the foot and top of the flag are 45° and 60°, respectively. on the ground is 45°. After flying for 15 s, the angle Find the height of the flag staff. of elevation changes to 30°. If the aeroplane is flying at a constant height of 2500 m, then find the 5. If the length of the shadow of a tower is increasing, average speed of the aeroplane. [CBSE 2013] then the angle of elevation of the Sun is also increasing. Why or why not? 16. The shadow of a flag staff is three times as long as the shadow of the flag staff, when the Sun rays meet the 6. A window in a building is at a height of 10 m from ground at an angle of 60°. Find the angle between the the ground. The angle of depression of a point P on Sun rays and the ground at the time of longer shadow. the ground from the window is 30°. The angle of elevation of the top of the building from the point P 17. An aeroplane, when flying at a height of 4000 m is 60°. Find the height of the building. [CBSE 2007] from the ground, passes vertically above another aeroplane at an instant when the angles of elevation 7. A player sitting on the top of a tower of height 20 m of two planes from the same point on the ground observes the angle of depression of a ball lying on are 60° and 45°, respectively. Find the vertical the ground as 60°. Find the distance between the foot of the tower and the ball. distance between the aeroplanes at that instant. 8. If two towers of height x m and y m subtend angles 18. There is a small island in the middle of a 100 m wide river and a tall tree stands on the island. P and of 30° and 60°, respectively at the centre of the line Q are points directly opposite to each other on two banks and in line with the tree. If the angles of joining their feet, then find x : y. [CBSE 2015] elevation of the top of the tree from P and Q are respectively 30° and 45°, then find the height of the 9. If a man standing on a platform 3 m above the surface of a lake observes a cloud and its reflection tree. [take, 3 = 1.732] in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection. G Long Answer Type Questions State true or false. Justify. 19. An aeroplane is at an altitude of 1200 m. If two 10. From the top of a hill, the angles of depression of ships are sailing towards it in the same direction. two consecutive kilometre stones due East are The angles of depression of the ships as observed found to be 30° and 45°. Find the height of the hill. from the aeroplane are 60° and 30°, respectively. Find the distance between both ships. [CBSE 2015] 20. The angles of depression of two consecutive kilometre 11. The shadow of a tower is 30 m long, when the Sun’s stones on the road on right and left of an aeroplane angle of elevation is 30°. What is the length of the are 60° and 45°, respectively as observed from the shadow, when Sun’s elevation is 60°? aeroplane. Find the height of the aeroplane. 12. Two ships are there in the sea on either side of a 21. The angle of elevation of the top of a tower at a light house in such a way that the ships and the base distance of 120 m from a point A on the ground of the light house are in the same straight line. is 45°. If the angle of elevation of the top of a flag The angle of depression of two ships as observed staff fixed at the top of the tower, at A is 60°, then from the top of the light house are 60° and 45°. find the height of the flag staff. [use, 3 =1.73] If the height of the light house is 200 m, then find the distance between the two ships. [CBSE 2014] [CBSE 2014] 13. From the top of a tower of height 50 m, the angles 22. A balloon is connected to an electric pole. It is inclined at 60° to the horizontal by a cable of length of depression of the top and bottom of a pole are 30° 215 m. Determine the height of the balloon from the ground. Also, find the height of the balloon, if and 45°, respectively. Find [CBSE 2015] the angle of inclination is changed from 60° to 30°. (i) how far the pole is from the bottom of the tower. [CBSE 2015] (ii) the height of the pole. [take, 3 = 1.732] 23. A man in a boat rowing away from a light house 100 m high takes 2 min to change the angle of elevation of 14. A man standing on the deck of a ship, which is 10 m the light house from 60° to 45°. Find the speed of boat. above the water level. He observes that the angle of elevation of the top of a hill is 60° and the angle of depression of the base of the hill is 30°. Calculate the distance of the hill from the ship and height of the hill. [CBSE 2016]
CBSE Term II Mathematics X (Standard) 99 24. The angle of elevation of the top of a tower from the opposite side of the lane are found to be α and β, certain point is 30°. If the observer moves 20 m respectively. Prove that the height of the other house towards the tower, the angle of elevation of the top is h(1 + tan α cot β)m. increases by 15°. Find the height of the tower. 32. The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m 25. The shadow of a tower standing on a level plane is vertically above the lower window. At any instant the found to be 50 m longer when Sun’s elevation is 30° angles of elevation of a balloon from these windows than when it is 60°. Find the height of the tower. are observed to be 60° and 30°, respectively. Find the height of the balloon above the ground. 26. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a G Case Based Questions point on the plane, the angles of elevation of the 33. A girl 8 m tall spots a parrot sitting on the top of a bottom and the top of the flag staff are α and β respectively. Prove that the height of the tower is building of height 58 m from the ground. The angle of elevation of the parrot from the eyes of girl at ⎛ h tan α α ⎠⎞⎟. any instant is 60°. The parrot flies away ⎜⎝ tan β − tan horizontally in such a way that it remained at a constant height from the ground. After 8 s, the 27. The angle of elevation of the top of a tower 30 m angle of elevation of the parrot from the same point high from the foot of another tower in the same is 30°. plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 60° 30°. Find the distance between the two towers and 30° 8 m also the height of the tower. Based on the above information, answer the 28. From the top of a tower h m high, angles of following questions. (Take 3 =1.73)58 m depression of two objects, which are in line with the (i) Find the distance of first position of the parrot foot of the tower are α and β (β > α). Find the distance between the two objects. from the eyes of the girl. (ii) If the distance between the position of parrot 29. A ladder against a vertical wall at an inclination α to increases, then the angle of elevation decreases. the horizontal. Its foot is pulled away from the wall Justify with girl. (iii) Find the distance between the girl and the building. through a distance p, so that its upper end slides a (iv) How much distance covers parrot covers? (v) Find the speed of the parrot in 8s. distance q down the wall and then the ladder makes an angle β with the horizontal. Show that p = cos β − cos α. q sin α − sin β 30. The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower. 31. A window of a house is h m above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on
100 CBSE Term II Mathematics X (Standard) Objective Questions SOLUTIONS 1. (b) Let the angle of elevation of the Sun is θ. 4. (b) Let the distance of the foot of tower from the point be A QR = x m. P 30° h 60 m 30° R Qx Bθ √3h C Given, height of pole = h and height PR = 60 m and ∠PQR = 30° Now, in ΔABC, tan θ = AC = h In ΔPRQ, tan 30° = PR = 60 BC 3h QR x ⇒ 1 = 60 ⇒ x = 60 3 m ⇒ tan θ = 1 = tan 30° 3 3x ⇒ θ = 30° 5. (a) In right angled ΔABC, ∠B = 90° Hence, the angle of elevation of the Sun is 30°. DA θ 2. (a) Let BC = 6 m be the height of the pole and AB = 2 3 m 4m be the length of the shadow on the ground. Let the Sun makes an angle θ on the ground. CB 4√3m C Sun Let ∠DAC = θ Then, ∠ACB = ∠DAC = θ [alternate angles] 6m Now, tan θ = Perpendicular = AB = 4 = 1 = tan 30° θ Base BC 4 3 3 AB ∴ The angle of depression from A is 30°. 2√3 m 6. (b) Let AB be the vertical pole and CA be the 20 m long rope Now, in ΔBAC, tan θ = BC such that its one end A is tied from the top of the vertical pole AB AB and the other end C is tied to a point C on the ground. A ⇒ tan θ = 6 = 3 ⋅ 3 [by rationalising] 23 3 3 ⇒ tan θ = 3 3 = 3 = tan 60°[Q tan 60° = 3] 20 m 3 30° ∴ θ = 60° C Hence, the Sun’s elevation is 60°. 3. (d) Let AB be the tower whose height is 300 3 m, B i.e. AB = 300 3 m. Again, let C be the point at a distance of In ΔABC, we have 300 m from the foot of the tower, i.e. AC = 300 m. sin 30° = AB Here, the angle of elevation is unknown, so let it be θ. AC Since, here base and perpendicular are given. B ⇒ 1 = AB 2 20 300√3 m ⇒ AB = 10 m Hence, the height of the pole is 10 m. θC 7. (d) Let the length of ladder AB = h m. 300 m A B So, in right angled ΔBAC, tan θ = Perpendicular = AB = 300 3 h Base AC 300 ⇒ tan θ = 3 = tan 60° 60° C A 9.5 m ∴ θ = 60° Hence, the required angle of elevation is 60°.
CBSE Term II Mathematics X (Standard) 101 Here, AC = 9. 5 m, ∠BAC = 60° ⇒ sin θ = 15 In ΔABC, cos 60° = AC ⇒ 1 = 9.5 17 AB 2 h B ⇒ h = 2 × 9.5 = 19 m 85 m Hence, length of ladder is 19 m. 8. (a) Let XZ be the length and YZ be the height of the ramp. Z 1m θ C A In ΔABC, sin θ = BC AB 30° X Y ⇒ 15 = BC 17 85 Then, ∠ZXY = 30° and YZ = 1 m ⇒ BC = 75 m In right angled ΔXYZ, sin 30° = Perpendicular = YZ Hence, height of kite is 75 m Hypotenuse XZ 11. (b) Let BC = h m be the height of the tower and A be the ⇒ 1 1 ⇒ XZ = 2 m point on the ground such that, ∠BAC = θ. = C 2 XZ Hence, the length of ramp is 2 m. 9. (b) Let C be the position of the kite and AC be the length of hm the string which makes an angle of 60° with the ground. The height of the kite from the ground is BC = 80 m. θ C A 120 m B Then, AB = 120 m 80 m In right angled ΔABC, tan θ = Perpendicular = BC = h Base AB 120 60° A B 5h ⎣⎡⎢Q tan θ = 5 , given ⎤ ⇒ 12 = 120 12 ⎥⎦ In right angled ΔABC, sin 60° = Perpendicular = BC ⇒ h = 50 m Hypotenuse AC Hence, height of the tower is 50 m. ⇒ 3 = 80 12. (a) Here, CD = 20 m [height of big pole] 2 AC AB = 14 m [height of small pole] AC = 80 × 2 × 3 D ⇒ [by rationalising] 33 = 160 3 = 160 × 1.732 = 92.37 m 6m 33 E 20 m Hence, the length of the string is 92.37 m. B 30° x 10. (a) Given, length of the string of the kite, AB = 85 m 14 m 14 m and tan θ = 15 AC 8 ⇒ cotθ = 8 ∴ DE = CD − CE 15 ⇒ DE = CD − AB [Q AB = CE] Q cosec2θ − 1 = cot2θ ∴ cosec2θ − 1 = 64 ⇒ DE = 20 − 14 = 6 m 225 In ΔBDE, sin 30° = DE ⇒ cosec2θ = 1 + 64 = 289 ⇒ BD 225 225 1 = 6 ⇒ BD = 12 m ⇒ cosec θ = 289 17 2 BD = 225 15 ∴ Length of wire = 12 m
102 CBSE Term II Mathematics X (Standard) 13. (b) Let BE = 22 m be the height of the tower and ⇒ 1 = 150 ⇒ BC = 150 m BC AD = 1. 5 m be the height of the observer. The point D be the observer’s eye. Draw DC ||AB. Hence, the distance of the point from foot of the building is 150 m. E 16. (c) Let AB =150 m be the height of the tower and angle of depression is ∠DAC = 30°. 20.5 m 22 m A D 30º Dθ C 150 m B 1.5 m 20.5 m A 30º B 20.5 m C Then, AB = 20. 5 m = DC Then, ∠ACB = ∠DAC = 30° [alternate angles] and EC = BE − BC = BE − AD Now, in right angled ΔABC, = 22 − 1. 5 = 20. 5 m [Q BC = AD ] tan 30° = Perpendicular = AB Base BC Let θ be the angle of elevation made by observer’s eye with the top of the tower i.e. ∠EDC = θ. ⇒ 1 = 150 3 BC In right angled ΔDCE, tan θ = Perpendicular = CE = 20. 5 ⇒ BC = 150 3 m Base DC 20. 5 Hence, the distance of the car from the tower is 150 3 m. 17. (a) Let the height of the building be, BC = 20 m ⇒ tan θ = 1 ⇒ tan θ = tan 45° and height of the tower be CD = x m ⇒ θ = 45° D 14. (c) In the figure, let AB be a tower which has height h m. The angle of elevation from point C at a distance of 40 m xm from point B is ∠ACB = 45° and BC = 40 m. A C 20 m hm 45° 60° 45° B C A ym 40 m B Let the point A be at a distance y m Then, in right angled ΔABC, from the foot of the building. tan 45° = Perpendicular = AB Now, in ΔABC, BC = tan 45° = 1 Base BC ⇒ AB 20 = 1 ⇒ tan 45° = h ⇒ 1 = h [Q tan 45° = 1] 40 40 y ∴ h = 40 m ⇒ y = 20 m Hence, the height of tower is 40 m. i.e. AB = 20 m Now, in ΔABD, BD = tan 60° = 15. (d) Let AB = 150 m be the height of building and C be a AB point on the ground such that ∠ACB = 45°. 3 ⇒ BD = 3 A AB 150 m ⇒ 20 + x = 3 20 45° ⇒ 20 + x = 20 3 C B ⇒ x = 20 3 − 20 In right angled ΔABC, = 20( 3 − 1) tan 45° = Perpendicular = AB = 20(1.732 − 1) Base BC ⇒ x = 20 × 0.732 = 14.64 m
CBSE Term II Mathematics X (Standard) 103 18. (d) Let BC be the width of the river and A, B be the ends of 20. (i) (b) Let in ΔABC, AC will be rope and AB be a vertical river such that AB = 150 m = Length of the bridge [given] pole. and ∠BAC = 45°. A River B River edge of river 17 m 20 m edge of river 45° AC 60° C B In right angled ΔACB, sin 45° = Perpendicular Then, AC = 20 m, ∠C = 60°, ∠B = 90° Hypotenuse In ΔABC, sin 60° = AB AC ⇒ 1 = BC = BC [Qsin 45° = 1 ] 2 AB 150 2 3 AB = BC = 150 × 2 2 20 ∴ 22 [by rationalising] AB = 10 3 m (ii) (c) cos 60° = BC ⇒ 1 = BC = 150 2 = 75 2 2 AC 2 20 = 75 × 1.414 [Q 2 = 1.414] BC = 10 m = 106.05 m (approx.) (iii) (d) sin 45° = AB [∠C = 45°] Hence, width of the river is 106.05 m. AC 19. (i) (b) Radha is more closer to the Parachute because the 1 = AB ⇒ 20 = AB 2 20 2 angles of elevation of Parachute from these balcony are observed to be 45°. So, Radha is more closer to the ⇒ AB = 20 × 2 [by rationalising] 22 Parachute than Ankit. (ii) (d) In ΔDEF, ∠D = 90°, ∠DFE = 45°, tan 45° = DE = 10 2 m DF (iv) (a) Length of rope = 20 − 3 = 17 m ⇒ 1 = DE DF sin 45° = AB [∠C = 45°] AC ⇒ DE = DF = h m 1 = AB ⇔ AB = 17 m (iii) (b) In ΔEGC, ∠EGC = 60°, ∠C = 90° 2 17 2 tan 60° = CE A CG 3 = CD + DE ⇒ 3 = h + 6 [Q CG = DF] 45° CG DF C ⇒ 3=h+6 h ⇒ 3 =1 + 6 ⇒( 3 − 1) = 6 B h h (v) (c) in this, mathematical concept trigonometric ratio is ∴ h = 6 × 3+1 [by rationalising] used here, which is application of trigonometry. ( 3 − 1) 3 + 1 21. (i) (b) Let AB be the monument of height 42 m and C is the = ( 6( 3 + 1) [Q (a + b)(a − b) = a2 − b2] point where they are standing, such that BC = 42 m. 3)2 − (1)2 A = 6( 3 + 1) = 6( 3 + 1) = 3( 3 + 1) m 3−1 2 42 m (iv) (d) Height of the Parachute from the ground is BE, then BE = BC + CD + DE θ B BE = 3 + 6 + 3( 3 + 1) C θ = 45° 42 m = 9 + 3 ( 3 + 1) Now, in ΔABC, =9+ 3 3+ 3 tan θ = AB BC = 12 + 3 3 = 3(4 + 3) m ⇒ tan θ = 42 =1 42 (v) (c) If the Parachute is moving towards the building, then both angles of elevation will decreases. ⇒ tan θ = 1 ⇒
104 CBSE Term II Mathematics X (Standard) (ii) (d) In ΔABC, 22. As it is given that satellite is the mid-point of the two mountain hills i.e. I is the mid-point of DS. A F 30° 60° 42 m A 30° G 60° B 60° C 968500 m tan 60° = AB 7816 m H P BC 968500 m 1930 m 3 = 42 DIS BC BC = 42 = 42 × 3 [by rationalising] 1937000 m 3 33 (i) (b) We have, AG = DI = 968500 m = 42 3 = 14 3 Now, in ΔFAG, 3 cos 30° = AG AF = 14 × 1.732 = 24.24 m ⇒ 3 = 968500 2 AF (iii) (a) Let AB = h be the height of the tower. ⇒ AF = 968500 × 2 = 1937000 3 1.73 A = 1119653.18 m hm = 1119.65 km (ii) (c) We have, HP = IS = 968500 m 60° B C Now, in ΔFHP, cos 60° = HP 20 m FP 1 968500 Now, in ΔABC, tan 60° = AB ⇒ 3 = h ⇒ h = 20 3m 2 = FP BC 20 FP = 968500 × 2 = 1937000 m = 1937 km (iv) (b) Let h and x be the height and length of shadow of the (iii) (b) In ΔFAG, tan 30° = FG ⇒ 1 = FG vertical tower. AG 3 968500 A ⇒ FG = 968500 = 559826. 59 m 3 hm = 559.82 km θ B ∴Height of satellite from ground = FI = FG + GI C = 559.82 + 7.816 xm [Q GI = AD = 7816m = 7.816 km ] Now, in ΔABC, = 567.64 km (iv) (b) Let E be the position of man. tan θ = AB ⇒ tan θ = h BC x A ⇒ tan θ = 1 [Q h : x = 1 : 1] ⇒ θ = 45° (v) (c) The angle of depression of the object viewed, is the angle formed by the line of sight with the horizontal, when it is below the horizontal level. Horizontal line D θE O 7816 m Angle of depression Then, DE = 7816 m In ΔADE, tan θ = AD = 7816 =1 Line of sight DE 7816 P (object) [Q height of mountain AD = 7816 m] ∴ θ = 45°
CBSE Term II Mathematics X (Standard) 105 (v) (c) Let T be the point where mile stone is kept. 2. Let AB be the tree whose part AC breaks and touches the P ground at D. Then, BD = 2 3 m [given] 1930 m and AC = CD A T 45° S So, In ΔPST, tan 45° = PS C TS ⇒ 1 = 1930 ⇒ TS = 1930 m TS Subjective Questions B 60° D 2√3 m 1. Case I Let the height of a tower be h and the distance of the point of observation from its foot be x. In right angled ΔCBD, cos 60° = BD In ΔABC, CD tan θ1 = AC = h ⇒ 1=2 3 BC x 2 CD ⇒ θ1 = tan −1 ⎜⎛⎝ h ⎠⎞⎟ ...(i) ⎢⎣⎡Q cos 60° = 1 and BD = 2 3 m ⎤ x 2 ⎦⎥ A ⇒ CD = 2 × 2 3 = 4 3 = 4 × 1.732 = 6.928 m [Q 3 = 1.732] h ∴ AC = CD = 6.928 m B θ1 C Again, in right angled ΔCBD, x tan 60° = BC BD Case II Now, the height of the tower increased by 10% 3 and BD = 2 3 m] ⇒ 3 = BC [Q tan 60° = 10 11 h 23 100 10 = h + 10% of h = h + h × = ⇒ BC = 3 × 2 3 = 6 m and the distance of the point of observation from its foot Now, AB = AC + BC = x + 10% of x = 6.928 + 6 = 12.928 m (approx.) = x + x × 10 = 11x Hence, whole height of the tree is 12.928 m. 100 10 3. Let AB = h km be the height of the mountain. Let C be a P point at a distance of x km from the base of the mountain such that ∠ACB = 30° and let D be a point at a distance of 11h 10 km from C along the same line. Then, ∠ADB = 15° 10 and AD = AC + DC = (x + 10) km θ2 11x B Q R 10 h km In ΔPQR, tan θ2 = PR = ⎜⎝⎛ 11h ⎟⎞⎠ D 15° 30° QR 10 C ⎝⎛⎜ 1110x⎠⎞⎟ A ⇒ tan θ2 = h In right angled ΔBAC, x tan 30° = AB AC tan −1 ⎝⎛⎜ h ⎞⎠⎟ ⇒ θ2 = x ...(ii) 1h ⎢⎡⎣Q tan 30° = 1⎤ ⇒ 3=x 3 ⎦⎥ From Eqs. (i) and (ii), ⇒ x=h 3 ...(i) θ1 = θ2 In right angled ΔBAD , tan 15° = AB Hence, the required angle of elevation of its top remains AD unchanged.
106 CBSE Term II Mathematics X (Standard) ⇒ 0.27 = h [given , tan 15° = 0.27] ⇒ tan θ = 3 = tan 60° x + 10 ∴ θ = 60° ⇒ 0.27 (x + 10) = h ...(ii) II. A same hight of tower casts a shadow 4m long from On putting x = 3h from Eq. (i) in Eq. (ii), we get preceding shadow, when the Sun’s elevation is 30°. 0.27( 3h + 10) = h In ΔAPB, tan θ = AB PB ⇒ 0.27 × 3h + 0.27 × 10 = h = AB ⇒ h(1 − 0.27 × 3) = 0.27 × 10 PC + CB ⇒ h(1 − 0.27 × 1.732 ) = 2.7 [Q 3 = 1.732] ⇒ tan θ = 2 3 = 2 3 4+2 6 ⇒ h (1 − 0.47) = 2.7 A ⇒ 0. 53h = 2.7 ⇒ h = 2.7 = 5.09 ≈ 5 km 0.53 Hence, the height of mountain is 5 km. 2√3 m 4. Let AB be the tower and AC be the flag staff on the tower. Pθ 60° Let D be a point on the ground such that the angles of 4m B elevation of foot A and top C of the flag staff are 45° and 60°, respectively. C 2m C ⇒ tan θ = 3 ⋅ 3 = 3 A 3 3 33 60° ⇒ tan θ = 1 = tan 30° B 45° D 3 Then, we have AB = 20 m , ∠ADB = 45° and ∠CDB = 60° ∴ θ = 30° Hence, we conclude from above two examples that if the In right angled ΔABD, ⎣⎡⎢Q tan θ = Perpendicular ⎤ length of the shadow of a tower is increasing, then the tan 45°= AB Base ⎦⎥ angle of elevation of the Sun is decreasing. BD 6. Let QS be the building and R be the position of window. ⇒ 1 = 20 Given, height of the window, QR = 10 m BD ∠QPR = ∠XRP = 30° [alternate angles] and ∠SPQ = 60° S ⇒ BD = 20 m [Q tan 45° = 1] and in right angled ΔCBD, [Q tan 60° = 3] X 30° R (Window) tan 60°= BC ⇒ 3 = BC BD 20 ⇒ BC = 20 3 = 20 × 1.732 [Q 3 = 1.732] P 30° 60° 10 m = 34. 64 m (approx.) Q Now, AC = BC − AB = 34.64 − 20 In right angled ΔPQR , tan 30° = QR = 14.64 m (approx.) PQ Hence, the height of the flag staff is 14.64 m. ⇒ 1 = 10 3 PQ 5. To understand the fact of this question, consider the ⎢⎡⎣Q tan 30°= 1⎤ following example ⇒ PQ = 10 3 m 3 ⎥⎦ In right angled ΔPQS, I. A tower 2 3 m high casts a shadow 2 m long on the ...(i) tan 60° = QS ground, when the Sun’s elevation is 60°. PQ In ΔACB, tan θ = BC = 2 3 ⇒ 3 = QS [Q tan 60° = AB 2 10 3 C ⇒ QS = 10 × 3 = 30 m Hence, height of the building is 30 m. 3 and from Eq. (i)] θ 2√3 m A 2m B
CBSE Term II Mathematics X (Standard) 107 7. Let AB = 20 m be the height of tower and let the ball lying So, angle of depression is different in the lake from the angle of elevation of the cloud above the surface of a lake. on the ground at point C. C Given, angle of depression, ∠TAC = 60° = ∠ACB [alternate angles] TA P θ1 hm 60° θ2 M 20 m Q 3m O 60° h C B In right angled ΔABC, θR tan 60° = AB CM h BC PM PM In ΔMPC, tan θ1 = = ⇒ 3 = 20 ⇒ tan θ1 = 1 …(i) BC h PM ⇒ BC = 20 = 20 = 11.55 m In ΔRPM, tan θ2 = RM = OR + OM = h+3 3 1.732 PM PM PM Hence, the distance between the foot of the tower and the ⇒ tan θ2 = 1 …(ii) ball is 11.55 m. h + 3 PM 8. Let AB be the tower of height x m , and CD be the tower of From Eqs. (i) and (ii), height y m. D tan θ1 = tan θ2 ⇒ tan θ2 = ⎜⎛⎝ h + 3⎠⎞⎟ tan θ1 h h+3 h B xm ym So, θ1 ≠ θ 2 C Hence, it is a false statement. 30° 60° 10. Let AB = h km be the height of the hill and C, D be two AaE a consecutive stones such that CD = 1 km . Let BC be x km, then BD = BC + CD = (x + 1)km Let E be the mid-point of the line AC. Then, ∠AEB = 30° and ∠CED = 60°. A 30° Also, AE = EC = a m (let) X In right angled ΔBAE, 45° tan 30° = AB = x AE a ⇒ 1 = x ⇒x = a …(i) h km 3a 3 and in right angled ΔDCE, 45° 30° tan 60° = DC = y CE a B x km C 1 km D (x+1) km ⇒ 3 = y ⇒ y = 3 a …(ii) a Now, ∠ ADB = ∠XAD = 30° [alternate angles] [Q Eq. (i) divide by Eq. (ii)] and ∠ ACB = ∠XAC = 45° [alternate angles] a In right angled ΔABC, ∴ x = 3 = 1 =1 y 3a 3 × 3 3 tan 45° = Perpendicular = AB Base BC Hence, x : y = 1 : 3 ⇒ 1= h ⇒x = h …(i) x 9. From figure, we observe that, a man standing on a platform at point P, 3 m above the surface of a lake observes a cloud at Now, in right angled ΔABD, tan 30° = AB point C. Let the height of the cloud from the surface of the BD platform is h and angle of elevation of the cloud is θ1. Now at same point P, a man observes a cloud reflection in ⇒ 1= h ⎣⎢⎡Q tan 30°= 1⎤ the lake at this time the height of reflection of cloud in lake 3 x+1 3 ⎦⎥ is (h + 3) because in lake platform height is also added to reflection of cloud. ⇒ 1= h [from Eq. (i)] 3 h+1
108 CBSE Term II Mathematics X (Standard) ⇒ h + 1 = 3h In right angled ΔBMP, ⇒ h( 3 − 1) = 1 tan 45° = PM BM ⇒ h= 1× 3 + 1 = ⎛ 3+ 1⎞ km 3 −1 ⎜ 2 ⎟ ⇒ 1 = 200 3+1 ⎝ ⎠ y [Q tan 45° = 1] Hence, height of the hill is 3 + 1 km. ⇒ y = 200 m 2 Now, distance between the two ships = AB = x + y 11. Let AB be the tower, BC be the shadow of tower, when angle of elevation of Sun is 30° and BD be the shadow of tower, = 115.47 + 200 when angle of elevation of Sun is 60°. = 315.47 m Then, we have 13. Let distance of the pole, say AE, from the bottom of the tower, BC = 30 m, ∠ACB = 30° and ∠ADB = 60° say BD, be x m and let the height of the pole, AE = y m Now, let AB = h m and BD = x m Now, draw EC ||AB. D A hm E 30º 50 m xm C 30° 60° ym C D xm B ym 30 m 45º A xm B Clearly, in ΔABC, we have Then, ∠DEC = 30°, ∠DAB = 45° tan 30° = Perpendicular = AB = h and DC =DB −BC = DB − AE [Q BC = AE] Base BC 30 …(i) ⇒ DC =(50− y)m ⇒ 1 = h ⇒ h = 30 3 30 3 (i) In right angled ΔABD, 30 3 30 3 tan 45° = Perpendicular = BD ⇒ × 3 = 3 = 10 3m Base AB 3 ⇒ 1 = 50 ⇒ x = 50 m x Also, in ΔABD , tan 60° = AB = h BD x ∴ The pole is 50 m away from the foot of the tower. ⇒ 3 = 10 3 ⇒ x = 10 m (ii) In right angled ΔECD , x tan 30° = Perpendicular = DC Hence, length of shadow is 10 m, when angle of elevation is Base EC 60°. 1 50− y ⎡⎣⎢tan 30° = 1⎤ 12. Let PM be the light house of height 200 m and let A and B be ⇒ 3 = x 3 ⎥⎦ two ships on either sides of light house such that the angles of depression of A and B are 60° and 45°, respectively. ⇒ 1 = 50 − y [Q x = 50 m from Eq. (i)] 3 50 Let AM = x m and BM = y m ⇒ 3 (50 − y) = 50 Then, ∠XPB = ∠MBP = 45° [alternate angles] and ∠YPA = ∠MAP = 60° [alternate angles] ⇒ 50 − y = 50 3 X PY 45° 60° ⇒ y = 50 ⎛⎝⎜1 − 1 ⎠⎞⎟ 3 200 m = 50⎛⎝⎜1 − 1 ⎠⎞⎟ 1.732 45° 60° A B M xm = 50(1 − 0. 57737) ym In right angled ΔAMP, = 50 × 0.4226 tan 60° = Perpendicular = PM = 21.13 m Base AM ∴ Height of the pole = 21.13 m ⇒ 3 = 200 [Q tan 60° = 3] 14. Let a man is standing on the deck of a ship at point A such x that AB = 10 m and let CD be the hill. ⇒ x = 200 m = 200 m = 115.47 m Then, ∠EAD = 60° 3 1.732 and ∠CAE = ∠BCA = 30° [alternate angles]
CBSE Term II Mathematics X (Standard) 109 Let BC = x m = AE and DE = h m 16. Let AB be the flag staff of height h units and AC = x units be D length of its shadow, when the Sun rays meet the ground at an angle of 60°. hm Also, let θ be the angle between the Sun rays and the ground, when the length of the shadow of the flag staff is AD = 3x units. B A 60° E h 10 m 30° 10 m C Dθ 60° A B 30° 2x C xm x In right angled ΔAED , tan 60° = Perpendicular = DE = h In right angled ΔCAB, Base EA x tan 60° = Perpendicular = AB ⇒ 3= h ⇒h = 3x Base AC x ⇒ 3=h [Q tan 60° = 3] In right angled ΔABC, x tan 30° AB ⇒ 1 10 ⎡⎢⎣Q tan 30° = 1⎤ ⇒ h = 3x ...(i) = BC = 3 ⎥⎦ 3 x Now, in right angled ΔDAB, ⇒ x = 10 3 m tan θ = AB = AB [Q AD = DC + CA ] AD DC + CA ⇒ h = 10 3 × 3 = 30 m ∴The height of hill, CD = h + 10 = 30 + 10 = 40 m ⇒ tan θ = h = h Hence, The distance of the hill from the ship is 10 3 m and 2x + x 3x height of the hill is 40 m. = 3x [from Eq. (i)] 15. Let OX be the horizontal ground; A and B be the two 3x positions of the plane and O be the point of observation. = 1 = tan 30° ⎣⎡⎢Q tan 30° = 1⎤ AB 3 3 ⎥⎦ 2500 m ∴ θ = 30° Hence, the angle between the sun rays and the ground at the time of longer shadow is 30°. 45° X 17. Let P and Q be the positions of two aeroplanes, where P is OC D vertically above Q and OP = 4000 m. P Here, AC = BD = 2500 m , ∠AOC = 45° and ∠BOD = 30° Q 4000 m In right angled ΔOCA, cot 45°= Base = OC perpendicular AC A 45° 60° ⇒ 1 = OC O AC [Q cot 45° = 1] Here, ∠PAO = 60° and ∠QAO = 45° ⇒ OC = AC = 2500 m In right angled ΔODB, Now, in right angled ΔAOP, cot 30°= OD ⇒ 3 = OD tan 60° = Perpendicular = OP BD 2500 Base AO ⇒ OD = 2500 3 m ⇒ 3 = 4000 [Q tan 60° = 3] AO ...(i) Now, CD = OD − OC = 2500 3 − 2500 ⇒ AO = 4000 [∴tan 45° = 1] = 2500( 3 − 1) = 2500(1.732 − 1) 3 ...(ii) = 2500 × 0.732 = 1830 m In right angled ΔAOQ, tan 45° = OQ ⇒ 1 = OQ Thus, distance covered by plane in 15 s is 1830 m. OA OA ∴Speed of plane = 1830 × 60 × 60 = 439.2 km/h ⇒ OA = OQ 15 1000
110 CBSE Term II Mathematics X (Standard) From Eqs. (i) and (ii), we get We have, AB = 1200 m OQ = 4000 m 3 Let AC = x m and CD = y m. In right angled ΔBAC, we have ∴vertical distance between the aeroplanes tan 60° = Perpendicular = AB = PQ = OP − OQ = 4000 − 4000 Base AC 3 ⇒ 3 = 1200 [Q tan 60° = 3] = 4000 ⎝⎜⎛1 − 1 ⎟⎞⎠ 4000⎛⎝⎜1 − 1 ⎞⎟⎠ x 3 = 1.732 ⇒ x = 1200 × 3 = 4000 (1 − 0.577) 33 [rationalising] = 4000 × 0.423 = 1692 m ⇒ x = 1200 3 = 400 3m ...(i) 3 18. Let OA be the tree of height h m. Given, PQ = 100 m and angles of elevation are ∠APO = 30° In right angled ΔBAD , we have [Q AD = DC + CA] and ∠OQA = 45°. tan 30° = AB = AB AD DC + CA A ⇒ 1 = 1200 hm 3 x +y 30° O 45° ⇒ x + y = 1200 3 ⎢⎡⎣Q tan 30° = 1⎤ P 100 m Q 3 ⎦⎥ Perpendicular OA ⇒ y = 1200 3 − x …(ii) Base OP In right angled ΔPOA, tan 30° = = On putting the value of x from Eq. (i) in Eq. (ii), we get ⇒ 1h ⎣⎡⎢Q tan 30° = 1⎤ y = 1200 3 − 400 3 3 = OP 3 ⎦⎥ = 800 3 ⇒ OP = 3h ...(i) = 800 × 1.732 [Q 3 = 1.732] Now, in right angled ΔQOA, = 1385.6 m tan 45° = OA OQ Hence, the distance between both ships is 1385.6 m. ⇒ 1= h [Q tan 45° = 1] 20. Let A be the aeroplane and AD be its height. Again, let B OQ and C be two consecutive kilometre stones on the road on the left and right of plane A and the angles of depression of ⇒ OQ = h ...(ii) C and B from plane A are 60° and 45°, respectively. On adding Eqs. (i) and (ii), we get A OP + OQ = 3 h + h PQ 45° 60° ⇒ PQ = ( 3 + 1)h [Q OP + OQ = PQ] ⇒ 100 = ( 3 + 1)h [QPQ = 100 m, given] ⇒ h = 100 × 3 − 1 [by rationalising] 45° 60° 3 +1 3 −1 B x km D (1 – x) km C = 100 (1.732 − 1) 2 1 km = 50 × 0.732 = 36.6 m Then, ∠ABC = ∠PAB = 45° [alternate angles] Hecne, height of the tree is 36.6 m. and ∠ACB = ∠QAC = 60° [alternate angles] 19. Let the aeroplane be at B and two ships be at C and D such that their angles of depression from B are 60° and 30°, Also, BC = 1 km respectively. Then, the angles of elevation of B from D and C are 30° and 60°, respectively. Let BD = x km , then B DC = BC − BD = (1 − x)km …(i) In right angled ΔADB, 30° tan 45° = Perpendicular = AD 60° Base BD ⇒ 1 = AD [Qtan 45° = 1] x 1200 m ⇒ AD = x 30° 60° A and in right angled ΔADC, D C xm tan 60° = AD DC ym
CBSE Term II Mathematics X (Standard) 111 ⇒ 3= x 22. Let D be the position of the balloon, when it is inclined at 1−x angle of 60° and AB be the height of the pole. [Qtan 60° = 3 and from Eq. (i)] D (Balloon) ⇒ 3 − 3x = x ⇒ 3 = 3x + x C B ⇒ ( 3 + 1) x = 3 ⇒ x = 3 = 3 × 3 −1 60° 30° 3+1 3 +1 3 −1 E [by rationalising] A = 3− 3 Given, length of cable, DE = 215 m ( 3)2 − (1)2 In right angled ΔEAD, [Q (a + b)(a − b) = a2 − b2] sin 60° = Perpendicular = AD Hypotenuse ED = 3 − 3 = 3 − 1.732 [Q 3 = 1.732] ⇒ 3 = AD ⎡ sin 60° = 3⎤ 22 2 215 ⎢Q ⎣ 2 ⎥ = 1.268 = 0.634 km ⎦ 2 ⇒ AD = 215 3 m Hence, the height of the aeroplane is 0.634 km. 2 21. Let height of the tower, BC = h m and height of the flagstaff Hence , initial height of the balloon from the ground is CD =H m. 215 3 m. 2 ∴ BD =BC + CD =(h + H)m …(i) Again, in right angled ΔEAD, Given, AB =120 m, ∠CAB = 45° and ∠DAB = 60° Base AE AE D cos 60°= == Hm Hypotenuse DE 215 C ⇒ 1 AE ⎡⎣⎢Q cos 60° = 1 ⎤ 2 = 215 2 ⎦⎥ (h+H) m ⇒ AE = 215 m 2 …(i) hm Now, the angle of inclination is changed, say ∠CEA = 30°. A 45° 60° In right angled ΔEAC, 120 m B tan 30° = Perpendicular = AC In right angled ΔABC, we get Base EA tan 45° = BC AB ⎣⎢⎡Q tan θ = Perpendicular ⎤ ⇒ 1 = AC × 2 Base ⎦⎥ 3 215 ⇒ 1= h 120 [Q tan 45° = 1] ⇒ 2 3AC = 215 ⇒ h = 120 m …(ii) ⇒ AC = 215 m …(iii) 23 Now, in right angled ΔABD, we get tan 60°= BD AB 23. Let the height of the light house AB be 100 m. C and D be the positions of man when angle of elevation changes from ⇒ 3= h+H 120 60° to 45°, respectively. The man has covered a distance CD in 2 min. [Q tan 60° = 3 and from Eq. (i)] Q Speed = Distance ⇒ Speed = CD ...(i) From Eqs. (ii) and (iii), Time 2 3 = 120 + H 120 In right angled ΔABC, A ⇒ 120 3 = 120 + H ⇒ H = 120 ( 3 − 1) 100 m Light house = 120(1.732 − 1) = 120 × 0.732 = 87.84 m D 45° 60° B Hence, height of flag staff is 87.84 m. C
112 CBSE Term II Mathematics X (Standard) tan 60° = Perpendicular = AB ⇒ 20 = h 3 − h Base BC ⇒ h ( 3 − 1) = 20 ⇒ 3 = 100 [Q tan 60° = 3] ∴ h = 20 ⋅ 3 + 1 [by rationalisation] BC 3 −1 3 +1 ⇒ BC = 100 3 m ...(ii) ⇒ = 20 ( 3 + 1) 3 3−1 In right angled ΔABD, tan 45° = AB = 20 ( 3 + 1) BD 2 ⇒ 1 = 100 [Q tan 45° = 1] ⇒ = 10 ( 3 + 1) m BD Hence, the required height of tower is 10 ( 3 + 1) m. ⇒ BD = 100 m 25. Let the height of the tower be h and RQ = x m CD = BD − BC = 100 − 100 3 Now, 3 Given that, PR = 50 m 100 ⎛ 3 − 3⎞ and ∠SPQ = 30° , ∠SRQ = 60° ⎜ 3 ⎟ and Speed = CD = ⎝ ⎠ = 50 (3 − 3) m/min Now, in ΔSRQ, tan 60° = SQ RQ 2 23 ⇒ 3=h ⇒ x= h 24. Let the height of the tower be h. x3 …(i) Also, SR = x m, ∠PSR = θ and in ΔSPQ, tan 30° = SQ = SQ = h PQ PR + RQ 50 + x Given that, QS = 20 m and ∠PQR = 30° P S Sun h hm Q 30° θ R P 30° 60° 20 m S xm 50 m R x m Q Now, in ΔPSR, 1h 3 = 50 + x tan PR = h θ = SR x ⇒ 3 ⋅ h = 50 + x ⇒ 3 ⋅ h = 50 + h [from Eq. (i)] ⇒ tan θ = h x 3 ⇒ ( 3 − 1 ) h = 50 ⇒ x= h ...(i) tan θ 3 ⇒ (3 − 1) h = 50 Now, in ΔPQR, tan 30° = PR = PR 3 QR QS + SR ∴ h = 50 3 ⇒ tan 30° = h x 2 20 + h = 25 3 m ⇒ 20 + x = h = h Hence, the required height of tower is 25 3 m. tan 30° 1 / 3 26. Let the height of the tower be H and OR = x ⇒ 20 + x = h 3 Given that, height of flag staff = h = FP and ∠PRO = α, ∠FRO = β ⇒ 20 + h = h 3 [from Eq. (i)] …(ii) tan θ F Since, after moving 20 m towards the tower the angle of flag staff h elevation of the top increases by 15°. P i.e. ∠PSR = θ = ∠PQR + 15° β x H Rα O ⇒ θ = 30° + 15 = 45° ∴From Eq. (i) 20 + h = h 3 tan 45° ⇒ 20 + h = h 3 1
CBSE Term II Mathematics X (Standard) 113 Now, in ΔPRO, tan α = PO = H 28. Let the distance between two objects is x m. ⇒ RO x and CD = y m. x= H tan α …(i) A …(ii) X Yα β and in ΔFRO, tan β = FO RO ⇒ ⇒ = FP + PO hm RO tan β = h + H α β x x Cy x= h +H B D tan β Given that, ∠BAX = α = ∠ABD, [alternate angle] From Eqs. (i) and (ii), ∠CAY = β = ∠ACD [alternate angle] H =h+H and the height of tower, AD = h m …(i) tan α tan β Now, in ΔACD, ⇒ H tan β = h tan α + H tan α tan β = AD = h ⇒ H tan β − H tan α = h tan α CD y ⇒ H (tan β − tan α) = h tan α ⇒ y= h ⇒ H = h tan α tan β tan β − tan α and in ΔABD, Hence, the required height of tower is h tan α tan α = AD = AD BD BC + CD tan β − tan α ⇒ tan α = h Hence proved. x+y 27. Let distance between the two towers = AB = x m ⇒ x+y= h tan α and height of the other tower = PA = h m Given that, height of the tower = QB = 30 m and ⇒ y = h α − x …(ii) ∠QAB = 60°, ∠PBA = 30° tan Q From Eqs. (i) and (ii), h = h −x P hm tan β tan α 30 m ∴ x= h − h tan α tan β A 60° 30° B = h ⎛1 − 1⎞ xm ⎝⎜ tan α tan β⎟⎠ Now, in ΔQAB, tan 60° = QB = 30 = h (cotα − cotβ) ⎡ cot θ = 1 θ ⎤ ⇒ AB x ⎣⎢Q tan ⎦⎥ ∴ 3 = 30 which is the required distance between the two objects. x Hence proved. x = 30 ⋅ 3 33 [by rationalising] 29. Let OQ = x and OA = y = 30 3 = 10 3 m Given that, BQ = q, SA = P and AB = SQ = Length of ladder 3 Also, ∠BAO = α and ∠QSO = β and in ΔPBA, (Wall) ⇒ B tan 30° = PA = h q AB x Q 1= h [Q x = 10 3 m] (Ladder) x 3 10 3 ⇒ h = 10 m βα Hence, the required distance and height are 10 3 m and 10 m, respectively. SA y O P
114 CBSE Term II Mathematics X (Standard) Now, in ΔBAO, cos α = OA ⇒ H − H = 10 ⇒ H ⎝⎛⎜1 − 1 ⎟⎞⎠ = 10 AB 3 3 ⇒ and OA = AB cos α ⇒ H ⎛ 3 − 1⎞ = 10 ⇒ sin α = OB ⎜ ⎟ Now, in ΔQSO, …(i) ⎝ 3⎠ AB …(ii) OB = BA sin α ∴ H = 10 3 ⋅ 3 + 1 [by rationalisation] 3 −1 3 +1 cos β = OS SQ = 10 3 ( 3 + 1) = 10 3 ( 3 + 1) 3−1 2 ⇒ = 5 3 ( 3 + 1) = 5( 3 + 3) m. ⇒ OS = SQ cos β = AB cos β Hence, the required height of the tower is 5 ( 3 + 3) m. [Q AB = SQ] …(iii) 31. Let the height of the other house = OQ = H and sin β = OQ and OB = MW = x m SQ Given that, height of the first house = WB = h = MO and ∠QWM = α, ∠OWM = β = ∠WOB [alternate angle] ⇒ OQ = SQ sin β = AB sin β [Q AB = SQ] …(iv) Now, SA = OS − AO Q P = AB cos β − AB cos α 10 m ⇒ 10 m (H – 10) mP = AB (cos β − cos α)…(v) and BQ = BO − QO (H – h) ⇒ q = BA sin α − ABsin β W α x MH (Window) β ⇒ q = AB (sin α − sin β) …(vi) On dividing Eq. (v) by Eq. (vi), we get hh p = AB (cos β − cos α ) = cos β − cos α β q AB (sin α − sin β) sin α − sin β B xO ⇒ p = cos β − cos α Hence proved. WB h q sin α − sin β Now, in ΔWOB, tan β = OB = x ⇒ 30. Let the height of vertical tower be, And in ΔQWM, x= h …(i) OT = H and OP = AB = x m ⇒ tan β ...(ii) ⇒ Given that, AP = 10 m tan α = QM = OQ − MO and ∠TPO = 60°, ∠TAB = 45° WM WM T tan α = H − h x x=H−h tan α 45° H From Eqs. (i) and (ii), A xm B h =H−h tan β tan α 60° O ⇒ h tan α = (H − h ) tan β P ⇒ h tan α = H tan β − h tan β xm Now, in ΔTPO, ⇒ H tan β = h(tan α + tan β) tan 60° = OT = H OP x ∴ H = h ⎛ tan α+ tan β⎞ ⎝⎜ tan β ⎟⎠ ⇒ 3=H ⇒ x= H x3 ...(i) = h ⎜⎝⎛1 + tan α⋅ 1 ⎞ [from Eq. (i)] tan β⎠⎟ and in ΔTAB, tan 45° = TB = H − 10 = h (1 + tan α ⋅ cot β) AB x ⎡ cot θ = 1 θ ⎤ ⇒ 1 = H − 10 ⇒ x = H − 10 ⎣⎢Q tan ⎥⎦ x Hence, the required height of the other house is ⇒ H = H − 10 3 h (1 + tan α ⋅ cot β) m. Hence proved.
CBSE Term II Mathematics X (Standard) 115 32. Let the height of the balloon above the ground is H ⇒ 2H = 16 and OP = W2R = W1Q = x ⇒ H=8 Given that, height of lower window from above the ground So, the required height is 8 m. = W2P = 2 m = OR Hence, the required height of the balloon above the ground Height of upper window from above the lower window is 8 m. = W1 W2 = 4 m = QR ∴ BQ = OB − (QR + RO) 33. (i) Distance of first position of parrot from the eyes of girl =H −(4 + 2) =H − 6 = AC and ∠BW1 Q = 30° EC ⇒ ∠BW2R = 60° Balloon B (H – 6) D 60° A 58 m B 30° 8 m GH F w1 30° xm Q In ΔABC, Upper 4m sin 60° = BC window Hm AC w2 60° x ⇒ AC = CH − BH Lower sin 60° window R 2m = 58 − 8 = 100 m P 3/2 3 xm O Now, in ΔBW2R, (ii) If the distance increases, then the angle of elevation decreases. tan 60° = BR = BQ+ QR (iii) Distance between girl and building = AB W2R x Now, in ΔABC, ⇒ 3 = (H − 6)+ 4 tan 60° = BC ⇒ 3 AB = 50 ⇒ AB = 50 m x AB 3 ⇒ x = H −2 ...(i) (iv) In ΔAED, tan 30° = DE 3 ...(ii) AD and in ΔBW1Q, ⇒ AD = 3 BC = 50 3 m tan 30° = BQ W1Q [QED = BC = 58 − 8 = 50] Now, distance between two position of parrot = EC ⇒ tan 30° = H − 6 = 1 x3 = BD = AD − AB = ⎜⎝⎛ 50 3 − 503⎟⎠⎞ m ⇒ x = 3(H − 6) = 50(3 − 1) = 100 = 57.80 m From Eqs. (i) and (ii), 1.73 1.73 3(H − 6) = (H −2) 3 (v) Speed of parrot = Distance covered Time taken 3(H − 6) = H − 2 = ⎝⎛⎜ 578.80⎠⎞⎟ m/s = 7.225 m/s ⇒ 3H − 18 = H − 2
Chapter Test Multiple Choice Questions Based on the above information, answer the following questions. 1. A circus artist is climbing from the ground along a rope stretched from the top of a vertical pole and tied (i) Measure of ∠ACD is equal to at the ground. The height of the pole is 12 m and the (a) 30° (b) 45° angle made by the rope with ground level is 30°. The (c) 60° (d) 90° distance covered by the artist in climbing to the top of (ii) If ∠YAB = 45°, then ∠ABD is also 45°, Why? (a) vertically opposite angles the pole is (b) alternate interior angles (c) alternate exterior angles (a) 12 m (b) 6 m (d) corresponding angles (c) 24 m (d) 32 m 2. A ladder 15 m long just reaches the top of a vertical (iii) Length of CD is equal to wall. If the ladder makes an angle of 60° with the wall, (a) 90 m (b) 50 3 m then the height of the wall is (c) 50 / 3 m (d) 100 m (a) 30 m (b) 15 m (iv) Length of BD is equal to 2 (c) 15 m (d) 25 m (a) 50 m (b) 100 m 3. A kite is flying at a height of 30 m from the ground. (c) 100 2 m (d) 100 3 m The length of string from the kite to the ground is (v) Length of AC is equal to 60 m. Assuming that there is no slack in the string, (a) 100 / 3 m (b) 100 3 m then the angle of elevation of the kite at the ground is (c) 50 m (d) 100 m (a) 30° (b) 45° Short Answer Type Questions (c) 60° (d) None of these 7. The angle of elevation of the top of a tower is 30°. If 4. The tops of two poles of height 30 m and 24 m are the height of the tower is doubled, then the angle of elevation of its top will also be doubled. State true or connected by a wire. If the wire makes an angle of 45° false. Explain. with the horizontal, then find the length of the wire. (a) 14 m (b) 3 m 8. A peacock is sitting on the top of a tree. It observes a (c) 4 m (d) 6 2 m serpent on the ground making an angle of depression of 30°. The peacock catches the serpent in 12 s with the 5. An observer 3.5 m tall is 38.5 m away from a tower speed of 300 m/min. What is the height of the tree? 42 m high. The angle of elevation of the top of the [CBSE 2015] tower from the eye of the observer is 9. The angles of elevation and depression of the top and (a) 30° (b) 90° bottom of a light house from the top of a 60 m high (c) 45° building are 30° and 60°, respectively. Find the (d) 60° difference between the heights of the light house and building. Case Based MCQs 10. As observed from the top of a 100 m high light house 6. A boy is standing on the top of mountain. He from the sea-level, the angles of depression of two observed that boat P and boat Q are approaching towards mountain from opposite directions. He finds ships are 30° and 45°. If one ship is exactly behind the that angle of depression of boat P is 60° and angle of other on the same side of the light house, find the depression of boat Q is 45°. He also knows that height of the mountain is 50 m. distance between the two ships. [CBSE 2018] Long Answer Type Questions A Y 11. Two ships are sailing in the sea on either side of the X light house. The angles of depression of two ships as 60° 45° observed from the top of the light house are 60° and 45°, respectively. If the distance between the ships is 50 m P QC D B 100 ⎛ 3 + 1⎞ m, then find the height of the light house. ⎜ ⎟ ⎝ 3⎠ Answers For Detailed Solutions Scan the code 1. (c) 2. (b) 3. (a) 4. (d) 5. (c) 6. (i) (c) (ii) (b) (iii) (c) (iv) (a) (v) (a) 7. False 8. 30 m 9. 20 m 10. 100 ( 3 − 1) m 11. 100 m
CBSE Term II Mathematics X (Standard) 117 CHAPTER 06 Surface Areas and Volumes In this Chapter... l Solid Figures l Surface Area l Volume l Combination of Two Figures l Conversion of Solid from One Shape to Another Solid Figures Different Types of Solid Figures The objects having definite shape, size and occupies a fixed 1. Cuboid amount of space in three dimensions are called solids such as A cuboid is a solid figure having 6 rectangular faces. Let its cube, cuboid, cylinder, cone, sphere and hemisphere, etc. length = l units, breadth = b units and height = h units. Surface Area (SA) b Surface area of a solid body is the area of all of its surfaces h together and it is always measured in square unit. e.g. A cube has 6 surfaces and each surface is in a square l shape. Therefore, its surface area will be 6a 2 sq units, where a 2 is the area of each surface of the cube. Then, Volume (i) Total surface area of cuboid (TSA) Space occupied by an object/solid body is called the volume = 2 (lb + bh + hl) sq units of that particular object/solid. Volume is always measured in cube unit. (ii) Lateral surface area of cuboid = 2(l + b)h sq units e.g. Suppose, a cube has edge of length a units. Volume of a cube is equal to the product of area of base and height of or Lateral surface area = Area of the 4 vertical faces a cube i.e. a 2 × a = a 3 cu units. (iii) Diagonal of the cuboid = 2 + b2 + h2 units l (iv) Volume of cuboid = l × b × h cu units
118 CBSE Term II Mathematics X (Standard) 2. Cube Then, Cube is a special case of cuboid which has 6 equal square faces. (i) Curved surface area (CSA) a = CSA of outer cylinder + CSA of inner cylinder a = 2πRh + 2πrh a = 2π(R + r )h sq units (ii) Total surface area (TSA) Let its length = breadth = height = a units ∴ Each edge of cube = a units = CSA of hollow cylinder + Area of both ends Then, = 2π(R + r)h + 2π(R 2 − 2 ) (i) Total surface area (TSA) of a cube = 6 × (Edge)2 = 6a 2 sq units r (ii) Lateral surface area of cube = 4 × (Edge)2 = 4a 2 sq units = 2π(R + r )h + 2π(R + r ) (R − r ) (iii) Diagonal of a cube = 3 × Edge = 3 a units (iv) Volume of a cube = (Edge)3 = a 3 cu units = 2π(R + r ) [h + R − r ] sq units 3. Right Circular Cylinder (iii) Total outer surface area = 2πRh + 2π(R 2 − r 2 ) sq units Cylinder is a solid figure obtained by revolving the rectangle, say ABCD, about its one side, say BC. Let base radius of right (iv) Volume of hollow cylinder circular cylinder be r units and its height be h units. Then, = Volume of outer cylinder AB − Volume of inner cylinder = πR 2h − πr2h = π(R 2 − r 2 )h cu units 5. Sphere A sphere is a solid generated by the revolution of a semi-circle about its diameter. Let radius of sphere be r units. A r hr O r B DC Then, (i) Curved surface area (CSA) = Circumference of the base × Height = 2πrh sq units (i) Surface area (SA) of sphere = 4πr 2 sq units (ii) Total surface area (TSA) (ii) Volume of sphere = 4 πr 3 cu units = Curved surface area (CSA) + Area of two ends 3 = 2πrh + 2πr 2 = 2πr(h + r ) sq units 6. Spherical Shell (iii) Volume of the cylinder = Area of base × Height = πr 2 h cu units If R and r are respectively the outer and inner radii of a spherical shell, then 4. Right Circular Hollow Cylinder Let R units and r units be the external and internal radii of the (i) Outer surface area = 4πR 2 sq units hollow cylinder, respectively and h units be its height. (ii) Inner surface area = 4πr 2 sq units R (iii) Volume of a hollow sphere = 4 π(R 3 − r 3 ) cu units h 3 B A r OR r
CBSE Term II Mathematics X (Standard) 119 7. Hemisphere e.g. A combined solid is formed by joining hemisphere and A plane passing through the centre, cuts the sphere in two right circular cone. equal parts, each part is called a hemisphere. Let radius of hemisphere be r units. Then, rO r (i) Curved surface area (CSA) of hemisphere = 2πr 2 sq units (i) Surface area of combined solid figure (ii) Total surface area (TSA) of hemisphere = CSA of cone + CSA of hemisphere = CSA of hemisphere + Area of one end (ii) Volume of combined solid figure = 2πr 2 + πr 2 = 3πr 2 sq units. = Volume of cone + Volume of hemisphere (iii) Volume of hemisphere = 2 πr 3 cu units 3 While calculating the surface area, we have not added the surface areas of the two individual solids, rather we have 8. Right Circular Cone added curved surface area because some part of the surface area disappeared in the process of joining them. But this will A right circular cone is a solid generated by the revolution of not be in the case, when we calculate the volume. a right angled triangle about one of its sides containing the right angle as axis as shown in figure. Let height of a right Conversion of Solid from One Shape to circular cone be h units and its radius be r units. Then, Another (i) Slant height of the cone, Sometimes, we need to convert solid figure of one shape to l = AC = r 2 + h 2 units another. When we come across objects which are converted from one shape to another or when a liquid which is (ii) Curved surface area (CSA) of cone = πrl sq units originally filled in one container of a particular shape is poured into another container of a different shape or size, the (iii) Total surface area (TSA) of a cone volume remains same. e.g. = Curved surface area (CSA) + Area of the base = πrl + πr 2 = πr(l + r ) sq units (i) If a solid metallic sphere is melted and recast into more than one spherical balls, then volume of metallic sphere A = Sum of volumes of all spherical balls. hl (ii) If the Earth taken out by digging a well and spreading it uniformly around the well to form an embankment in the B rO C shape of a cylindrical shell from its original shape of right circular cylinder, then volume of embankment (iv) Volume of cone = 1 πr 2 h cu units = Volume of Earth taken out by digging a well. 3 Important Results or Formulae Combination of Two Solids If a solid of one shape is converted into solid (or solids) of Sometimes, we have to find the curved surface area and another shape, then volume of a solid, which is a combination of two solids. Then, for finding the surface area, we add the curved surface areas (i) Volume of the solid to be converted = Total volume of the of individual solids and for finding the volume of this solid, solids into which the given solid is to be converted we add the volumes of individual solids. (ii) Number of solids of a given shape in which a given solid is to be converted Volume of the solid to be converted = Volume of one converted solid
120 CBSE Term II Mathematics X (Standard) Solved Examples Example 1. Three metallic solid cubes whose edges are We know that, Volume of cylinder = πr2h 3 cm, 4 cm and 5 cm are melted and formed into a single cube. Find the edge of the cube so formed. 66 cm 10cm Sol. Given, edges of three solid cubes are 3 cm, 4 cm and 5 cm, respectively. 13 cm ∴ Volume of first cube = (3)3 = 27 cm 3 240 cm [Q volume of cube = (side)3] ∴ Total volume of iron pole = Volume of first cylinder Volume of second cube = (4)3 = 64 cm 3 + Volume of second cylinder and volume of third cube = (5)3 = 125 cm 3 = π (13)2 × 240 + π (10)2 × 66 ∴ Sum of volume of three cubes = (27 + 64 + 125) = 216 cm 3 Let the edge of the resulting cube = R cm Then, volume of the resulting cube, R 3 = 216 ⇒ R = 6 cm Example 2. The volume of a right circular cylinder with its height equal to the radius is 25 1 cm 3. Find the 7 = π [169 × 240 + 100 × 66] height of the cylinder. (Use π = 22 ) = 3.14 [40560 + 6600] 7 = 3.14 × 47160 Sol. Let h and r be the height and radius of right circular = 148082.4 cm 3 cylinder, respectively. Hence, total mass of the iron pole Given, height of cylinder = Radius of cylinder = 148082.4 × 8 g = 1184659.2 g [given, 1 cm3 ≈ 8 g] i.e. h = r Q Volume of cylinder = πr2h = 1184659.2 kg 1000 ∴ 25 1 = 22 × h 2 × h [Q h = r and V = 25 1 , given] 77 7 = 1184.66 kg ⎡Q 1g = 1 kg⎤⎥⎦ ⎣⎢ 1000 ⇒ 176 = 22 × h 3 77 Example 4. A spherical metal ball of radius 8 cm is ⇒ h3 = 8 melted to make 8 smaller identical balls. The radius of each new ball is ……… cm. ⇒ h3 = 23 ⇒ h = 2 [taking cube root] Sol. Let radius of larger sphere be R = 8 cm Hence, height of cylinder is 2 cm. and radius of smaller sphere be r cm Example 3. An iron pole consists of a cylinder of height Let number of smaller sphere be n = 8 According to the given condition, 240 cm and base diameter 26 cm, which is Volume of larger sphere = n × volume of smaller sphere surmounted by another cylinder of height 66 cm ∴ 4 πR 3 = n × 4 πr3 and radius 10 cm. Find the mass of the pole given 33 that 1 cm3 of iron has approximately 8 g mass. ∴ (8)3 = 8 × r3 [take, π = 3.14] ⇒ r3 = 82 ⇒ r3 = 64 = (4)3 Sol. Here, solid iron pole is a combination of two cylinders. ⇒ r = 4 cm [taking cube root] For first cylinder, Hence, radius of new ball is 4 cm. Height = 240 cm Example 5. A solid is in the shape of a cone mounted Base diameter = 26 cm ∴ Base radius = 26 cm = 13 cm on a hemisphere of same base radius. If the curved surface areas of the hemispherical part and the 2 conical part are equal, then find the ratio of the For second cylinder, radius and the height of the conical part. Height = 66 cm Radius = 10 cm
CBSE Term II Mathematics X (Standard) 121 Sol. Let radius, height and slant height of a cone are r, h and l, Example 7. In figure, a tent is in the shape of a cylinder respectively. Then, radius of hemisphere will be r. surmounted by a conical top. The cylindrical part is hl r 2.1 m high and conical part has slant height 2.8 m. Both the parts have same radius 2 m. Find the area of the canvas used to make the tent. ⎢⎣⎡Use π = 22 ⎤ 7 ⎦⎥ Now, curved surface area of cone C1 = πrl …(i) 2.8 m and curved surface area of hemisphere, C 2 = 2πr2 2.1 m According to the question, 2m C1 = C2 Sol. Given radius of conical and cylindrical part is r = 2 m. ∴ πrl = 2πr2 ⇒ l = 2r Slant height of cone is l = 2.8 m Also, l = r2 + h 2 And height of cylinder is h = 2.1 m ⇒ (2r) = r2 + h 2 [Q from Eq. (i)] 2.8 m On squaring both sides, we get 2.1 m (2r)2 = ( r2 + h 2 )2 ⇒ 4r2 = r2 + h 2 2m ∴The area of the canvas used, to make the tent = curve ⇒ 3r2 = h 2 ⇒ ( 3r)2 = h 2 surface area of cone + curve surface area of cylinder Taking square root both sides, we get = πrl + 2πrh 3r = h ⇒ r = 1 = 22 × 2 × 2.8 + 2 × 22 × 2 × 2.1 h3 77 Hence, the ratio of the radius and height of the conical part = 17.6 + 26.4 = 44 cm 2 Hence, the area of the canvas used to make the tent is is 1 : 3. 44 cm 2. Example 6. A solid is in the shape of a hemisphere Example 8. From a solid right circular cylinder of surmounted by a cone. If the radius of hemisphere height 14 cm and base radius 6 cm, a right circular cone of same height and same base radius is and base radius of cone is 7 cm and height of cone removed. Find the volume of the remaining solid. is 3.5 cm, find the volume of the solid. Sol. Given radius and height of cylinder are r = 6 cm and h = 14 cm ⎡⎢⎣take, π = 22 ⎤ 7 ⎦⎥ 14 cm Sol. Given, radius of hemisphere and cone is r = 7 cm. 6 cm Also, radius and height of cone will be And height of cone (h ) = 3.5 cm A r1 = 6 cm and h1 = 14 cm 3.5 cm B 7 cm C 7 cm Now, volume of cone V1 = 1 πr2h = 1 × 22 × (7)2 × 3.5 3 3 7 = 179.67 cm3 and Volume of hemisphere , V2 = 2 πr3 = 2 × 22 × (7)3 3 3 7 = 718.67 cm3 ∴ The volume of solid figure = Volume of cone + Volume of hemisphere = V1 + V2 = 179.67 + 718.67 = 898.34 cm3 Hence, volume of solid shape is 898.34 cm3.
122 CBSE Term II Mathematics X (Standard) Now, volume of cylinder, Example 10. Two cones with same base radius 8 cm and V1 = πr12h height 15 cm are joined together along their bases. = 22 × (6)2 × 14 = 1584 cm3 Find the surface area of the shape so formed. 7 Sol. If two cones with same base and height are joined together Volume of cone, V2 = 1 πr12h1 along their bases, then the shape so formed is look like as 3 figure shown. = 1 × 22 × (6)2 × 14 = 528 cm3 8 cm 37 8 cm ∴Volume of remaining solid = Volume of cylinder − Volume of cone 15 cm 16 cm = V1 − V2 = 1584 − 528 = 1056 cm3 Hence, volume of the remaining solid is 1056 cm3. 30 cm Example 9. An ice-cream cone full of ice-cream having Given that, radius of cone, r = 8 cm and height of cone, h = 15 cm radius 5 cm and height 10 cm as shown in figure So, surface area of the shape so formed 5 cm = Curved area of first cone + Curved surface area of second cone 10 cm = 2 ⋅ Surface area of cone [since, both cones are identical] = 2 × πrl = 2 × π × r × r2 + h 2 Calculate the volume of ice-cream, provided that its = 2 × 22 × 8 × (8)2 + (15)2 = 2 × 22 × 8 × 64 + 225 1 part is left unfilled with ice-cream. 77 6 = 44 × 8 × 289 = 44 × 8 × 17 Sol. Given, ice-cream cone is the combination of a hemisphere 77 and a cone. = 5984 = 854.85 cm 2 7 = 855 cm2 (approx.) Hence, the surface area of shape so formed is 855 cm2. Also , radius of hemisphere = 5 cm Example 11. The barrel of a fountain pen, cylindrical in ∴ Volume of hemisphere = 2 πr3 = 2 × 22 × (5)3 shape, is 7 cm long and 0.5 cm in diameter. A full 3 37 = 5500 = 261.90 cm 3 barrel of ink in the pen can be used for writing 275 21 words on an average. How many words would be written using a bottle of ink containing one-fourth Now, radius of the cone = 5 cm of a litre? [CBSE 2015, 14] and height of the cone = 10 − 5 = 5 cm Sol. Given, height of cylindrical pen = 7 cm ∴ Volume of the cone = 1 πr2h Radius = Diameter = 0. 5 cm 3 2 2 = 1 × 22 × (5)2 × 5 ∴Volume of barrel of a fountain pen = πr2h 37 = 22 × ⎛⎝⎜ 02. 5⎞⎠⎟ 2 ×7 = 22 cm 3 = 2750 = 130.95 cm 3 7 16 21 It is given that, a pen can write 275 words by using the ink Now, total volume of ice-cream cone 22 cm 3. 16 = 261.90 + 130.95 = 392.85 cm 3 ∴ Volume of ink = 275 words Since, 1 part is left unfilled with ice-cream. ⇒ 22 cm 3 = 275 words 6 16 ∴Required volume of ice-cream = 392.85 − 392.85 × 1 ⇒ 1 × 1000 cm 3 = 275 × 16 × 1 × 1000 = 50000 6 4 22 4 [Q he will use 1 L of ink to write words] = 392.85 − 65.475 4 = 327.4 cm 3 Hence, the pen can write 50000 words by 1 L of ink. 4
CBSE Term II Mathematics X (Standard) 123 Example 12. 500 persons are taking a dip into a Sol. It is clear, from the figure, length = 20 m cuboidal pond which is 80 m long and 50 m broad. and width = 1 m of each step. What is the rise of water level in the pond, if the 2 average displacement of the water by a person is 0.04 m 3? and height of Ist step which is in the bottom = 1 m 4 Sol. Let the rise of water level in the pond be h m when 500 persons are taking a dip into a cuboidal pond. ∴Height of second step = 2 × 1 = 1 m 42 h Height of third step = 3 × 1 = 3 m Water 44 Given that, MM Length of the cuboidal pond = 80 m Breadth of the cuboidal pond = 50 m Height of tenth step = 10 × 1 = 10 m Now, volume for the rise of water level in the pond 44 = Length × Breadth × Height Total volume of the concrete used = 80 × 50 × h = 4000 h m 3 =20 × 1 × 1 + 20 × 1 × 2 + 20 × 1 × 3 + ... + 20 × 1 × 10 and the average displacement of the water by a person 24 24 24 24 = 0.04 m3 So, the average displacement of the water by 500 persons [Q volume of cuboid = l × b × h] = 500 × 0.04 m3 = 20 × 1 × 1 [1 + 2 + 3 + ... + 10] Now, by given condition, Volume for the rise of water level in the pond = Average 24 displacement of the water by 500 persons ⇒ 4000 h = 500 × 0.04 = 20 × 1 × 1 × 10 × 11 ⎡⎢⎣Q 1 + 2 + ... + n = n (n + 1) ⎤ ∴ h = 500 × 0.04 = 20 = 1 m 24 2 2 ⎦⎥ 4000 4000 200 = 0. 005 m = 137.5 m 3 = 0. 005 × 100 cm Example 14. A wall 24 m long, 0.4 m thick and 6 m [Q 1 m = 100 cm ] = 0. 5 cm high is constructed with the bricks each of Hence, the required rise of water level in the pond is 0.5 cm. dimensions 25 cm × 16 cm × 10 cm. If the mortar occupies 1 th of the volume of the wall, then find Example 13. A small terrace at a hockey ground 10 comprises of 10 steps each of which 20 m long and the number of bricks used in constructing the wall. built of solid concrete. Each step has a rise of 1 m Sol. Given that, a wall is constructed with the help of bricks and 4 mortar. and a tread of 1 m. Calculate the total volume of ∴Number of bricks 2 concrete required to build the terrace. (Volume of wall) − ⎜⎝⎛ 1 th volume of wall ⎠⎞⎟ 10 = ...(i) Volume of a brick Also, given that Length of a wall (l) = 24 m , Thickness of a wall (b) = 0.4 m, Height of a wall (h ) = 6 m So, volume of a wall constructed with the bricks = l × b × h = 24 × 0.4 × 6 = 24 × 4 × 6 m 3 10 Now, 1 th volume of a wall = 1 × 24 × 4 × 6 10 10 10 = 24 ×4 × 6 m3 102 and Length of a brick (l ) = 25 cm = 25 m 100 1 Breadth of a brick (b1) = 16 cm = 16 m 100 21 m Height of a brick ( h1 ) = 10 cm = 10 m 100 20 m So, volume of a brick = l1 × b1 × h1 1 m 1 m 34 m = 25 × 16 × 10 = 25 × 16 m 3 4 2 100 100 100 105
124 CBSE Term II Mathematics X (Standard) From Eq. (i), Since, the water is flowing at the rate of 5 km/h. Number of bricks = ⎝⎛⎜ 24 ×4 × 6 − 24 × 4 × 6⎞⎟⎠ Therefore, length of the water flow in x h 10 100 = 5x km = 5000x m ⎛⎜⎝ 251×0516⎞⎠⎟ [Q 1 km = 1000 m] = 24 × 4 × 6 × 9 × 105 We have, diameter of cylindrical pipe = 14 cm 100 25 × 16 ∴ Radius of cylindrical pipe, r = 14 = 7 cm = 7 m 2 100 = 24 × 4 × 6 × 9 × 1000 Volume of the water flowing through the cylindrical 25 × 16 pipe in = 24 × 6 × 9 × 10 = 12960 xh = πr 2h = 22 × ⎜⎝⎛ 1700⎞⎟⎠ 2 × 5000x 7 Hence, the required number of bricks used in constructing the wall is 12960. = 77x m 3 Example 15. Water is flowing at the rate of 5 km/h Also, volume of the water that falls into the tank in x h through a pipe of diameter 14 cm into a rectangular = l×b×h tank which is 50 m long and 44 m wide. Determine the = 50 × 44 × 7 time in which the level of the water in the tank will rise by 7 cm. 100 = 154 m 3 14 cm ⎡ l = 50 m, b = 44 m and h = radius = 7 m ⎤ ⎣⎢Q 100 ⎥⎦ 7 cm Q Volume of the water flowing through the cylindrical pipe in x h = Volume of water that falls in the tank in x h 44 m ⇒ 77x = 154 ⇒ x =2 Hence, the level of water in the tank will rise by 7 cm in 2 h. 50 m Sol. Suppose, the level of the water in the tank will rise by 7 cm in x h.
CBSE Term II Mathematics X (Standard) 125 Chapter Practice PART 1 7. A cylindrical pencil sharpened at one edge is the Objective Questions combination of G Multiple Choice Questions (a) a cone and a cylinder (b) cube and a cylinder (c) a hemisphere and a cylinder (d) two cylinders 1. Three cubes each of side 5 cm are joined end to 8. A surahi is the combination of end, then the surface area of the resulting solid is (a) a sphere and a cylinder (b) a hemisphere and a cylinder (a) 250 cm2 (b) 180 cm2 (c) two hemispheres (d) a cylinder and a cone (c) 350 cm2 (d) None of these 2. A solid ball is exactly fitted inside the cubical box of 9. Two cones have their heights in the ratio 1 : 3 and side a. The volume of the ball is radii in the ratio 3: 1, then the ratio of their (a) 1 πa3 (b) 4 πa3 volumes is 6 3 (c) 1 πa3 (d) None of these (a) 1 : 3 (b) 3 : 1 3 (c) 2 : 3 (d) 3 : 2 3. A cubical icecream brick of edge 22 cm is to be 10. The shape of a gilli, in the gilli-danda game (see figure) is a combination of distributed among some children by filling icecream cones of radius 2 cm and height 7 cm upto its brim. How many children will get icecream cones? (a) two cylinders (b) a cone and a cylinder (a) 163 (b) 263 (c) two cones and a cylinder (d) two cylinders and a cone (c) 363 (d) 463 11. A plumbline (sahul) is the combination of 4. A right circular cylinder of radius r cm and height h (see figure) cm (where, h > 2r) just encloses a sphere of diameter (a) r cm (b) 2r cm (c) h cm (d) 2h cm 5. If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is (a) 4πr2 (b) 6πr2 (a) a cone and a cylinder (b) a hemisphere and a cone (c) cube and a cylinder (d) sphere and cylinder (c) 3πr2 (d) 8πr2 6. A solid cylinder of radius r and height h is placed 12. A solid cone of radius r and height h is placed over over other cylinder of same height and radius. The total surface area of the shape so formed is a solid cylinder having same base radius and height (a) 4πr(h 2 + r2) as that of a cone. The total surface area of the (b) 4πr[h + r] combined solid is (c) 4π (h 2 + r2) (a) πrl + 2πrh (b) πr2(l + 2h ) (d) None of the above (c) πr[ r2 + h 2 + 2h + r] (d) None of these
126 CBSE Term II Mathematics X (Standard) 13. The capacity of a cylindrical vessel with a 20. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as hemispherical portion raised upward at the bottom shown in figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm. Find the total as shown in the figure is πr 2 [3h − 2 r ]. surface area of the article. 3 r cm h cm (a) 1 πr2[2h − 3r] (b) 2 πr2[3h − 2r] (a) 374 cm 2 [CBSE 2018] 3 3 (c) 475 cm 2 (b) 370 cm 2 (c) 1 πr2[3h − 2r] (d) None of these (d) None of these 3 21. A heap of rice is in the form of a cone of base 14. The diameter of a sphere is 6 cm. It is melted and diameter 24 m and height 3.5 m. Find the volume of drawn into a wire of diameter 2 mm. the rice. How much canvas as cloth is required to The length of the wire is just cover the heap? [CBSE 2018] (a) 12 m (b) 18 m (a) 105.5 m 2 (b) 471.42 m 2 (c) 36 m (d) 66 m (c) 173.5 m 2 (d) None of these 15. During conversion of a solid from one shape to 22. A mason constructs a wall of dimensions another, the volume of the new shape will 270 cm × 300 cm × 350 cm with the bricks each of (a) increase (b) decrease (c) remain unaltered (d) be doubled size 22.5 cm × 11.25 cm × 8.75 cm and it is assumed that 1 space is covered by the mortar. Then, the 16. From a solid circular cylinder with height 10 cm 8 and radius of the base 6 cm, a right circular cone of number of bricks used to construct the wall is the same height and same base is removed, then the (a) 11100 (b) 11200 (c) 11000 (d) 11300 volume of remaining solid is (a) 280 π cm 3 (b) 330 π cm 3 G Case Based MCQs (c) 240 π cm 3 (d) 440 π cm 3 23. To make the learning process more interesting creative and innovative Shavya’s class teacher 17. A 20 m deep well, with diameter 7 m is dug and the brings clay in the classroom, to teach the topic. Surface Areas and Volumes. With clay, she forms a earth from digging is evently spread out to form a cylinder of radius 4 cm and height 18 cm. Then, she moulds the cylinder into a sphere and ask some platform 22 m by 14 m. The height of the platform question to students. is (a) 2.5 m (b) 3.5 m (c) 3 m (d) 2 m 18. If the radius of the base of a right circular cylinder is halved, keeping the height same, then find the ratio of the volume of the cylinder thus obtained to the volume of original cylinder. [CBSE 2009] (a) 1 (b) 1 (c) 1 (d) 1 34 2 5 19. Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water. The water level rises by 5.6 cm. When marble dropped into the beaker, then the number of marble is (i) The radius of the sphere so formed is (a) 150 (b) 160 (a) 4 cm (b) 6 cm (c) 175 (d) 235 (c) 7 cm (d) 8 cm
CBSE Term II Mathematics X (Standard) 127 (ii) The volume of the sphere so formed is 25. The Great Stupa at Sanchi is one of the oldest stone (a) 905.14 3 (b) 903.27 3 structures in India, and an important monument of cm cm (c) 1296.5 3 (d) 1156.63 3 Indian Architecture. It was originally cm cm (iii) Find the ratio of the volume of sphere to the commissioned by the emperor Ashoka in the 3rd volume of cylinder. century BCE. Its nucleus was a simple (a) 2 : 1 (b) 1 : 2 hemispherical brick structure built over the relics (c) 1 : 1 (d) 3 : 1 of the Buddha. It is a perfect example of (iv) Total surface area of the cylinder is combination of solid figures. A big hemispherical (a) 553.14 cm2 (b) 751.52 cm2 dome with a cuboidal structure mounted on it. (c) 625 cm2 (d) 785.38 cm2 (take π = 22 ) 7 (v) During the conversion of a solid from one shape to [CBSE Question Bank] another the volume of new shape will (a) be increase (b) be decrease (c) remain unaltered (d) be double 24. Geeta and Meena have 10 and 6 CD respectively, each of radius 4 cm and thickness 1 cm. They place their CD one above the other to form solid cylinders. Dome Chattra Toranas Harmika Balustrade Stairs Based on the above information, answer the following questions. (i) Curved surface area of the cylinder made by Geeta is (i) Calculate the volume of the hemispherical dome if (a) 308.17 cm2 (b) 132 cm2 the height of the dome is 21 m. (c) 154 cm2 (d) 251.42 cm2 (a) 19404 cu m (b) 2000 cu m (ii) The ratio of curved surface area of the cylinder (c) 15000 cu m (d) 19000 cu m made by Geeta and Meena is (ii) The formula to find the volume of sphere is (a) 3 : 5 (b) 3 : 2 (a) 2 π r3 (b) 4 πr3 3 3 (c) 5 : 3 (d) 5 : 7 (iii) The volume of the cylinder made by Meena is (c) 4π r2 (d) 2 πr2 (a) 301.44 cm3 (b) 144 cm3 (iii) The cloth require to cover the hemispherical dome (c) 132 cm3 (d) 208.42 cm3 if the radius of its base is 14m is (iv) The ratio of the volume of the cylinders made by (a) 1222 sq m (b) 1232 sq m Geeta and Meena is (c) 1200 sq m (d) 1400 sq m (a) 1 : 2 (b) 2 : 5 (iv) The total surface area of the combined figure (c) 3 : 5 (d) 5 : 3 i.e. hemispherical dome with radius 14 m and (v) When two CD Cassette are shifted from Geeta cuboidal shaped top with dimensions 8 m × 6 m cylinder to Meena’s cylinder, then × 4 m is (a) Volume of two cylinder become equal (a)1200 sq m (b) 1232 sq m (b) Volume of Geeta’s cylinder > Volume of Meena’s (c) 1392 sq m (d) 1932 sq m cylinder (v) The volume of the cuboidal shaped top is with (c) Volume of Meena’s cylinder > Volume of Geeta’s dimensions mentioned in question (iv). cylinder (a) 182.45 m3 (b) 282.45 m3 (d) None of the above (c) 292 m3 (d) 192 m3
128 CBSE Term II Mathematics X (Standard) PART 2 9. A building is in the form of a cylinder surmounted Subjective Questions by a hemispherical dome (see the figure). The base diameter of the dome is equal to 2 of the total G Short Answer Type Questions 3 height of the building. Find the height of the 1. Two identical cubes each of volume 64 cm 3 are building, if it contains 67 1 m 3 of air. joined together end to end. What is the surface area 21 of the resulting cuboid? r 2. How many shots each having diameter 3 cm can be r made from a cuboidal lead solid of dimensions 9 cm × 11 cm × 12 cm? H h 3. 16 glass spheres each of radius 2 cm are packed into 10. Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter a cuboidal box of internal dimensions 16 cm × 8 cm 2 cm and height 16 cm. Find the diameter of each × 8 cm and then the box is filled with water. Find sphere. the volume of water filled in the box. 4. If a solid piece of iron in the form of a cuboid of 11. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid dimensions 49 cm × 33 cm × 24 cm, is moulded to is 104 cm and the radius of each hemispherical end form a solid sphere. Then, find radius of the sphere. is 7 cm, then find the cost of polishing its surface at the rate of ` 2 per dm2 . 5. If volumes of two spheres are in the ratio 64 : 27, then find the ratio of their surface areas. 6. The decorative block shown in the following figure ⎢⎡⎣take, π = 22 ⎤ is made of two solids, a cube and a hemisphere. 7 ⎦⎥ The base of the block is a cube with edge 6 cm and the hemisphere fixed on the top has a diameter of 12. A solid metallic hemisphere of radius 8 cm is 2.1 cm, then find the total surface area of the melted and recasted into a right circular cone of block and find the total area to be painted. base radius 6 cm. Determine the height of the cone. ⎣⎡⎢take, π = 22 ⎤ 7 ⎦⎥ 13. A rectangular water tank of base 11 m × 6 m 2.1 cm contains water upto a height of 5 m. If the water in the tank is transferred to a cylindrical tank of 6 cm radius 3.5 m, find the height of the water level in the tank. 6 cm 14. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 8 m of uniform thickness. Find the thickness of the wire. 6 cm 15. The rain water from a roof of dimensions 22 m × 20 m drains into a cylindrical vessel having 7. From a solid cube of side 7 cm, a conical cavity of diameter of base 2 m and height 3.5 m. If the rain height 7 cm and radius 3 cm is hollowed out. Find water collected from the roof just fill the the volume of the remaining solid. cylindrical vessel, then find the rainfall (in cm). 8. A hemispherical bowl of internal radius 9 cm is full 16. A cylindrical bucket of height 32 cm and base of liquid. The liquid is to be filled into cylindrical radius 18 cm is filled with sand. This bucket is shaped bottles each of radius 1.5 cm and height emptied on the ground and a conical heap of sand 4 cm. How many bottles are needed to empty the is formed. If the height of the conical heap is 24 bowl? cm, find the radius and slant height of the heap.
CBSE Term II Mathematics X (Standard) 129 17. The barrel of a fountain pen, cylindrical in shape, is 26. A medicine-capsule is in the shape of a cylinder of 7 cm long and 5 mm in diameter. A full barrel of ink diameter 0.5 cm with two hemispheres stuck to in the pin is used up on writing 3300 words on an each of its ends. The length of entire capsule is average. How many words can be written in a bottle 2 cm. The capacity of the capsule is of ink containing one-fifth of a litre? 27. A rocket is in the form of a right circular cylinder 18. Water flows at the rate of 10 m min −1 through a closed at the lower end and surmounted by a cone cylindrical pipe 5 mm in diameter. How long would it with the same radius as that of the cylinder. The take to fill a conical vessel whose diameter at the diameter and height of the cylinder are 6 cm and base is 40 cm and depth 24 cm? 12 cm, respectively. If the slant height of the conical portion is 5 cm, then find the total surface 19. Water flows through a cylindrical pipe, whose inner area and volume of the rocket. [use π = 3.14] radius is 1 cm, at the rate of 80 cms −1 in an empty cylindrical tank, the radius of whose base is 40 cm. 28. A solid toy is in the form of a hemisphere What is the rise of water level in tank in half an hour? surmounted by a right circular cone. The height 20. A factory manufactures 120000 pencils daily. The pencils are cylindrical in shape each of length 25 cm of the cone is 3 cm and the diameter of the base is and circumference of base as 1.5 cm. Determine the cost of colouring the curved surfaces of the pencils 4 cm. Determine the volume of the solid toy. If a manufactured in one day at ` 0.05 per dm 2. right circular cylinder circumscribes the toy, then 21. A well of diameter 10 m is dug 14 m deep. The Earth taken out of it is spread evenly all around to a width find the difference of the volumes of the cylinder of 5 m to form an embankment. Find the height of embankment. and the toy. [take, π = 3. 14] 22. Marbles of diameter 1.4 cm are dropped into a 29. A wooden toy rocket is in the shape of a cone cylindrical beaker of diameter 7 cm containing some mounted on a cylinder, as shown in figure. The water. Find the number of marbles that should be height of the entire rocket is 24 cm, while the dropped into the beaker, so that the water level rises height of the conical part is 4 cm. by 5.6 cm. 4 cm G Long Answer Type Questions 24 cm 23. If a hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is 4 cm 6 cm assumed that 1 space of the cube remains unfilled. 8 6 cm Base of cylinder Then, the number of marbles that the cube can Base of cone accomodate is The base of the conical portion has a diameter of 24. A solid iron cuboidal block of dimensions 6 cm, while the base diameter of the cylindrical 4.4 m × 2.6 m × 1 m is recast into a hollow cylindrical portion is 4 cm. If the conical portion is to be pipe of internal radius 30 cm and thickness 5 cm. painted orange and the cylindrical portion yellow, Find the length of the pipe. then find the area of the rocket painted with each of these colours. [take, π = 3.14] 25. A building is in the form of a cylinder surmounted by 30. Two solid cones A and B are placed in a cylindrical a hemispherical vaulted dome and contains 41 19 m 3 tube as shown in the figure. The ratio of their 21 capacities is 2 : 1. Find the heights and capacities of air. If the internal diameter of dome is equal to its of cones. Also, find the volume of the remaining total height above the floor, find the height of the portion of the cylinder. building? 21 cm AB
130 CBSE Term II Mathematics X (Standard) 31. How many spherical lead shots of diameter 4 cm (i) Ram Mandir is constructed in the form of the can be made out of a solid cube of lead whose edge cubical base of 30 cm × 20 cm × 10 cm, then find measures 44 cm. the area covered. 32. A metallic spherical shell of internal and external (ii) If the radius of the cylinder is 7 cm and Height of diameters 4 cm and 8 cm, respectively is melted the cylinder is 60 cm and the radius of the cone is and recast into the form of a cone of base diameter similar to that of cylinder and Height of the cone is 8 cm. Find the height of the cone. 24 cm, then the ratio of curved surface area of cylinder to curved surface area of the cone. 33. How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular (iii) Given structure in based on the concept of lead piece with dimensions 66 cm, 42 cm and 21 cm? (a) Area and perimeter 34. Find the number of metallic circular disc with (b) Surface area and volume 1.5 cm base diameter and of height 0.2 cm to be melted to form a right circular cylinder of height (c) Both (a) and (b) 10 cm and diameter 4.5 cm. (d) None of the above 35. A heap of rice is in the form of a cone of diameter 9 m and height 3.5 m. Find the volume of the rice. 39. Adventure camps are the perfect place for the How much canvas cloth is required to just cover children to practice decision making for themselves heap? without parents and teachers guiding their every move. Some students of a school reached for 36. How many cubic centimetres of iron is required to adventure at Sakleshpur. At the camp, the waiters construct an open box whose external dimensions served some students with a welcome drink in a are 36 cm, 25 cm and 16.5 cm provided the cylindrical glass and some students in a thickness of the iron is 1.5 cm. If one cubic hemispherical cup whose dimensions are shown centimetre of iron weights 7.5 g, then find the below. weight of the box. d=7 cm d=7 cm 37. Water is flowing at the rate of 15 kmh −1 through a h=10.5 cm pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in pond rise by 21 cm? G Case Base Questions 38. Mathematics teacher of a school took her 10th standard students to show Ram Mandir. It was a part of their Educational trip. The teacher had interest in history as well. She narrated the facts of Ram Mandir to students. Ram mandir is a Hindu temple that is being built in Ayodhya, which is in Uttar Pradesh. The temple construction is being supervised by the Shri Ram Janmabhoomi Teerth Kshetra. Then the teacher said in this monuments one can find combination of solid figures. She pointed that there are cubical bases and in centre cylinder with the cone shape structure on the top is constructed.
CBSE Term II Mathematics X (Standard) 131 After that they went for a jungle trek. The jungle trek 40. On a Sunday, your Parents took you to a fair. You could see lot of toys displayed, and you was enjoyable but tiring. As dusk fell, it was time to take wanted them to buy a RUBIK’s cube and strawberry ice-cream for you. shelter. Each group of four students was given a canvas Observe the figures and answer the questions. of area 551m2. Each group had to make a conical tent to [CBSE Question Bank] accommodate all the four students. Assuming that all the stitching and wasting incurred while cutting, would amount to 1 m2, the students put the tents. The radius of the tent is 7 m. [CBSE Question Bank] Area = 551 m2 (i) Find the length of the diagonal if each edge measures 6cm. r=7m (ii) Find volume of the solid figure if the length (i) Find the volume of cylindrical cup. of the edge is 7cm. (ii) Find the volume of hemispherical cup. (iii) What is the curved surface area of hemisphere (ice-cream) if the base radius is 7 (iii) Find the height of the conical tent prepared to cm? accommodate four students. (iv) Find the slant height of a cone if the radius is 7 cm and the height is 24 cm (v) Find the total surface area of cone with hemispherical ice cream. SOLUTIONS Objective Questions 3. (c) Given, volume of brick = 223 cm3 1. (c) Here, on joining three cubes, we get a cuboid whose ∴Volume of 1 cone = 1 πr2h length, l = 5 + 5 + 5 = 15 cm, breadth, b = 5 cm and 3 height, h = 5 cm = 1 × 22 × 2 × 2 × 7 = 22 × 4 5 cm 5 cm 5 cm 37 3 5 cm Let number of cones = n I II III Then, n × 22 × 4 = 22 × 22 × 22 5 cm 3 ∴ Required surface area of the resulting solid ⇒ n = 22 × 22 × 3 4 = Surface area of new cuboid ∴ n = 121 × 3 = 363 = 2 (lb + bh + hl) = 2 (15 × 5 + 5 × 5 + 5 × 15) = 2 (75 + 25 + 75) = 2 (175) = 350 cm2 4. (b) Because the sphere encloses in the cylinder, therefore the diameter of sphere is equal to diameter 2. (a) Because solid ball is exactly fitted inside the cubical box of side of cylinder which is 2r cm. a. So, a is the diameter for the solid ball. 5. (a) Because curved surface area of a hemisphere is 2πr2 and here, we join two solid hemispheres along ∴ Radius of the ball = a their bases of radius r, from which we get a solid 2 sphere. 4 ⎛⎝⎜ 2a⎟⎠⎞ 3 1 πa3 Hence, the curved surface area of new solid π 6 = 2πr2 + 2πr2 = 4πr2 So, volume of the ball = = 3
132 CBSE Term II Mathematics X (Standard) 6. (d) Since, the total surface area of cylinder of radius r and and total surface area of a cylinder of base radius, r and height h = 2πrh + 2πr2 height, h When one cylinder is placed over the other cylinder of same = Curved surface area + Area of both base height and radius, = 2πrh + 2πr2 then height of the new cylinder = 2 h Here, when we placed a cone over a cylinder, then one base and radius of the new cylinder = r is common for both. ∴Total surface area of the new cylinder = 2πr(2h ) + 2πr2 So, total surface area of the combined solid = 4πrh + 2πr2 = πrl + 2πrh + πr2 = 2πr(2h + r) = πr [l + 2h + r] 7. (a) Because the shape of sharpened pencil is = πr ⎡ r2 + h2 + 2h + r⎤ ⎣ ⎦ 13. (c) We know that, capacity of cylindrical vessel = πr2h cm 3 = + and capacity of hemisphere = 2 πr3 cm = Cylinder + Cone 3 8. (a) Because the shape of surahi is From the figure, capacity of the cylindrical vessel = πr2h − 2 πr3 = 1 πr2[3h − 2r] 33 14. (c) We have, diameter of metallic sphere = 6 cm = + = Sphere + Cylinder ∴Radius of metallic sphere, r1 = 3 cm Also, diameter of cross-section of cylindrical wire = 0.2 cm ∴Radius of cross-sections of cylindrical wire, r2 = 0.1 cm Let the length of the wire be h cm. Since, metallic sphere is converted into a cylindrical shaped 9. (b) Let the radii of two cones are r1, r2 and their heights are wire of length h cm. h1 and h 2. ∴Volume of the metal used in wire = Volume of the sphere Given, r1 = 3 and h1 = 1 ⇒ πr22h = 4 πr13 r2 1 h2 3 3 1 πr12h1 ⎜⎛⎝ 110⎟⎞⎠ 2 4 3 3 V1 ⎛ r1 ⎞ 2⎛ h1 ⎞ ⇒ π × × h = × π × 27 V2 ⎝⎜ r2 ⎟⎠ ⎜ h2 ⎟⎠ Now, the ratio of their volumes, = 1 = ⎝ 3 πr22h 2 ⇒ π × 1 × h = 36π 100 = ⎜⎝⎛ 13⎠⎞⎟ 2 × ⎝⎛⎜ 13⎞⎟⎠ = 9× 1 = 3 = 3:1 31 ⇒ h = 36π × 100 π Hence, the ratio of their volumes is 3 : 1. = 3600 cm = 36 m 10. (c) [Q 1m = 100cm ] 15. (c) During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered. =+ + 16. (c) Volume of the remaining solid = Cone + Cylinder + Cone = Volume of the cylinder − Volume of the cone = Two cones and a cylinder = ⎧⎨⎩π × 62 × 10 − 1 × π × 62 × 10⎭⎬⎫ 3 11. (b) = (360π − 120π ) = 240π cm 3 17. (a)Q Radius of the well = 7 m = 3. 5 m 2 = + = Hemisphere + Cone ∴Volume of the earth dug out = 22 × (3. 5)2 × 20 7 12. (c) We know that, total surface area of a cone of radius, r and height, h = Curved surface Area + area of base = 22 × 3. 5 × 3. 5 × 20 = πrl + πr2 7 where, l = h 2 + r2 = 770m 3 Area of platform = (22 × 14) m 2 = 308 m 2 ∴ Height = 770 = 2.5 m 308
CBSE Term II Mathematics X (Standard) 133 18. (b) Let the radius of original right circular cylinder be ‘r’ cm. Now, volume of rice = volume of cone = 1 πr2h and height be ‘h’ cm. Then, 3 Volume of original cylinder V1 = (πr2h )cm3 = 1 × 22 × (12)2 × 3. 5 ⎢⎣⎡Q r = d = 24 = 12 m⎥⎦⎤ 37 2 2 Volume of circular cylinder, when radius is halved = 22 × 144 × 3.5 V2 = π ⎜⎛⎝ r ⎞⎠⎟ 2 cm3 = πr 2h cm3 21 2 4 h = 11088 = 528 m3 21 Thus, V2 = πr 2h × π 1 = 1. V1 4 r 2h 4 Now, slant height l = h 2 + r2 19. (a) So, volume of one spherial marble = 4 π (0.7)3 3 = (3. 5)2 + (12)2 = 12.25 + 144 ⎣⎢⎡Q volume of sphere = 4 πr 3 ⎤ = 156.25 = 12. 5 m 3 ⎦⎥ ∴The canvas required to cover the heap = πrl = 1.372 π cm 3 = 22 × 12 × 12. 5 = 3300 = 471. 42 m2 3 77 ∴ Volume of the raised water in beaker = π (3.5)2 × 5.6 22. (b) Volume of the wall = 270 × 300 × 350 = 28350000 cm 3 [Q volume of cylinder = πr2h] [Q volume of cuboid = length × breadth × height] = 68.6 π cm3 Since, 1 space of wall is covered by mortar. Now, required number of marbles 8 Volume of the raised water in beaker So, remaining space of wall = Volume of wall = Volume of one spherical marble = 68.6 π × 3 = 150 marbles − Volume of mortar 1.372 π = 28350000 − 28350000 × 1 8 20. (a) Given, wooden article is a combination of a cylinder and = 28350000 − 3543750 two hemispheres. = 24806250 cm 3 3.5 cm 10 cm Now, volume of one brick = 22. 5 × 11.25 × 8.75 = 2214.844 cm 3 [Q volume of cuboid = length × breadth × height] ∴ Required number of bricks = 24806250 = 11200 (approx.) 2214. 844 Hence, the number of bricks used to construct the wall is 11200. Here, height of the cylinder, h = 10 cm 23. (i) (b) Since, volume of sphere = Volume of cylinder ⇒ 4 πR 3 = πr2h, where R ,r are the radii of sphere and Q Radius of base of the cylinder 3 cylinder, respectively. = Radius of hemisphere, r = 3.5 cm ⇒ R3 = r2 × h × 3 4 Now, required TSA of the wooden article ⇒ R 3 = 4 × 4 × 18 × 3 = 8 × 27 ⇒ R 3 = (2 × 3)3 4 = 2 × CSA of one hemisphere + CSA of cylinder = 2 × (2πr2) + 2π r h ∴ R = 6 cm (ii) (a) Volume of sphere = 4 πR 3 = 4 × 22 × 6 × 6 × 6 = 2 π r (2r + h ) 3 37 =2 × 22 × 3.5 × (2 × 3.5 + 10) = 905.14 cm3 7 = 22 × 7 × (7 + 10) = 22 × 17 = 374 cm 2 (iii) (c) Since the volume of sphere is equal to volume of 7 21. (b) Given diameter d = 24 m cylinder, then the ratio of volume of the sphere to the A volume of cylinder = 1 : 1 3.5m (iv) (a) Total surface area of cylinder = 2πr(r + h ) = 2 × 22 × 4(4 + 18) = 2 × 22 × 4 × 22 77 = 553.14 2 cm B C (v) (c) During the conversion of a solid from one shape to 24 m another the volume of new shape will remain unaltered.
134 CBSE Term II Mathematics X (Standard) 24. We have radius of each CD cassette = 4 cm Subjective Questions and thickness of each cassette = 1 cm 1. Let the length of a side of a cube = a cm a So, height of cylindrical made by Geeta, h1 = 10 × 1 = 10 cm and height of cylindrical made by Meena, h 2 = 6 × 1 = 6 cm a (i) (d) Curved surface area of cylinder made by Geeta = 2πrh =2 × 22 × 4 × 10 = 251.42 cm2 a 7 ⎡Curved surface area of ⎤ (ii) (c) ∴Required ratio = ⎣⎢cylinder made by Geeta⎦⎥ 2a ⎡Curved surface area of ⎤ ⎣⎢cylinder made by Meena⎥⎦ Given, volume of the cube, a3 = 64 cm3 ⇒ a = 4 cm = 2 πrh1 = h1 10 = 5: 3 On joining two cubes, we get a cuboid whose = 2πrh 2 h 2 6 length, l = 2a cm (iii) (a) Volume of cylinder made by Meena breadth, b = a cm and height, h = a cm = πr2h 2 = 22 × 4 × 4 × 6 Now, surface area of the resulting cuboid 7 = 2 (lb + bh + hl) = 301.44 cm3 = 2 (2a ⋅ a + a ⋅ a + a ⋅ 2a) = 2 (2a2 + a2 + 2a 2) = 2 ( 5a2) (iv) (d) Required ratio = 10 a2 = 10 (4)2 = 160 cm 2 = Volume of the cylinder made by Geeta Volume of the cylinder made by Meena = πr 2h1 = h1 = 5: 3 2. Given, dimensions of cuboidal = 9 cm × 11 cm × 12 cm πr2h 2 h2 ∴Volume of cuboidal = 9 × 11 × 12 = 1188 cm 3 (v) (a) When two CD Cassette are shifted from Geeta’s and diameter of shot = 3 cm cylinder to Meena’s cylinder, then length of both cylinders become equal. ∴ Radius of shot, r = 3 = 1. 5 cm 2 So, volume of both cylinders become equal. Volume of shot = 4 πr3 = 4 × 22 × (1.5)3 25. (i) (a) As, we know that hemisphere is a type of solid in 3 37 which radius is the height. So, radius = 21 m = 297 = 14.143 cm 3 21 ∴Required volume = 2 πr3 3 ∴Required number of shots = 1188 = 84 (approx.) 14.143 = 2 × 22 × 21 × 21 × 21 3 7 = 19404 cu m 3. Given, dimensions of the cuboidal = 16 cm × 8 cm × 8 cm ∴Volume of the cuboidal = 16 × 8 × 8 = 1024 cm 3 (ii) (b) Volume of sphere = 4 πr3 3 Also, given radius of one glass sphere = 2 cm (iii) (b) Given, radius (r) = 14 m ∴Volume of one glass sphere = 4 πr3 = 4 22 × (2)3 ∴Curved surface area of hemisphere dome = 2πr2 × = 2 × 22 × 14 × 14 3 37 7 = 704 = 33.523 cm 3 21 = 1232 sq m Now, volume of 16 glass spheres = 16 × 33.523 = 536.37 cm 3 (iv) (c) Here, radius of hemispherical dome (r) = 14 m ∴ Required volume of water = Volume of cuboidal Surface area of dome = 2πr2 = 2 × 22 × 14 × 14 = 1232 m 2 − Volume of 16 glass spheres 7 = 1024 − 536.37 = 487.6 cm 3 and CSA of cuboidal shaped top = 2 × h(l + b) + lb 4. Given, dimensions of the cuboid = 49 cm × 33 cm × 24 cm ∴ Volume of the cuboid = 49 × 33 × 24 = 38808 cm 3 = 2 × 4(8 + 6) + 8 × 6 [Q volume of cuboid = length × breadth × height] = 8(14) + 48 Let the radius of the sphere is r, then = 112 + 48 = 160m 2 Volume of the sphere = 4 πr3 ∴Total surface area = 1232 + 160 = 1392 m 2 3 [Q volume of the sphere = 4 π × (radius)3] (v) (d) Volume of cuboidal shape = lbh = 8 × 6 × 4 = 192 m 3 3
CBSE Term II Mathematics X (Standard) 135 According to the question, Clearly, the total area to be painted = Total surface area of the decorative block − Area of base of cube Volume of the sphere = Volume of the cuboid ⇒ 4 πr3 = 38808 = 219.465 − 62 = 219.465 − 36 = 183.465 cm2 7. Given that, side of a solid cube (a) = 7 cm 3 ⇒ 4 × 22 r3 = 38808 × 3 Height of conical cavity i.e. cone, h = 7 cm 7 ⇒ r3 = 38808 × 3 × 7 = 441 × 21 4 × 22 ⇒ r3 = 21 × 21 × 21 ∴ r = 21 cm 7 cm Hence, the radius of the sphere is 21 cm. 5. Let the radii of the two spheres are r1 and r2, respectively. ∴ Volume of the sphere of radius, r1 = V1 = 4 πr13 …(i) 3 cm 3 [Q volume of sphere = 4 π (radius)3] Since, the height of conical cavity and the side of cube is 3 equal that means the conical cavity fit vertically in the cube. and volume of the sphere of radius, r2 = V2 = 4 π r23 …(ii) Radius of conical cavity i.e. cone, r = 3 cm 3 ⇒ Diameter = 2 × r = 2 × 3 = 6 cm 4 πr13 64 3 27 Since, the diameter is less than the side of a cube that means Given, ratio of volumes = V1: V2 = 64 : 27 ⇒ = the base of a conical cavity is not fit inhorizontal face of 4 πr23 cube. 3 Now, volume of cube = (side)3= a3 = (7)3 = 343 cm3 [using Eqs. (i) and (ii)] and volume of conical cavity i.e. cone = 1 π × r2 × h 3 ⇒ r13 = 64 ⇒ r1 = 4 …(iii) r23 27 r2 3 = 1 × 22 × 3 × 3 × 7 37 Now, ratio of surface area = 4πr12 4πr22 = 66 cm 3 [Q surface area of a sphere = 4π (radius)2] ∴Volume of remaining solid = Volume of cube − Volume of conical cavity = r12 r22 = 343 − 66 = 277 cm 3 Hence, the required volume of solid is 277 cm3. ⎛ r1 ⎞ 2 ⎛⎝⎜ 43⎟⎠⎞ 2 16 ⎜⎝ r2 ⎠⎟ 9 8. Given, radius of hemispherical bowl, r = 9 cm = = = [using Eq. (iii)] Hence, the required ratio of their surface area is 16 : 9. and radius of cylindrical bottles, R = 1. 5 cm and height, h = 4 cm 6. Here, the decorative block is a combination of a cube and a ∴Number of required cylindrical bottles hemisphere. = Volume of hemispherical bowl Volume of one cylindrical bottle For cubical portion, Each edge = 6 cm 2 πr3 For hemispherical portion, 3 = πR 2h Diameter = 2.1 cm ∴ Radius, r = 2.1 cm 2 ×π ×9×9×9 = 3 = 54 2 π × 1. 5 × 1. 5 × 4 Now, total surface area of the cube = 6 × (Edge)2= 6 × 6 × 6 = 216 cm2 9. Let r be the radius of the hemispherical dome Here, the part of the cube where the hemisphere is and total height of building be H m. attached, is not included in the surface area. It is given that diameter of dome So, the total surface area of the decorative block = 2 × Total height of the building 3 = Total surface area of cube − Area of base of hemisphere ⇒ r = 1H m + Curved surface area of hemisphere 3 = 216 − πr2 + 2πr2 = 216 + πr2 = 216 + 22 × 2.1 × 2. 1 Let h m be the height of the cylinder. 7 2 2 ∴ h = H − r = H − 1H = 2 Hm = 216 + 3.465 = 219.465 cm2 3 3
136 CBSE Term II Mathematics X (Standard) Volume of the air inside the building = Volume of air =2 × ⎡⎢⎣2 × 22 × (7 )2 ⎤ + 2 × 22 × (7) (90) 7 ⎦⎥ 7 inside the dome + Volume of air inside the cylinder = 2 πr3 + πr2h = 4 × 22 × 7 + 2 × 22 × 90 = 22 [28 + 180] 3 = 4576 cm2 ⎡2 ⎝⎛⎜ 1 H⎞⎟⎠ 3 ⎜⎝⎛ 13 H⎟⎞⎠ 2 ⎛⎜⎝ 2 H⎠⎞⎟ ⎤ 3 3 = π ⎢ 3 + ⎥ 7 cm ⎣ ⎦ 7 cm = π ⎡2H 3 + 2 H 3 ⎤ 27 ⎥ ⎢ 81 ⎦ ⎣ 104 cm = 8 πH3 m3 90 cm 81 7 cm Given, volume of the air inside the building = 67 1 m3 7 cm 21 ∴ 8 πH3 = 1408 81 21 ⇒ H3 = 1408 × 81 Then, the cost of polishing at the rate of ` 2 per dm2 21 8π = ` 4576 × 2 = ` 91.52 [Q1 dm2 = 100 cm2] ⇒ H3 = 1408 × 81 × 7 100 21 × 8 × 22 ⇒ H3 = 216 ⇒ H = 6 m 12. Let height of the cone be h. 10. Given, diameter of the cylinder = 2 cm Given, radius of the base of the cone = 6 cm ∴ Volume of circular cone = 1 πr2h = 1 π (6)2 h ∴Radius = 1 cm and height of the cylinder = 16 cm [Q diameter = 2 × radius] 33 ∴Volume of the cylinder = π × (1)2 × 16 = 16 π cm 3 = 36 π h = 12 π h cm 3 3 [Q volume of cylinder= π × (radius)2 × height] Also, given radius of the hemisphere = 8 cm Now, let the radius of solid sphere = r cm ∴Volume of the hemisphere = 2 π r3 = 2 π (8)3 Then, its volume = 4 πr3 cm 3 33 3 = 512 × 2π cm 3 [Q volume of sphere = 4 × π × (radius)3] 3 3 According to the question, According to the question, Volume of the cone = Volume of the hemisphere Volume of the twelve solid sphere = Volume of cylinder ⇒ 12πh = 512 × 2π ⇒ 12 × 4 πr3 = 16π 3 3 ∴ h = 512 × 2π = 256 = 28.44 cm ⇒ r3 = 1 ⇒ r = 1 cm 12 × 3 π 9 ∴ Diameter of each sphere, d = 2r = 2 × 1 = 2 cm 13. Given, dimensions of base of rectangular tank = 11 m × 6 m Hence, the required diameter of each sphere is 2 cm. and height of water = 5 m 11. Given, whole length of the solid = 104 cm Volume of the water in rectangular tank = 11 × 6 × 5 and the radius of each hemisphere = 7 cm = 330 m 3 Therefore, the length of the cylindrical part of the solid Also, given radius of the cylindrical tank = 3.5 m = (104 − 2 × 7) = 90 cm Let height of water level in cylindrical tank be h. For hemispherical portion, Then, volume of the water in cylindrical tank = πr2h Radius, r = 7 cm = π (3.5)2 × h = 22 × 3.5 × 3.5 × h = 11.0 × 3.5 × h = 38. 5 h m 3 For cylindrical portion, 7 Radius, r = 7 cm Height, h = 90 cm According to the question, ∴ Total surface area of the solid 330 = 38. 5 h = 2 × Curved surface area of a hemisphere [since, volume of water is same in both tanks] + Curved surface area of cylindrical part ∴ h = 330 = 3300 = 2 [2πr2] + 2πrh 38. 5 385 = 2 × [2π (7)2] + 2π (7) (90) ∴ = 8. 57 m or 8.6 m Hence, the height of water level in cylindrical tank is 8.6 m.
CBSE Term II Mathematics X (Standard) 137 14. Here, a rod of cylindrical shape is converted into a wire of ⇒ r2 = 10368 = 1296 ⇒ r = 36 cm 8 cylindrical shape. For rod, Again, let the slant height of the conical heap = l Diameter = 1 cm ⇒ Radius, r1 = 1 cm and length, Now, 2 = h2 + r2 = (24)2 + ( 36)2 2 l h = 8 cm = 576 + 1296 = 1872 ∴ Volume of the rod = πr12h = π × ⎛⎜⎝ 21⎟⎞⎠ 2 × 8 = 2 π cm 3 ∴ l = 43.267 cm Hence, radius of conical heap of sand = 36 cm For wire, and slant height of conical heap = 43.267 cm 17. Given, length of the barrel of a fountain pen = 7 cm Length = 8 m = 800 cm [Q1m = 100cm ] Let r be the radius (in cm) of cross-section of the wire, then and diameter = 5 mm = 5 cm = 1 cm ⎡⎢⎣Q 1mm = 1 cm ⎤ Volume of wire = π × r2 × 800 cm3 10 2 10 ⎦⎥ Since, the rod is converted into wire, so ∴ Radius of the barrel = 1 = 0.25 cm 2 ×2 Volume of wire = Volume of rod Volume of the barrel = πr2h [since, its shape is cylindrical] ⇒ π × r2 × 800 = 2π ⇒ r2 = 1 ⇒ r = 1 cm 400 20 = 22 × (0.25)2 × 7 7 Then, diameter = 2 = 1 cm 20 10 Hence, the diameter of the cross-section, i.e. the thickness = 22 × 0.0625 = 1.375 cm 3 of the wire is 1 cm, i.e. 1 mm [Q1cm = 10mm ] Also, given volume of ink in the bottle = 1 of litre 10 5 15. Given, length of roof = 22 m and breadth of roof = 20 m = 1 × 1000 cm 3 = 200 cm 3 5 Let the rainfall be a cm. Now, 1.375 cm 3 ink is used for writing number of words ∴Volume of water on the roof = 22 × 20 × a = 22a m 3 100 5 = 3300 ∴1 cm 3 ink is used for writing number of words = 3300 Also, we have radius of base of the cylindrical vessel, r = 1 m 1.375 and height of the cylindrical vessel, h = 3.5 m ∴200 cm 3 ink is used for writing number of words ∴Volume of water in the cylindrical vessel when it is just full = 3300 × 200 = 480000 1.375 = πr 2h = ⎛⎜⎝ 22 ×1×1× 7 ⎟⎞⎠ = 11 m 3 7 2 18. Given, speed of water flow = 10 m min−1 = 1000 cm/min Now, volume of water on the roof = Volume of water in and diameter of the pipe = 5 mm = 5 cm ⎢⎣⎡Q 1mm = 1 cm ⎤ 10 10 ⎦⎥ the vessel ⇒ 22 a = 11 ∴ Radius of the pipe = 5 = 0.25 cm 10 × 2 5 ∴ a = 11 × 5 = 2.5 cm ∴ Area of the face of pipe = πr2 = 22 × (0.25)2 = 0.1964 cm 2 22 7 [Q volume of cylinder=π × (radius)2 × height] Hence, the rainfall is 2.5 cm. 16. Given, radius of the base of the bucket = 18 cm Height of the bucket = 32 cm Also, given diameter of the conical vessel = 40 cm So, volume of the sand in cylindrical bucket = π r2h ∴Radius of the conical vessel = 40 = 20 cm = π (18)2 × 32 = 10368 π 2 and depth of the conical vessel = 24 cm Also, given height of the conical heap (h ) = 24 cm ∴Volume of conical vessel = 1 πr2h = 1 × 22 × (20)2 × 24 Let radius of heap be r cm. 3 37 Then, volume of the sand in the heap = 1 π r2h = 211200 = 10057.14 cm 3 3 21 = 1 πr2 × 24 = 8 π r2 ∴Required time = Area Volume of the conical vessel water 3 of the face of pipe × Speed of According to the question, = 10057.14 0.1964 × 1000 Volume of the sand in cylindrical bucket = Volume of the sand in conical heap = 51.20 min = 51 min 20 × 60 s ⇒ 10368 π = 8πr2 100 ⇒ 10368 = 8 r2 = 51 min 12 s
138 CBSE Term II Mathematics X (Standard) 19. Given, radius of tank, r1 = 40 cm The embankment is in the form of cylindrical shell, so area of embankment = π(R 2 − r2) = π(102 − 52) Let height of water level in tank in half an hour = h1 Also, given internal radius of cylindrical pipe, r2 = 1 cm = π (100 − 25) = 22 × 75 m2 and speed of water = 80 cm/s i.e. in 1s water flow = 80 cm 7 ∴ In 30 (min) water flow = 80 × 60 × 30 = 144000 cm According to the question, Since, Earth taken out from well is used to form embankment. Volume of water in cylindrical tank = Volume of water flow ∴ Volume of embankment from the circular pipe in half an hour = Volume of Earth taken out on digging the well ⇒ π r12 h1 = π r22 h 2 ⇒ Area of embankment × Height of embankment ⇒ 40 × 40 × h1 = 1 × 1 × 144000 = Volume of Earth dugout Volume of Earth dugout ∴ h1 = 144000 = 90 cm 40 × 40 ⇒ Height of embankment = Area of the embankment Hence, the level of water in cylindrical tank rises 90 cm in = 1100 = 4. 67 m half an hour. 22 × 75 7 20. Given, pencils are cylindrical in shape. 22. Given, diameter of a marble = 1.4 cm Length of one pencil = 25 cm ∴ Radius of marble (r) = 1.4 = 0.7 cm 2 and circumference of base, 2πr = 1.5 cm So, volume of one marble = 4 πr3 = 4 π (0.7)3 ⇒ r = 1.5 × 7 = 0.2386 cm 33 = 4 π × 0.343 = 1.372 π cm 3 22 × 2 33 Now, curved surface area of one pencil = 2πrh = 2 × 22 × 0. 2386 × 25 7 Also, given diameter of beaker = 7 cm = 262.46 = 37.49 cm 2 ∴ Radius of beaker = 7 = 3.5 cm 7 2 = 37.49 dm 2 ⎢⎣⎡Q 1 ⎤ 100 1 cm = 10 dm ⎥⎦ Height of water level raised = 5.6 cm = 0.375 dm 2 ∴Volume of the raised water in beaker = π (3.5)2 × 5.6 = 68.6π cm 3 ∴Curved surface area of 120000 pencils = 0.375 × 120000 Now, required number of marbles = 45000 dm 2 = Volume of the raised water in beaker Volume of one spherical marble Now, cost of colouring 1 dm 2 curved surface of the pencils manufactured in one day = ` 0.05 = 68.6 π × 3 = 150 ∴Cost of colouring 45000 dm 2 curved surface 1.372 π = 45000 × 0.05 = ` 2250 23. Given, edge of the cube = 22 cm ∴ Volume of the cube = (22)3 = 10648 cm 3 21. Here, a well is dug and Earth taken out of it is used to form [Q volume of cube = (side)3] an embankment. Given, Diameter of well = 10 m Also, given diameter of marble = 0.5 cm ∴ Radius = 10 = 5 m ∴ Radius of a marble, r = 0.5 = 0.25 cm 2 2 Also, depth = 14 m [Q diameter =2 × radius] Volume of one marble = 4 πr3 = 4 × 22 × (0.25)3 ∴Volume of Earth taken out on digging the well = πr2h = 22 ×(5)2 ×14 = 1100 m3 3 37 7 10 m [Q volume of sphere = 4 × π × (radius)3] 5m 5m 3 h = 1.375 = 0.0655 cm 3 21 14 m Filled space of cube = Volume of the cube − 1 8 × Volume of cube = 10648 − 10648 × 1 8 = 10648 × 7 = 9317 cm 3 8
CBSE Term II Mathematics X (Standard) 139 ∴ Required number of marbles ∴ Total volume of the building = Volume of the cylinder = Total space filled by marbles in a cube Volume of one marble + Volume of hemispherical dome = 9317 = 142244 (approx.) = ⎝⎜⎛ πr3 + 2 πr 3⎟⎠⎞ m3 = 5 πr3 m3 0.0655 3 3 According to the question, Hence, the number of marbles that the cube can accomodate Volume of the building = Volume of the air is 142244. ⇒ 5 πr3 = 41 19 24. Given that, a solid iron cuboidal block is recast into a hollow 3 21 cylindrical pipe. ⇒ 5 πr3 = 880 Length of cuboidal pipe (l) = 4.4 m 3 21 ⇒ r3 = 880 × 7 × 3 Breadth of cuboidal pipe (b) = 2.6 m and height of cuboidal pipe (h ) = 1 m 21 × 22 × 5 = 40 × 21 = 8 So, volume of a solid iron cuboidal block = l ⋅ b ⋅ h = 4.4 × 2.6 × 1 = 11.44 m 3 21 × 5 ⇒ r3 = 8 ⇒ r = 2 m Also, internal radius of hollow cylindrical pipe (ri ) = 30 cm = 0.3 m ∴ Height of the building = 2r = 2 × 2 = 4 m and thickness of hollow cylindrical pipe = 5 cm = 0.05 m 26. Given, diameter of cylinder = Diameter of hemisphere = 0.5 cm So, external radius of hollow cylindrical pipe [since, both hemispheres are attach with cylinder] (re) = ri + Thickness = 0.3 + 0. 05 = 0. 35 m ∴ Radius of cylinder (r) = radius of hemisphere (r) = 0.5 = 0.25 cm ∴Volume of hollow cylindrical pipe 2 = Volume of cylindrical pipe with external radius [Q diameter = 2 × radius] − Volume of cylindrical pipe with internal radius = πre2h1 − πri2h1 = π (re2 − ri2) h1 0.25 0.5 cm = 22 [( 0. 35)2 − ( 0. 3)2 ] ⋅ h1 0.25 7 = 22 [ 0. 1225 − 0.09]⋅ h1 = 22 [ 0. 0325] ⋅ h1 2 cm 7 7 and total length of capsule = 2 cm = 0.715 × h1 / 7 ∴Length of cylindrical part of capsule, where, h1 be the length of the hollow cylindrical pipe. Now, by given condition, h = Length of capsule − Radius of both hemispheres = 2 − (0.25 + 0.25) = 1.5 cm Volume of solid iron cuboidal block = Volume of hollow cylindrical pipe Now, capacity of capsule = Volume of cylindrical part + 2 × Volume of hemisphere ⇒ 11.44 = 0.715 × h / 7 ∴ h = 11.44 × 7 = 112 m = πr2h + 2 × 2 πr3 3 0.715 [Q volume of cylinder = π × (radius)2 Hence, required length of pipe is 112 m. × height and volume of hemisphere = 2 π(radius)3] 3 25. Let total height of the building = Internal diameter of the = 22 [(0.25)2 × 1.5 + 4 × (0.25)3] dome = 2r m 73 = 22 [0.09375 + 0.0208] rr r 7 2r r = 22 × 0.11455 = 0.36 cm 3 7 ∴ Radius of building (or dome) = 2r = r m 2 Hence, the capacity of capsule is 0.36 cm3. Height of cylinder = 2r − r = r m 27. Since, rocket is the combination of a right circular cylinder ∴ Volume of the cylinder = π r2 ( r) = πr3 m 3 and a cone. and volume of hemispherical dome on cylinder = 2 πr3 m 3 Given, diameter of the cylinder = 6 cm 3 ∴ Radius of the cylinder = 6 = 3 cm 2 and height of the cylinder = 12 cm ∴ Volume of the cylinder = πr2h = 3.14 × (3)2 × 12 = 339.12 cm 3 and curved surface area = 2πrh = 2 × 3.14 × 3 × 12 = 226.08
140 CBSE Term II Mathematics X (Standard) A = 2 πr3 + 1 πr2h 33 5 cm h = ⎡2 × 3.14 × (2 )3 + 1 × 3.14 × (2)2 × 3⎦⎤⎥ ⎢⎣ 3 3 B O 3 cm C = 16.75 + 12. 56 = 29.31 cm3 ED Hence, volume of the solid toy is 29.31 cm3. 6 cm Now, in right angled ΔAOC,12 cm Now, let the right circular cylinder EFGH circumscribe the given solid toy. Then, radius of the base of the right circular h = 52 − 32 cylinder = HP = BO = 2 cm and its height, = 25 − 9 = 16 = 4 EH = AP = AO + OP = 3 + 2 = 5 cm ∴Height of the cone, h = 4 cm So, volume of the cylinder = π r 2h and radius of the cone, r = 3 cm Now, volume of the cone = 3.14 × 22 × 5 = 1 πr2h = 1 × 3.14 × (3)2 × 4 = 62.8 cm3 33 Now, required difference of the volume of the cylinder and = 113.04 = 37.68 cm 3 3 the solid toy = Volume of cylinder − Volume of solid toy = 62.8 − 29.31 = 33.49 cm 3 and curved surface area = πrl = 3.14 × 3 × 5 = 47.1 Hence, total volume of the rocket 29. Here, the given wooden toy rocket is combination of a cone and a cylinder. = 339.12 + 37.68 = 376.8 cm 3 and total surface area of the rocket For conical portion, = CSA of cone + CSA of cylinder + Area of base of cylinder = 47.1 + 226.08 + 28.26 Diameter = 6 cm = 301.44 cm2 28. Here, given solid toy is a combination of a right circular ∴ Radius, r1 = 6 cm = 3 cm cone and a hemisphere. 2 Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere as shown in the figure. Height, h1 = 4 cm [Q l = r2 + h 2 ] For conical portion, height, h = 3 cm Then, slant height, l = (3)2 + 42 Diameter, d = 4 cm = 9 + 16 = 25 = 5 cm E AF For cylindrical portion, B OC Diameter = 4 cm H PG ∴Radius, r = 4 cm = 2 cm ∴ Radius, r2 = 4 cm = 2 cm 2 2 For hemispherical portion, Height, h 2 = Total height of rocket − Height of cone Radius, r = 2 cm [Q radii of hemisphere and cone are same] = 24 − 4 [Q total height of rocket = 24 cm] So, volume of the solid toy = 20 cm = Volume of hemisphere + Volume of cone Here, we have to find the area of the rocket painted with orange and yellow colours separately. Since, radius of base of cone is larger than radius of base of cylinder and cone is mounted on cylinder. ∴ Area to be painted orange = Curved surface area of cone + Area of base of cone − Area of base of cylinder [Q area of base of cylinder is common in area of base of cone] = πr1l + πr12 − πr22 = 3.14 × 3 × 5 + 3.14 × (3)2 − 3.14 × (2)2 = 3.14 [15 + 9 −4] = 3.14 × 20 = 62.8 cm 2 Now, area to be painted yellow = Curved surface area of cylinder + Area of base of the cylinder = 2πr2h 2 + π r22 = 2 × 3.14 × 2 × 20 + 3.14 × (2)2 = 3.14 [80 + 4] = 3.14 × 84 = 263.76 cm 2
CBSE Term II Mathematics X (Standard) 141 30. Let volume of cone A be 2 V and volume of cone B be V. Now, since edge of a solid cube (a) = 44 cm Again, let height of the cone A = h1 cm, then height of cone So, volume of a solid cube = (a)3 B = (21 − h1) cm 21 cm = (44)3 AB = 44 × 44 × 44 cm3 6 cm From Eq. (i), 6 cm Number of spherical lead shots = 44 × 44 × 44 × 21 4 × 22 × 8 h1 21 – h1 = 11 × 21 × 11 = 121 × 21 Given, diameter of the cone = 6 cm = 2541 Hence, the required number of spherical lead shots is 2541. ∴ Radius of the cone = 6 = 3 cm 2 Now, volume of the cone, A = 2V = 1 πr 2h = 1 π( 3)2 h1 32. Given, internal diameter of spherical shell = 4 cm 3 3 and external diameter of shell = 8 cm 1 3 ⇒ V = 6 π 9 h1 = 2 h1π …(i) ∴ Internal radius of spherical shell, r1 = 4 cm =2 cm 2 and volume of the cone, [Q diameter = 2 × radius] B =V = 1 π ( 3)2 (21 − h1 ) = 3π (21 − h1 ) …(ii) and external radius of shell, r2 = 8 = 4 cm 3 2 From Eqs. (i) and (ii), [Q diameter = 2 × radius] 3 h1π = 3π (21 − h1 ) 8 cm 2 ⇒ h1 = 2 (21 − h1) 4 cm ⇒ 3h1 = 42 ⇒ h1 = 42 = 14 cm 3 ∴ Height of cone, B = 21 − h1 = 21 − 14 = 7 cm Spherical shell Now, volume of the cone, A = 3 × 14 × 22 = 132 cm 3 Now, volume of the spherical shell = 4 π [ r23 − r13 ] 7 3 [using Eq. (i)] [Q volume of the spherical shell = 4 π and volume of the cone, B = 1 × 22 × 9 × 7 = 66 cm 3 3 37 {(external radius)3 − (internal radius)3}] [using Eq. (ii)] = 4 π (43 − 23) Now, volume of the cylinder = πr2h = 22 (3)2 × 21 = 594 cm 3 3 7 = 4 π (64 − 8) ∴ Required volume of the remaining portion 3 = Volume of the cylinder = 224 π cm 3 − (Volume of cone A + Volume of cone B) 3 = 594 − (132 + 66) = 396 cm 3 31. Given that, lots of spherical lead shots made out of a solid Let height of the cone = h cm cube of lead. Diameter of the base of cone = 8 cm ∴ Number of spherical lead shots ∴ Radius of the base of cone = 8 = 4 cm = Volume of a solid cube of lead 2 Volume of a spherical lead shot ...(i) [Q diameter = 2 × radius] Given that, diameter of a spherical lead shot i.e. sphere = 4 cm According to the question, ⇒ Radius of a spherical lead shot (r) = 4 Volume of cone = Volume of spherical shell 2 ⇒ 1 π(4)2h = 224 π r = 2 cm [Q diameter = 2 × radius] 33 ⇒ h = 224 = 14 cm So, volume of a spherical lead shot i.e. sphere 16 = 4 π r3 [Q volume of cone = 1 × π × (radius)2 × (height)] 3 3 = 4 × 22 × (2)3 = 4 × 22 × 8 cm3 37 21 Hence, the height of the cone is 14 cm.
142 CBSE Term II Mathematics X (Standard) 33. Given that, lots of spherical lead shots made from a solid ⇒ Radius of a right circular cylinder (r) = 4. 5 cm 2 rectangular lead piece. ∴ Number of spherical lead shots ∴ Volume of right circular cylinder = π r2h = Volume of solid rectangular lead piece Volume of a spherical lead shot ...(i) = π ⎛⎝⎜ 42.5⎠⎟⎞ 2 × 10 Also, given that diameter of a spherical lead shot = π × 4. 5 × 4. 5 × 10 4 i.e. sphere = 4.2 cm ∴ Radius of a spherical lead shot, r = 4.2 = 2.1 cm ∴ Number of metallic circular disc = Volume of a right circular cylinder 2 Volume of a metallic circular disc ⎣⎡⎢Q radius = 1 diameter ⎤⎥⎦ π × 4. 5 × 4. 5 × 10 2 So, volume of a spherical lead shot i.e. sphere = 4 = 4 π r3 π 3 × 1. 5 × 1. 5 × 0. 2 = 4 × 22 × (2.1)3 4 37 = 4 × 22 × 2.1 × 2.1 × 2.1 = 3 × 3 × 10 = 900 = 450 37 0.2 2 = 4 × 22 × 21 × 21 × 21 3 × 7 × 1000 Hence, the required number of metallic circular disc is 450. 35. Given that, a heap of rice is in the form of a cone. Height of a heap of rice i.e. cone (h ) = 3. 5 m Now, length of rectangular lead piece, l = 66 cm and diameter of a heap of rice i.e. cone = 9 m Breadth of rectangular lead piece, b = 42 cm Radius of a heap of rice i.e. cone (r) = 9 m Height of rectangular lead piece, h = 21 cm 2 ∴Volume of a solid rectangular lead piece i.e. cuboid So, volume of rice = 1 π × r2h =l×b×h = 66 × 42 × 21 3 = 1 × 22 × 9 × 9 × 3. 5 From Eq. (i), 3 7 22 Number of spherical lead shots = 6237 = 74.25 m3 = 66 × 42 × 21 × 3 × 7 × 1000 4 × 22 × 21 × 21 × 21 84 = 3 × 22 × 21 × 2 × 21 × 21 × 1000 4 × 22 × 21 × 21 × 21 Now, canvas cloth required to just cover heap of rice = Surface area of a heap of rice = 3 × 2 × 250 = πrl = 22 × r × r2 + h 2 7 = 6 × 250 = 1500 = 22 × 9 × ⎜⎛⎝ 29⎠⎞⎟ 2 + (3.5)2 72 Hence, the required number of spherical lead shots is 1500. 34. Given that, lots of metallic circular disc to be melted to form = 11 × 9 × 81 + 12. 25 a right circular cylinder. Here, a circular disc work as a 7 4 circular cylinder. = 99 × 130 = 99 × 32. 5 Base diameter of metallic circular disc = 1. 5 cm 7 47 ∴ Radius of metallic circular disc = 1. 5 cm = 14.142 × 5.7 = 80. 61 m2 2 Hence, 80.61 m2 canvas cloth is required to just cover heap. [Q diameter = 2 × radius] 36. Let the length(l), breadth (b) and height (h) be the external dimension of an open box and thickness be x. and height of metallic circular disc i.e. = 0.2 cm ∴Volume of a circular disc = π × (Radius)2 × Height = π × ⎜⎛⎝ 12. 5⎞⎠⎟ 2 × 0.2 = π × 1. 5 × 1. 5 × 0. 2 xx 4 x Now, height of a right circular cylinder (h ) = 10 cm and diameter of a right circular cylinder = 4. 5 cm
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