Exaguru

3. What is the value tan 60°? (a) 1 (b) 3 (c) 3 (d) 1 3 22 4. The value of 1 + sin B when ∠B = 60° is cos B (a) 3 3 (b) 2 – 3 (c) 2 3 (d) 2 + 3 5. The value of 1+ tan 2 B when ∠B = 60° is 1+ cot 2 B (a) 3 (b) 4 (c) 3 3 (d) 4 3 3 1 3. (b) 3 4. (d) 2 + 3 5. (a) 3 2. (c) Ans. 1. (a) 2 2 Experts’ Opinion Questions based on following types are very important for Exams. So, students are advised to revise them thoroughly. 1. To use trigonometric ratios of some specific angles. 2. To use trigonometric identities. COMMON ERRORS Errors Corrections (i) Incorrectly interpreting that sin A means the (i) sin A is an abbreviation for ‘the sine of product of ‘sin’ and ‘A’. angle A’. It is not the product of ‘sin’ and A. Similar interpretations follow for all the other trigonometric ratios. (ii) Interpreting incorrectly that trigonometric ratios (ii) Trigonometric ratios are numerical quantities. represent lengths. Each one of them represents the ratio of one length to another. They must themselves never be considered as lengths. (iii) Interpreting incorrectly that trigonometric ratios ( iii) The trigonometric ratios depend on the magnitude depend on the lengths of the sides of the triangle. of the angle and not upon the lengths of the sides of the triangle. (iv) Drawing other types of triangles instead of right- (iv) Any right-angled triangles are to be used and angled triangles for calculating trigonometric hypotenuse is related to right triangles only. ratios. (v) Writing trigonometric ratios without angle. (v) It is not correct. Must write ratios with correct angles and do sufficient practice. (vi) Considering sin2 q + cos2 q = 1 and taking its (vi) Remember that square root is taken only for square root as sin q + cos q = ± 1. whole term i.e., not with + and – sign. 100  Mathematics–10

QUICK REVISION NOTES •• Trigonometry deals with the relationships between sides and angles of a triangle. •• Trigonometric ratios of the angle A in a right triangle ABC, right-angled at B are defined as: C B BC AB (i) sin A = AC (ii) cos A = AC BC 1 AC (iii) tan A = AB (iv) cosec A = sin A = BC 1 AC 1 AB (v) sec A = cos A = AB (vi) cot A = tan A = BC •• Trigonometric Ratios of some specific angles: A q 0° 30° 45° 60° 90° sin q 0 1 1 3 1 2 2 2 cos q 1 31 1 0 2 22 tan q 0 11 3 ND* 3 cosec q ND* 2 22 1 3 sec q 1 2 2 2 ND* 3 cot q ND* 3 1 10 3 * ND stands for ‘Not defined’. •• Trigonometric Relations/Identities: sin θ cos θ (i) tan q = cos θ (ii) cot q = sin θ (iii) sin2 q + cos2 q = 1, 0° £ q £ 90° (iv) sec2 q = tan2 q + 1, 0° £ q < 90° (v) cosec2 q = 1 + cot2 q, 0° < q £ 90° IMPORTANT FORMULAE •• In a right triangle ABC, right-angled at B, (i) sin A = side opposite to angle A (ii) cos A = side adjacent to angle A hypotenuse hypotenuse side opposite to angle A sin A 1 A (iii) tan A = side adjacent to angle A = cos A (iv) cosec A = sin A 1 1 cos A C (v) sec A = cos A (vi) cot A = tan A = sin A • (i) sin2 A + cos2 A = 1, for 0° ≤ A ≤ 90°. (ii) sec2 A – tan2 A = 1, for 0° ≤ A B < 90°. (iii) cosec2 A – cot2 A = 1, for 0° < A ≤ 90°. Introduction to Trigonometry  101

7 Areas Related to Circles Topics Covered 1. Perimeter and Area of a Circle 2. Areas of Sector and Segment of a Circle 1. Perimeter and Area of a Circle As we know that a circle is a closed curve consisting of a set of all those points of the plane which are at a constant distance from a fixed point in the plane. The fixed point is called its centre. The constant distance is called its radius. It is the line segment joining any point on the boundary (circumference) to centre. The boundary (or perimeter) of a circle is called its circumference. • Circumference of a circle = 2pr (unit) • Area of a circle = pr2 unit2 • Area of the circular path formed by two concentric circles of radii r1 and r2 (r1 > r2) = pr12 – pr22 = p (r12 – r22) (unit)2, • The distance travelled (covered) by a wheel in 1 round = its circumference = 2pr (unit) • Total distance covered by a wheel = its circumference × number of rounds taken by it. • Number of rounds made by a wheel = Total distance covered Its circumference • Speed of the wheel = Total distance covered Time taken 5 18 • If speed = x km/hr, then speed = x × m/s and if speed = x m/s, then speed = x × km/hr. 18 5 22 Note: Unless stated otherwise, the value of p is to be taken as . 7 Example 1. The area of the circle, the circumference of which is equal to the perimeter of a square of side 11 cm is (b) 144 cm2 (c) 154 cm2 (d) 180 cm2 (a) 122 cm2 Solution. Since, side of the square = 11 cm \\ Perimeter of the square = 4 × 11 cm = 44 cm Given: Circumference of circle = perimeter of square 22 ⇒ 2pr = 44  ⇒  2 × × r = 44  ⇒  r = 7 cm 7 \\ Area of the circle = pr2 = 22 × (7)2 = 154 cm2 7 Hence, option (c) is the correct answer. Example 2. The area of a ring shaped region enclosed between two concentric circles of radii 20 cm and 15 cm is (b) 415 cm2 (c) 520 cm2 (d) 550 cm2 (a) 330 cm2 102

Solution. For the given ring shaped region; R = 20 cm and r = 15 cm \\ Required area = pR2 – pr2 = p (R + r) (R – r) = 22 (20 + 15) (20 – 15) cm2 = 550 cm2 7 Hence, option (d) is the correct answer. Example 3. If the perimeter and area of a circle are numerically equal; its radius will be (a) 1 unit (b) 2 units (c) 4 units (d) None of these Solution. Let the radius of the circle be r units; then given that: 2pr = pr2 ⇒ r = 2 units Hence, option (b) is the correct answer. Example 4. A wheel has diameter 84 cm. Number of complete revolutions must it make to cover 792 metres will be (a) 100 (b) 160 (c) 220 (d) 300 22 Solution. Since distance covered by the wheel in 1 round = p × diameter = × 84 cm = 264 cm 7 And, total distance covered = 792 m = 792 × 100 cm \\  No. of complete revolutions made = Total distance covered = 792 × 100 cm = 300 Distance covered by wheel in 1 round 264 cm Hence, option (d) is the correct answer. Example 5. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour? (a) 2275 (b) 2650 (c) 3815 (d) 4375 Solution.  Speed = 66 km per hour = 66 × 5 m/s = 66 × 5 × 100 cm/s 18 18 and, time taken = 10 minutes = 10 × 60 sec = 600 sec \\ Distance covered in 10 min =S peed × time = 66 × 5 × 100 × 600 cm = 11,00,000 cm 18 Given, diameter of each wheel = 80 cm \\  Distance covered by each wheel in 1 revolution = its circumference. 22 1760 = p × diameter = × 80 cm = cm 77 ⇒  No. of revolutions made by each wheel in 10 min = Total distance covered Distance covered in 1 revolution 11,00,000 cm = 4375 = 1760 cm 7 Hence, option (d) is the correct answer. Exercise 7.1 A. Multiple Choice Questions (MCQs) . Choose the correct answer from the given options: 1. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is (a) 22 : 7 (b) 14 : 11 (c) 7 : 22 (d) 11 : 14 Areas Related to Circles  103

2. The area of the square that can be inscribed in a circle of radius 8 cm is (a) 256 cm2 (b) 128 cm2 (c) 64 2 cm2 (d) 64 cm2 3. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is (a) 31 cm (b) 25 cm (c) 62 cm (d) 50 cm 4. If the area of circle is numerically equal to twice its circumference, then the diameter of the circle is (a) 4 units (b) 6 units (c) 8 units (d) 12 units 5. The perimeter (in cm) of a square circumscribing a circle of radius a cm is (a) 2a (b) 4a (c) 6a (d) 8a 6. What is the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm? (a) 20 cm (b) 30 cm (c) 50 cm (d) 80 cm 7. What is the area of the circle that can be inscribed in a square of side 6 cm? (a) 9 p cm2 (b) 11 p cm2 (c) 16 p cm2 (d) 15 p cm2 8. The area of a quadrant of a circle whose circumference is 25 cm is [Imp.] (a) 24 cm2 (b) 28 cm2 (c) 32.5 cm2 (d) 38.5 cm2 9. The areas of two circles are in the ratio 9 : 4, then what is the ratio of their circumferences? (a) 1 : 2 (b) 2 : 1 (c) 3 : 2 (d) 2 : 3 10. The cost of fencing a circular field at the rate of ` 24 per metre is ` 5280. The radius of the field is (a) 15 m (b) 35 m (c) 25 m (d) 30 m 11. The radii of two circles are 8 cm and 6 cm respectively. The radius of the circle having area equal to the sum of the areas of the two circles is (a) 5 cm (b) 10 cm (c) 12 cm (d) 15 cm 12. An athlete runs on a circular track of radius 49 m and covers a distance of 3080 m along its boundary. How many rounds has he taken to cover this distance? Take π = 22  7  (a) 5 (b) 8 (c) 10 (d) 15 13. The area of the largest triangle that can be inscribed in a semi-circle of radius r units will be (a) r sq. units (b) r sq. units (c) r2 sq. units (d) 2r sq. units 2 14. The area of circle whose circumference is 22 cm is (a) 32 cm2 (b) 45 cm2 (c) 55 cm2 (d) 77 cm2 2222 15. A road which is 7 m wide surrounds a circular park whose circumference is 88 m. The area of the road is (a) 220 m2 (b) 340 m2 (c) 550 m2 (d) 770 m2 B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): If the circumference of a circle is 176 cm, then its radius is 28 cm. Reason (R): Circumference = 2p × radius. 2. Assertion (A): If the outer and inner diameter of a circular path is 10 m and 6 m respectively, then area of the path is 16 p m2. Reason (R): If R and r be the radius of outer and inner circular path respectively, then area of circular path = p (R2 – r2). 104  Mathematics–10

Answers and Hints A. Multiple Choice Questions (MCQs) 1 0. (b) 35 m 1. (b) 14 : 11 2. (b) 128 cm2 Length of the fence = Total cost Rate 3. (d) 50 cm 4. (c) 8 units ` 5280 5. (d) 8a =    ` 24/metre = 220 m 6. (c) 50 cm According to the question, pR2 = πr12 + πr22 So, length of fence = Circumference of the field ⇒ p(R2) = π(r12 + r22 ) ⇒ R2 = r12 + r22 220 m = 2pr = 2 × 22 × r = (24)2 + (7)2 r1 = 24 cm \\ 7 = 576 + 49 = 625 r2 = 7 cm  So, r = 220 × 7 m = 35 m 2 × 22 ⇒ R = 625   ⇒ R = 25 ∴ Diameter = 2R = 2 × 25 = 50 cm. 1 1. (b) 10 cm pr2 = 64p + 36p 7. (a) 9p cm2  pr2 = 100p  ⇒  r2 = 100 Diameter of the circle inscribed in a square = side of square ⇒ ∴ 2r = a ⇒ r = 10 cm ⇒ r = a = 6 = 3 cm ∴  Radius of the required circle is 10 cm. 2 2 12. (c) 10 ∴ Area of circle = pr2 = p(3)2 = 9p cm2 1 3. (c) r2 square units. 8. (d) 38.5 cm2 2πr + 2r = 25 cm ⇒ 4 pr + 4r = 50 ⇒ r  22 + 4 = 50  7 ⇒ 7 Base AB of triangle ABC in semicircle is r = 50 × = 7 cm constant, i.e., equal to 2r, and maximum altitude may be equal to r. 50 \\A rea of quadrant = πr2 = 22 × 7 × 7 1 4 7×4 ∴ Area of triangle = base × altitude = 38.5 cm2 2 9. (c) 3 : 2 1 AB × OC = 1 (2r) × r = r2 = Let r1 and r2 be the radii of two circles, A1 22 and A2 be areas of two circles and C1 and C2 be circumferences of two circles. ∴  Area of triangle in semicircle A1 = 9 = r2 square units. A2 4 Then, A1 : A2 = 9 : 4  ⇒  1 4. (d) 77 cm2 πr12 2 2pr = 22 cm πr22 ⇒ = 9 ⇒  r1 = 3 r = 22 = 22 × 7 ⇒ r = 7 Now,   r2 2 4 C1 : C2 = 2pr1 : 2.pr2 2π 2 × 22 2 = 2πr1 = 3 Area of circle = pr2 = 22 ×  7 2 = 77 cm2 2πr2 2 7  2 2 So, C1 : C2 = 3 : 2 Areas Related to Circles  105

15. (d) 770 m2 22 Circumference of circular = (21 + 14)(21 – 14) 7 park = 88 m 22 × 35 × 7 = 770 m2 = 2pr1 = 88 m 7 ⇒ r1 = 88 = 88 × 7 = 14 m B. Assertion-Reason Type Questions 2×π 2 × 22 1. (a) Both assertion (A) and reason (R) are true Width of road = 7 m and reason (R) is the correct explanation of assertion (A). So, r2 = r1 + 7 m = 14 + 7 = 21 m 2. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of So, area of the road = πr22 − πr12 = 22 (212 – 142) assertion (A). 7 2. Areas of Sector and Segment of a Circle A sector is a part of the circular region which is enclosed by two radii and the corresponding arc. Hence, OABC is a minor sector and OCDA is a major sector. ∠AOC is called the angle of sector. • Area of the sector of a circle of radius r with central angle θ = θ × πr2 , where 360° q is measured in degrees. OR 11 Area of the sector = 2 × length of arc × radius = 2 lr • Length of the arc of the sector of a circle of radius r with central angle θ = θ × 2πr , where q is measured in degrees. 360° • Area of the minor segment APB of the circle in the given figure = area of sector OAPB – area of DOAB = i × rr2 – 1 r2 × sin q 360° 2 • Area of the major sector of a circle of radius r = pr2 – area of the corresponding minor sector. Example 1. The area of a sector of a circle with radius 6 cm if angle of the sector is 60° is (a) 15 2 cm2 (b) 16 1 cm2 (c) 18 6 cm2 (d) 19 3 cm2 3 2 7 8 Solution. Given: r = 6 cm and q = 60° \\ Area of sector = πr 2 × θ = 22 60° cm2 = 132 cm2 6 cm2 360° ×6×6× 7 = 18 7 360° 7 Hence, option (c) is the correct answer. Example 2. A chord AB of a circle of radius 10 cm subtends a right angle at the centre. The area of the minor-sector is (b) 42 cm2 (c) 78.5 cm2 [Take p = 3.14] [Imp.] (a) 38.5 cm2 (d) 82 cm2 Solution:  q = 90° and r = 10 cm \\  Area of the minor-sector OACB = πr 2 ×θ O 10 cm 360° B = 3.14 × 10 × 10 × 90° cm2 10 cm C 360° A = 78.5 cm2 Hence, option (c) is the correct answer. 106  Mathematics–10

Areas of Combinations of Plane Figures Example 3. A square ABCD is inscribed in a circle of radius 10 units. The area of the circle, not included in the square is (b) 108 cm2 (c) 114 cm2 (Take p = 3.14) (a) 84 cm2 (d) 122 cm2 Solution. When a square or a rectangle is inscribed in a circle then diagonal of the square is the diameter of the circle. \\  The diagonal AC of the square ABCD = Diameter of the circle = 2 × 10 cm = 20 cm Since, the diagonals of a square are equal and bisect each other at 90°; therefore; AC = BD = 20 cm 1 and area of the square = 2 × AC × BD = 1 × 20 × 20 cm2 = 200 cm2 2 Also, area of the circle = pr2 = 3.14 × (10)2 cm2 = 314 cm2 \\ The required area = Area of circle – Area of the square = 314 cm2 – 200 cm2 = 114 cm2. Hence, option (c) is the correct answer. Example 4. The area of an equilateral triangle is 17320.5 cm2. With each vertex as centre, a circle is described with radius equal to half the length of the side of the triangle. The area of the triangle not included in the circles is (b) 1810.25 cm2 (Use p = 3.14 and 3 = 1.73205). (a) 1620.51 cm2 (c) 2430.60 cm2 (d) None of these Solution. We know that area of an equilateral triangle = 3 × (side)2 4 \\ 3 × (side)2 = 17320.5 A 4 ⇒ (side)2 = 17320.5 × 4 = 17320.5 × 4 60° 3 1.73205 ⇒ Side = 200 cm. 200 BC ⇒  Radius of each circle drawn = 2 cm = 100 cm For each minor sector; q = 60° ×θ 360° \\  Area of each sector inside the triangle = πr 2 = 3.14 × (100)2 × 60° cm2 = 5233.33 cm2 360° \\  Area of the triangle not included in the circles = Area of triangle – 3 × area of each sector inside the triangle = (17320.5 – 3 × 5233.33) cm2 = 1620.51 cm2. Hence, option (a) is the correct answer. Example 5. In the given figure, PQ = 24 cm, PR = 7 cm and O is the centre of the circle. The area of the shaded portion is (b) 148.20 cm2 (a) 132.58 cm2 (d) 161.54 cm2 (c) 154.36 cm2 Solution. We know that the angle subtended by semi-circle is 90° ⇒ ∠RPQ = 90°, i.e., DRPQ is a right-angled triangle with hypotenuse = RQ \\ RQ2 = PR2 + PQ2 Areas Related to Circles  107

⇒ RQ2 = (7)2 + (24)2 = 49 + 576 = 625 ⇒ RQ = 25 cm = diameter of the circle. RQ 25 \\ Radius of the circle = 2 = 2 cm = 12.5 cm Area of given shaded portion = Area of semicircle – Area of right DRPQ = 1 πr2 − 1 × RP × PQ = 1 × 22 × (12.5)2 − 1 × 7 × 24 = 1718.75 – 84 22 27 2 7 = 1718.75 - 588 1130.75 cm2 = 161.54 cm2 7 = 7 Hence, option (d) is the correct answer. Example 6. In a circular table cover of radius 32 cm, a design is formed having an equilateral triangle ABC in the middle, as shown below. The area of the design is  [Imp.] (a) 777.36 cm2 (b) 1888.11 cm2 (c) 2010.54 cm2 (d) None of these Solution. Let O be the centre of the circle. Join OB and OC, also draw OD ⊥ BC. Since, ABC is an equilateral triangle, ∠BAC = 60°. Also, angle subtended by an arc at the centre of the circle is twice the angle at remaining circumference. \\ ∠BOC = 2 ∠BAC = 2 × 60° = 120° Now, by R.H.S., DOBD ≅ DOCD  ⇒  ∠BOD = ∠COD = 60° In DOBD, OB = radius of the circle = 32 cm ⇒ BD 3 BD ⇒  BD = 16 3 = CD sin 60° = OB ⇒ 2 = 32   ∴ BC = 2 BD = 2 × 16 3 cm = 32 3 cm Area of the design = Area of the circle – Area of equilateral DABC = πr2 − 3 × (side)2 =  22 × (32)2 − 3 3)2  cm2 4  7 × (32    4 =  22528 − 768 3 cm2 = 1888.11 cm2  7 Hence, option (b) is the correct answer. Example 7. The area of the shaded portion in the figure, given below, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre is [Imp.] (a) 156.64 cm2 (b) 188.46 cm2 (c) 256.64 cm2 (d) 310.25 cm2 Solution. Since, DOAB is equilateral, ∠AOB = 60° Required area = Area of circle + Area of DOAB – Area of sector with angle 60° = πr2 + 3 × (side)2 − πr2 × θ 4 360° 108  Mathematics–10

= 22 × 62 + 3 × (12)2 − 22 × 62 × 60 cm2 7 4 7 360 22 3 ×144 cm2 = 7 (36 − 6) + 4 =  660 + 36 3 = 156.64 cm2  7 Hence, option (a) is the correct answer. Example 8. ABCD is a square of side 4 cm. At each corner of the square, a quarter circle of radius 1 cm, and at the centre, a circle of radius 1 cm, are drawn, as shown in the given figure. The area of the shaded region is (b) 7.25 cm2 (c) 9.71 cm2 (d) 10.43 cm2 (a) 8.46 cm2 Solution. Area of the shaded portion AB = Area of given square ABCD 2 cm – 4 × area of each quarter circle – area of the circle at the centre. = 4 × 4 − 4  π(1)2  − π(1)2 sq. cm 1 cm  4    = 16 − 22 22  5 D C 7 7  9 − cm2 = cm2 = 9.71 cm2 4 cm 7 Hence, option (c) is the correct answer. Exercise 7.2 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options. 1. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The length of the arc is (a) 11 cm (b) 22 cm (c) 27 cm (d) 44 cm 2. An arc of length 15.7 cm subtends a right angle at the centre of the circle. Then the radius of the circle is (a) 20 cm (b) 10 cm (c) 15 cm (d) 12 cm 3. The angle described by a minute hand in 5 minutes is (a) 30° (b) 60° (c) 90° (d) None of these 4. The area of the sector in the following figure showing a chord AB of [Imp.] a circle of radius 18 cm subtending an angle of 60° at the centre O is  [Take p = 3.14] O (a) 151.31 cm2 (b) 169.56 cm2 60° (c) 173.33 cm2 (d) None of these A B 5. The given figure is a sector of circle of radius 10.5 cm. The perimeter [Imp.] of the sector is [Take p = 22 ] (a) 32 cm 7 (b) 44 cm (c) 54 cm (d) None of these Areas Related to Circles  109

6. In a circle of diameter 42 cm,if an arc subtends an angle of 60° at the centre where p = 22 , then what 7 will be the length of arc? (a) 11 cm (b) 20 cm (c) 22 cm (d) 28 cm 7. A horse is tied to a pole with 28 m long rope. The perimeter of the field where the horse can graze is (Take p = 22/7) (a) 60 cm (b) 85 cm (c) 124 cm (d) 176 cm 8. A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through 22 an angle 120°. The total area cleaned at each sweep of the blades is [Take p = 7 ] (a) 360 cm2 (b) 448 cm2 (c) 556 cm2 (d) 924 cm2 9. The area of the shaded region in the given figure, if AC = 24 cm, BC = 10 A cm and O is the centre of the circle is [Take p = 3.14] (a) 128.56 cm2 (b) 145.33 cm2 O (c) 248.16 cm2 (d) None of these BC 10. A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its 22 centre. The radius of the circle is [Take p = 7 ] (a) 7 cm (b) 14 cm (c) 21 cm (d) 28 cm 11. The short and long hands of a clock are 4 cm and 6 cm long respectively. The sum of distances travelled by their tips in 2 days is (a) 1148 cm (b) 1426.35 cm (c) 1910.85 cm (d) None of these 12. The area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm is (a) 3.25 cm2 (b) 8.75 cm2 (c) 4.60 cm2 (d) 5.50 cm2 13. The area of the largest circle that can be drawn inside the given rectangle of length ‘a’ cm and breadth ‘b’ cm (a > b) is (a) 1 πb2 cm2 (b) 1 πb2 cm2 (c) 1 πb2 cm2 (d) pb2 cm2 234 14. All the vertices of a rhombus lie on a circle. The area of the rhombus, if the area of the circle is 1256 cm2 is [Use p = 3.14] (a) 300 cm2 (b) 600 cm2 (c) 800 cm2 (d) 900 cm2 15. The difference of the areas of two segments of a circle formed by a chord of radius 5 cm subtending an angle of 90° at the centre is [Imp.] (a)  25π − 25 cm2 (b)  15π − 7 cm2 (c)  7π − 3 cm2 (d) None of these  4 2   4 2  4 2 B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): In a circle of radius 6 cm, the angle of a sector is 60°. Then the area of the sector is 6 cm2. 18 7 110  Mathematics–10

Reason (R): Area of the circle with radius r is pr2. 2. Assertion (A): The length of the minute hand of a clock is 7 cm. Then the area swept by the minute hand in 5 minute is 5 cm2. 12 6 θ × 2πr . Reason (R): The length of an arc of a sector of angle q and radius r is given by l = 360° Answers and Hints A. Multiple Choice Questions (MCQs) = 120° 22 × ×21×21 1. (b) 22 cm 2. (b) 10 cm 3. (a) 30° 360° 7 4. (b) 169.56 cm2 = 462 cm2 Area of sector = 36600°° × 3.14 × 18 × 18 ∴  Total area cleaned by two wipers = 169.56 cm2 = 2 × 462 = 924 cm2 5. (a) 32 cm 9. (b) 145.33 cm2 Radius (r) = 10.5 cm Here, AB is diameter, Angle (q) = 60° AC = 24 cm, BC = 10 cm and –ACB = 90° Perimeter = θ × 2πr + 2r  [Angle in a semicircle is 90°] 360° \\ AB2 = AC2 + BC2 = 60° 22 × 10.5 + 2 × 10.5 [By Pythagoras theorem] ×2× 360° 7 ⇒ AB = (24)2 + (10)2 cm = 11 + 21 = 32 cm = 576 +100 cm 6. (c) 22 cm = 676 cm = 26 cm Length of arc = θ ( 2πr ) ⇒ AB 360° OB = OA = = 13 cm 60°  22 21 2 360°  7 = 2 × × \\  Area of shaded region = Area of semicircle = 22 cm  – Area of DACB 7. (d) 176 cm =  1 π(13)2 1 × 24 ×10 cm2  2 − Horse can graze in the field which is a circle 2 of radius 28 cm. =  1 × 3.14 × 169 − 120 cm2  2 So, required perimeter = [265.33 – 120] cm2 = 145.33 cm2 = 2pr = 2p (28) cm 22 10. (c) 21 cm = 2 × 7 × 28 cm = 176 cm Length of wire = Length of Arc 8. (d) 924 cm2 θ 22 Here, r = 21 cm, q = 120° ⇒ 22 cm = ×2× ×r 360° 7 ⇒ 22 = 60° 22 ×2× ×r 360° 7 ⇒ r = 22 × 360 × 7 = 21 cm 60 × 2 × 22 Area of a sector = θ ×πr2 360° Thus, the radius of the circle = 21 cm Areas Related to Circles  111

11. (c) 1910.85 cm 14. (c) 800 cm2 1 2. (b) 8.75 cm2 Diagonalofarhombusareperpendicularbisector Here, l = 3.5 cm, r = 5 cm of each other. Length of arc l = 2rri 360° Q ⇒ 3.5 = 2×r×5×i rθr 360° ri ⇒ 36 = 3.5 3.5 × 36 l ⇒ q = r \\  Each diagonal is diameter of the circle. Now, Area of sector Now, area of circle = 1256 cm2 = rr2 i = r × 5 × 5 × 35 × 36 360° 360° × r ×10 pr2 = 1256  r2 1256 fi fi  = π 25 × 35 875 = 100 = 100 1256 fi r2 = 3.14 = 400 = 8.75 cm2 ∴  Area of sector = 8.75 cm2 fi r = 20 cm 13. (c) 1 πb2 cm2 \\  Diameter of the circle = 40 cm 4 = Each diagonal of the rhombus The diameter of circle that can be drawn Area of rhombus inside the rectangle is equal to the breadth of 11 = 2 (d1 × d2) = 2 × 40 × 40 rectangle. = 800 cm2 The length of the rectangle = a cm 15. (a)  25π − 25  cm 2  4 2  The breadth of the rectangle = b cm ∴ Diameter of circle = b cm B. Assertion-Reason Type Questions ⇒ r = b cm 1. (b) Both assertion (A) and reason (R) are true but 2 reason (R) is not the correct explanation of ∴ Area of circle A = pr2 assertion (A). b 2 1 πb2 2. (b) Both assertion (A) and reason (R) are true but 2 4 = π   = cm2 reason (R) is not the correct explanation of assertion (A). Case Study Based Questions I. A student of class X standard finds in a circular table cover with radius 32 cm, a design which is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Some questions arises in his mind which he shares with you. Help him to solve these questions. 1. What is the area of the circular table? (a) 22528 cm2 20528 (b) cm2 7 7 (c) 22028 cm2 (d) 28022 cm2 7 7 112  Mathematics–10

2. What is the measure of angle subtended by each side of the equilateral triangle at the centre of the circle? (a) 30° (b) 60° (c) 120° (d) 90° 3. Each side of the equilateral triangle is (a) 64 3 cm (b) 8 3 cm (c) 16 3 cm (d) 32 3 cm 4. What is the area of equilateral triangle? (a) 700 3 cm2 (b) 768 3 cm2 (c) 678 3 cm2 (d) 876 3 cm2 5. What is the area of the design? (a) 1888 cm2 (b) 1688 cm2 (c) 1988 cm2 (d) 1588 cm2 Ans. 1. (a) 22528 cm2 2. (c) 120° 3. (d) 32 3 cm 7 4. (b) 768 3 cm2 5. (a) 1888 cm2 II. Some students of class-Xth get together and decide to form a design by combining different mathematical shapes. First of all they select a triangle ABC with AB = 3 cm, AC = 4 cm and ∠BAC = 90°. They draw semicircles on sides AB, AC and BC but they have some doubts which they want to clarify with you. So, answer their questions given below: A 3 cm 4 cm BC 1. The length of side BC is (a) 5 cm (b) 25 cm (c) 9 cm (d) 16 cm 2. Area of semicircle on side AB is (a) 3 p cm2 (b) 9 p cm2 (c) 9p cm2 (d) 4p cm2 28 3. Area of semicircle on side AC is (a) 4p cm2 (b) 2p cm2 (c) 9p cm2 (d) 16p cm2 4. Area of given figure is (a)  9 π + 6 cm2 (b)  16π + 6 cm2 (c)  25π + 6 cm2 (d)  4π + 6 cm2 4 9 8 25 5. Find the area of the shaded region. (a) 6 cm2 (b) 8 cm2 (c) 10 cm2 (d) 12 cm2 Ans. 1. (b) 25 cm 2. (c) 9p cm2 3. (d) 16p cm2 4. (b)  16π + 6 cm2 5. (a) 6 cm2 9 Areas Related to Circles  113

III. A human chain is to be formed in concentric form of radius 56 m and 63 m at India Gate. But the organiser has some problems, he wants you to solve his problems. Give answer to his questions by using information given in the questions: 1. The circumference of circular chain of radius 56 m is (a) 352 m (b) 176 m (c) 704 m (d) 88 m 2. The area covered by circular chain of radius 56 m is (a) 5544 m2 (b) 3850 m2 (c) 9856 m2 (d) 12474 m2 3. If each person is given two metres of space to stand, find how many persons can be accomodated in the chain of radius 56 m. (a) 126 persons (b) 176 persons (c) 156 persons (d) 186 persons 4. He wants to include one more concentric circle with a radius of 63 m, the circumference of the new circle will be (a) 156 m (b) 396 m (c) 176 m (d) 288 m 5. How many persons can accomodate on the new circle if each person needs 2 metres of space? (a) 198 persons (b) 156 persons (c) 176 persons (d) 186 persons Ans. 1. (a) 352 m 2. (c) 9856 m2 3. (b) 176 persons 4. (b) 396 m 5. (a) 198 persons IV. A square park has each side of 100 m. At each corner of the park, there is a flower bed in the form of a quadrant of radius 14 m as shown in given figure. Answer the question from (1) to (5): 1. Area of each quadrant is (a) 145 m2 (b) 154 m2 (c) 415 m2 (d) 514 m2 2. Area of 4 quadrants is (a) 616 m2 (b) 166 m2 (c) 661 m2 (d) 510 m2 114  Mathematics–10

3. Area of park with bed roses is (a) 10,000 m2 (b) 1000 m2 (c) 100 m2 (d) 10 m2 4. Area of park without bed roses is (a) 9834 m2 (b) 9384 m2 (c) 9483 m2 (d) 9348 m2 5. Angle of a quadrant is equal to (a) 70° (b) 80° (c) 90° (d) 45° Ans. 1. (b) 154 m2 2. (a) 616 m2 3. (a) 10,000 m2 4. (b) 9384 m2 5. (c) 90° V. A student of class-Xth decided to create a model of a circular wall clock and try to paste the numbers from 1 to 12 on its dial. But he is facing some problems. Give solutions to his problems by looking at the figure. 1. The angle between points showing hour is (a) 15° (b) 30° (c) 45° (d) 60° 2. What is the angle made at the centre between 12 and 3? (a) 30° (b) 45° (c) 60° (d) 90° 3. Find the area covered by minute hand of a clock from 12 to 3 if the radius of the clock is 14 cm. (a) 164 sq. cm (b) 77 sq. cm (c) 154 sq. cm (d) 308 sq.cm 4. The formula used in finding the area of sector with radius r and angle at the centre q is (a) πr2θ (b) πr2θ (c) 2πrθ (d) πrθ 360° 180° 180° 360° 5. The arc length between the region 12 and 3 if the radius of the clock is 14 cm. (a) 22 cm (b) 44 cm (c) 66 cm (d) 88 cm Ans. 1. (a) 15° 2. (d) 90° 3. (c) 154 sq. cm 4. (a) πr2θ 5. (a) 22 cm 360° Experts’ Opinion Questions based on following types are very important for Exams. So, students are advised to revise them thoroughly. 1. To find the area of sector of a circle. 2. To find the area of segment. 3. To find area of combination of plane figures. IMPORTANT FORMULAE •• Circumference of a circle = 2pr • Circumference of a semicircle = pr + 2r 1 •• Area of a circle = pr2 • Area of semicircle = 2 pr2 1 •• Area of quadrant = 4 pr2 •• Area of the ring = p (R2 – r2), R = outer radius and r = inner radius. θ 360° •• Length of an arc of a sector of a circle with radius r and angle with degree measure q = × 2πr •• Area of a sector of circle with radius r and angle with degrees measure q = θ × πr2 360° •• Area of the major sector = pr2 – Area of minor sector •• Area of segment of a circle = Area of the corresponding sector – Area of the corresponding triangle •• Area of major segment = pr2 – Area of minor segment Areas Related to Circles  115

COMMON ERRORS Errors Corrections (i) Interpreting the concept of sector and segment (i) Make concept of both term clear by understanding incorrectly. them. (ii) Finding perimeter of sector, without taking the (ii) Draw the figure and note the data given to get radius into consideration. the answer correctly. (iii) Using formula for area of sector and length of (iii) Remember all the formulae and use them arc of sector incorrectly. properly. (iv) While finding the area of shaded region students (iv) Be careful in such problems. Do enough practice forget to subtract the common area overlapped. to avoid the errors. QUICK REVISION NOTES •• A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point always remains the same. The fixed point is called as centre and constant distance is known as radius. •• Area and perimeter of a circle with radius r  Circumference (perimeter) = 2pr unit or pd unit, where d (diameter) = 2r unit  Area = pr2 or πd2 (unit)2 4  Area of semicircle = πr2 or πd2 (unit)2 2 8  Area of a of a quadrant = πr2 (unit)2 4 •• Area of a sector of circle with radius r and an angle q at the centre =p r2 × θ (unit)2 360° OR = 12 × length of arc × radius = 1 lr 2 •• Area of segment of a circle with radius r and angle with degree measure q •• = Area of the sector – Area of the triangle = θ × πr2 – 1 r2 × sin θ (unit)2 360° 2 •• Area enclosed by two concentric circles: If R and r are radii of two concentric circles, then area enclosed by the two circles = pR2 – pr2 = p (R2 – r2) = p (R – r) (R + r) (unit)2 116  Mathematics–10

8 Probability Topics Covered 1. Probability–A Theoretical Approach 1. Probability–A Theoretical Approach In theoretical approach of probability, predictions about the happenings, on the basis of certain assumptions, are made without actually performing the experiment. Probability of an event E is defined as P(E) and is given by the formula: P(E) = Number of trials in which the event happened Total number of trials Example 1. A card is drawn from a deck of 52 cards. The event E is that the card is not an ace of hearts. The number of outcomes favourable to E is (a) 52 (b) 53 (c) 51 (d) 31 Solution. There is only one ace of hearts. Therefore, number of outcomes favourable to E = 52 – 1 = 51. H ence, option (c) is the correct answer. Example 2. A card is selected from a deck of 52 cards. The probability of it being a red face card is 57 3 5 (a) (b) (c) (d) 52 52 26 26 Solution. There are 6 red face cards in a deck of 52 cards. \\  No. of favourable outcomes = 6 out of 52 possible outcomes Thus, No. of favourable outcomes 6 3 P(E) = No. of possible outcomes = 52 = 26 Hence, option (c) is the correct answer. Example 3. A card is drawn at random from a well-shuffled pack of 52 playing cards. The probability of getting neither a red card nor a queen is 6 7 11 9 (a) (b) (c) (d) 13 13 13 13 Solution. Total number of possible outcomes = 52 Number of red cards = 26 Number of queens = 2 So, number of red cards and queens = 28 Number of cards which are neither red card nor queen = 52 – 28 = 24 \\  P (getting neither a red card nor a queen) = 24 6 = 52 13 H ence, option (a) is the correct answer. 117

Example 4. Two dice are thrown at the same time and the product of numbers appearing on them is noted. The probability that the product is a prime number is 11 1 5 (a) (b) (c) (d) 36 5 6 Solution. Now for the product of the numbers on the dice is prime number can be have in these possible ways—(1, 2), (2, 1), (1, 3), (3, 1), (5, 1), (1, 5) So, number of possible ways = 6 61 \\ required probability = 36 = 6 H ence, option (b) is the correct answer. Example 5. Rahim tosses two different coins simultaneously. The probability of getting at least one tail is 13 3 1 (a) (b) (c) (d) 44 5 6 Solution. Number of possible outcomes = 4 as possible outcomes are HH, HT, TH, TT. Favourable outcomes for getting at least one tail are HT, TH, TT No. of favourable outcomes = 3 3 \\ P (getting at least one tail) = 4 Hence, option (b) is the correct answer. Example 6. A jar contains 24 marbles, some are green and other are blue. If a marble is drawn at 2 random from the jar, the probability that it is green is . The number of blue marbles in the jar is 3 (a) 5 (b) 6 (c) 4 (d) 8 Solution. Let E be the probability of getting green marbles. They, 2 P(E) = 3 ⇒ No. of green marbles 2 = No. of green marbles P(E) = No. of total marbles in jar   ⇒  3 24 ⇒ 2 × 24 = No. of green marbles 3 ⇒ Number of green marbles = 16 \\ Number of blue marbles = 24 – 16 = 8 Hence, option (d) is the correct answer. Example 7. A game consists of tossing a 10 rupee coin 3 times and noting its outcome each time. Sudhir wins if all the tosses give the same result, i.e., three heads or three tails and loses otherwise. The probability that Sudhir will not win the game is 11 3 (a) (b) (c) (d) 10 46 4 Solution. The possible outcomes are HHH, HHT, HTH, THH, TTH, THT, HTT, TTT Number of possible outcomes = 8 Number of favourable outcomes when he will not win = 8 – 2 = 6 63 \\ P (Sudhir will not win) = 8 = 4 H ence, option (c) is the correct answer. 118  Mathematics–10

Eatwrxrioacwme.spTtlohepe8sf.irnIantchtthieoefnigrsibavt sepnisinffiaognrumdre‘e,bda’,idwsitshhceeirsneus‘hmao’bwiesnrtohofenthnweuhmsiecbchteoarropinflaswyeehcrtiocshrpoitnhnsewaarrhrrioocwwh 2 3 stops in the second spin. On each spin, each sector haas equal chance of selection 1 by the arrow. The probability that the fraction b > 1 is 6 4 15 5 (a) (b) 12 6 5 (d) 1 (c) 12 a Solution. For b > 1, when a = 1, b cannot take any value. a = 2, b can take 1 value a = 3, b can take 2 values a = 4, b can take 3 values a = 5, b can take 4 values a = 6, b can take 5 values Thus, the favourable outcomes are: (2,1), (3,1), (3,2), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3), (6,4), (6,5) Total no. of possible outcomes = 36 PKKJKL > 1OOONP = \\ a 15 or 5 b 36 12 H ence, option (c) is the correct answer. Exercise 8.1 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. One card is drawn from a well shuffled deck of 52 cards. The probability that it is black queen is 11 12 (a) (b) (c) (d) 26 13 52 13 2. The probability of an impossible event is (a) 1 1 (d) 0 (b) (c) not defined 2 3. If P(A) denotes the probability of an event A, then (a) P(A) < 0 (b) P(A) > 1 (c) 0 ≤ P(A) ≤ 1 (d) –1 ≤ P(A) ≤ 1 4. If the probability of an event is p, the probability of its complementary event will be (a) p – 1 (b) p (c) 1 – p (d) 1 − 1 p 5. Someone is asked to take a number from 1 to 100. The probability that it is a prime is 1 61 13 (a) (b) (c) (d) 5 25 4 15 6. A number is chosen at random from the numbers –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5. Then the probability that square of this number is less than or equal to 1 is (a) 1 (b) 2 (c) 3 (d) 3 11 11 11 26 7. If the probability of an event E happening is 0.023, then P(E) = (a) 0.245 (b) 0.977 (c) 0.678 (d) 0.5 Probability  119

8. A card is drawn at random from a well shuffled pack of 52 playing cards. The probability of getting a red face card is 352 1 (a) (b) (c) (d) 26 26 13 26 9. A die is thrown once. What is the probability of getting a number greater than 4? 1111 (a) (b) (c) (d) 2345 10. Two different dice are tossed together. The probability that the product of the two numbers on the top of the dice is 6 is 4581 (a) (b) (c) (d) 9999 11. A number is chosen at random from the numbers −3, −2, −1, 0, 1, 2, 3. What will be the probability that square of this number is less than or equal to 1? 3456 (a) (b) (c) (d) 7777 12. A letter of English alphabet is chosen at random. The probability that the chosen letter is a consonant is 7 5 11 21 (a) (b) (c) (d) 26 26 26 26 13. A die is thrown once. What is the probability of getting a number less than 3? 1111 (a) (b) (c) (d) 2359 14. If the probability of winning a game is 0.07, what is the probability of losing it? (a) 0.33 (b) 0.63 (c) 0.93 (d) 0.57 15. The probability of getting a doublet in a throw of a pair of dice is 1111 (a) (b) (c) (d) 2456 16. The probability of getting a black queen when a card is drawn at random from a well-shuffled pack of 52 cards is 1 38 (a) (b) (c) (d) 1 26 26 13 17. A game consists of tossing a coin 3 times and noting the outcomes each time. If getting the same result in all the tosses is a success, the probability of losing the game is 313 (a) (b) (c) (d) 1 448 18. A ticket is drawn at random from a bag containing tickets numbered from 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is 134 (a) (b) (c) (d) 1 555 19. The king, queen and jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. The probability of getting a card of queen is 1235 (a) (b) (c) (d) 49 49 49 49 20. A coin is tossed two times. Find the probability of getting at least one head is 1313 (a) (b) (c) (d) 4488 120  Mathematics–10

21. The probability of guessing the correct answer to a certain test is p . If the probability of not guessing 1 12 the correct answer to this question is 3 , then the value of p is (a) 2 (b) 4 (c) 6 (d) 8 22. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3. What is probability that x2 ≤ 4? 5346 (a) (b) (c) (d) 7777 23. A letter is selected at random from the set of English alphabets. What is the probability that it is a vowel? 1357 (a) (b) (c) (d) 26 26 26 26 B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 5 1. Assertion (A): If a pair of dice is thrown once, then the probability of getting a sum of 8 is . 36 1 Reason (R): In a simultaneous toss of two coins, the probability of getting exactly one head is . 2 2. Assertion (A): The probability of a sure event is 1. Reason (R): Let E be an event. Then 0 ≤ P (E) ≤ 1. Answers and Hints 1 2. (d) 0 A = Product of the numbers on the top of the 1. (a) dice is 6. 26 \\  Favourable outcomes (1,6), (2,3), (3,2), (6,1), i.e. 4. 3. (c) 0 ≤ P(A) ≤ 1 4. (c) 1 – p Favourable outcomes 1 3 P(A) = Total number of possible outcomes 5. (c) 6. (c) 41 4 11 = = 7. (b) 0.977 36 9 11. (a) 3 8. (a) 3 26 6 3 7 Favourable outcomes are –1, 0, 1 Required probability = 52 = 26 3 1 \\  Required Probability = 7 9. (b) 21 3 12. (d) Total number of outcomes = 6, 26 i.e.,{1, 2, 3, 4, 5, 6} Total number of possible outcomes = 26 No. of favourable outcomes = 21 No. of fabourable outcomes = 2, i.e.{5, 6} 21 21 \\  Required probability = 26 \\  Required probability = 6 = 3 1 10. (d) 9 Total number of possible outcomes = 36 Probability  121

1 18. (a) 1 3 13. (b) 5 19. (c) 49 3 3 20. (b) 4 Total number of outcomes = 6, i.e., {1, 2, 3, 4, 5, 6} Number of favourable outcomes 21. (d) 8 = 2, i.e., {1, 2} p +1 = 1  ⇒  p 1 = 2 , \\p=8 21 12 3 = 1− 12 3 3 \\  Required probability = = 63 22. (a) 5 7 14. (c) 0.93 We know that for any event E, Total number of possible outcomes = 7 P(E) + P(E) = 1  ⇒  0.07 + P(E) = 1 No. of favourable outcomes to x2 ≤ 4 are (–1)2 ⇒ P(E) = 1 – 0.07 = 0.93 = 1, 02 = 0, 12 = 1, (–2)2 = 4 and (2)2 = 4 15. (d) 1 16. (a) 1 \\  Favourable outcomes = 5 6 26 3 5 17. (a) \\  Required probability = 7 4 5 23. (c) 26 Possible outcomes = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} B. Assertion-Reason Type Questions 1. (b) Both assertion (A) and reason (R) are true Total possible outcomes = 8, but reason (R) is not the correct explanation There are two cases of the same result, i.e. of assertion (A). {HHH, TTT} 2. (b) Both assertion (A) and reason (R) are true Number of outcomes in which the game is lost but reason (R) is not the correct explanation =8–2=6 of assertion (A). 63 ∴  Probability of losing the game = 8 = 4 Case Study Based Questions I. Five friends make up their minds to play a game by using a spinner. 2 3 4 They put a spinner down on the ground and set around themselves to 1 6 5 play. A spinner contains eight regions, numbered through 8, as shown in the figure, the arrow has an equally likely chance of landing on any 8 of the eight regions. If the arrow lands on a line, the result is not counted 7 and the arrow spun again. 1. How many possible outcomes are in the sample spaces? (a) 4 (b) 6 (c) 8 (d) 10 2. What is the probability that the arrow lands on 4? 1111 (a) (b) (c) (d) 4628 3. The set of possible outcomes for event A, in which the arrow lands on an even number is (a) 2, 4, 6, 8 (b) 2, 4, 6, 10 (c) 2, 6, 8, 10 (d) 4, 6, 8, 10 122  Mathematics–10

4. The probability that the arrow lands on an even number is (a) 1 (b) 1 (c) 1 (d) 1 468 2 5. The probability that the arrow will point at any factor of 8 is (a) 1 (b) 1 (c) 1 (d) 1 2468 Ans. 1. (c) 8 1 3. (a) 2, 4, 6, 8 4. (d) 1 5. (a) 1 2. (d) 2 2 8 II. Some boys are playing with numbers. There is a box which contains 90 discs. They are numbered from 1 to 90. The boys are drawing disc one by one from the box at random and want to find the probability of a particular number. 1. If one disc is drawn at random from the box, the probability that it bears a two-digit number is 9 8 89 79 (a) 10 (b) 90 (c) 90 (d) 90 2. Then they put the disc into the box and draw another disc at random. They want to know the probability that it bears a perfect square number. So, the probability of perfect square number is 8 3 11 (a) 90 (b) 10 (c) 10 (d) 9 3. They put the disc into the box and another boy draws another disc at random, and wants to know the probability that it bears a number divisible by 5. So the probability is 2341 (a) 5 (b) 5 (c) 5 (d) 5 4. After putting the disc into the box, the next boy draws a disc at random and wants to know the probability that it bears a prime number. The probability is (a) 5 4 2 13 (b) 15 (c) 15 (d) 18 45 5. The boys put all the disc together into the box and next boy draws a disc at random and find the probability that the disc bears a number divisible by 7. 1 2 41 (a) 15 (b) 15 (c) 15 (d) 5 9 1 1 2 2 A ns. 1. (a) 10 2. (c) 10 3. (d) 5 4. (c) 15 5. (b) 15 Experts’ Opinion Questions based on following types are very important for Exams. So, students are advised to revise them thoroughly. 1. Problems based on tossing a coin. 2. Problems based on throwing a die. 3. Problems based on playing cards. 4. Problems based on selecting of an object from bag/box. IMPORTANT FORMULAE • P(E) = No. of outcomes favourable to E No. of all possible outcomes of the experiment • For any event E, P (E) + P (E) = 1, where E stands for ‘not E’. Probability  123

COMMON ERRORS Errors Corrections (i) Interpreting incorrectly that all the experiments (i) An experiment whose outcome is known is not are random experiment. a random experiment. (ii) Writing incorrectly the possible outcomes when a (ii) Take care of the order of occurrence. Here, coin is tossed one by one or two coins are tossed (HH, HT, TH, TT). together, say (HH, HT, TT). (iii) Using negative values and a number greater than (iii) The probability of an event lies between 0 and 1 one for probability. (both 0 and 1 inclusive). So, a negative value and a number greater than 1 cannot be used for probability. QUICK REVISION NOTES • An experiment which has a number of possible outcomes is known as a random experiment. • The theoretical probability or classical probability of an event E, written as P(E) is defined as No. of outcomes favourable to E P(E) = No. of all possible outcomes of the experiment where, we assume that the outcomes of the experiment are equally likely. • The probability of a sure event (or a certain event) is 1. • The probability of an event E is a number P(E) such that 0 ≤ P(E) ≤ 1 • An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1. • For any event E, P(E) + P(E) = 1, where E stands for ‘not E’. E and E are called complementary events. • When each outcome of a random experiment is likely to occur as the other, then they are termed as equally likely outcomes. A deck of Playing Cards (52) Red Cards (26) Black Cards (26) Hearts (13) Diamonds (13) Spades (13) Clubs (13) Ace (1 × 4) Number Cards (9 × 4) Face Cards (3 × 4) 2, 3, 4, 5, 6, 7, 8, 9, 10 Jack, Queen, king 124  Mathematics–10

Sample Paper - 1 (Issued by CBSE on 2nd September, 2021) Time Allowed: 90 Minutes Maximum Marks: 40 General Instructions: 1. The question paper contains three sections A, B and C. 2. Section A consists of 20 questions of 1 mark each. Attempt any 16 questions. 3. Section B consists of 20 questions of 1 mark each. Attempt any 16 questions. 4. Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions. 5. There is no negative marking. SECTION-A Section A consists of 20 questions. Any 16 questions are to be attempted. 1. A box contains cards numbered 6 to 50. A card is drawn at random from the box. The probability that the drawn card has a number which is a perfect square like 4, 9, .... is 1 (a) 1 (b) 2 (c) 4 (d) 1 45 15 45 9 2. In a circle of diameter 42 cm, if an arc subtends an angle of 60° at the centre where p = 22/7, then the length of the arc is 1 (a) 22 cm (b) 11 cm (c) 22 cm (d) 44 cm 7 3. If sin q = x and sec q = y, then tan q is 1 (a) xy (b) x (c) y (d) 1 y x xy 4. The pair of linear equations y = 0 and y = –5 has 1 (a) One solution (b) Two solutions (c) Infinitely many solutions (d) No solution 5. A fair die is thrown once. The probability of even composite number is 1 (a) 0 (b) 1 (c) 3 (d) 1 3 4 6. 8 chairs and 5 tables cost ` 10,500 while 5 chairs and 3 tables cost ` 6,450. The cost of each chair will be 1 1 (a) ` 750 (b) ` 600 (c) ` 850 (d) ` 900 7. If cos q + cos2 q = 1, the value of sin2 q + sin4 q is (a) –1 (b) 0 (c) 1 (d) 2 8. The decimal representation of 23 will be 1 23 × 52 (a) Terminating (b) Non-terminating (c) Non-terminating and repeating (d) Non-terminating and non-repeating 9. The LCM of 23 × 32­ and 22 × 33 is 1 (a) 23 (b) 33 (c) 23 × 33 (d) 22 × 32 10. The HCF of two numbers is 18 and their product is 12,960. Their LCM will be 1 (a) 420 (b) 600 (c) 720 (d) 800 125

11. In the given figure, DE || BC. Which of the following is true? A 1 (a) x = a+b a ay xE (b) y = ax b a+b yC (c) x = ay D a+b B (d) x = a y b 12. The co-ordinates of the point P dividing the line segment joining the points A(1, 3) and B(4, 6) internally in the ratio 2 : 1 are 1 (a) (2, 4) (b) (4, 6) (c) (4, 2) (d) (3, 5) 13. The prime factorisation of 3825 is 1 (a) 3 × 52 × 21 (b) 32 × 52 × 35 (c) 32 × 52 × 17 (d) 32 × 25 × 17 14. In the figure given below, AD = 4 cm, BD = 3 cm and CB = 12 cm, then cot q equals 1 (a) 3 A 4 (b) 5 D 12 B (c) 4 Cq 3 (d) 12 5 D x+y C 15. If ABCD is a rectangle, the values of x and y are given by 1 (a) x = 10, y = 2 x–y 8 1 (b) x = 12, y = 8 1 1 (c) x = 2, y = 10 A 12 B (d) x = 20, y = 0 16. In an isosceles triangle ABC, if AC = BC and AB2 = 2AC2, then the measure of angle C will be (a) 30° (b) 45° (c) 60° (d) 90° 17. If –1 is a zero of the polynomial p(x) = x2 – 7x – 8, then the other zero is (a) –8 (b) –7 (c) 1 (d) 8 18. In a throw of a pair of dice, the probability of the same number on each die is (a) 1 (b) 1 (c) 1 (d) 5 6 3 2 6 19. The mid-point of (3p, 4) and (–2, 2q) is (2, 6). The value of p + q is equal to 1 1 (a) 5 (b) 6 (c) 7 (d) 8 20. The decimal expansion of 147 will terminate after how many places of decimals? 120 (a) 1 (b) 2 (c) 3 (d) 4 SECTION-B Section B consists of 20 questions. Any 16 questions are to be attempted. 21. The perimeter of a semicircular protractor whose radius is ‘r’ is given by 1 1 (a) p + 2r (b) p + r (c) pr (d) pr + 2r 22. If P(E) denotes the probability of an event E, then (a) 0 < P(E) ≤ 1 (b) 0 < P(E) < 1 (c) 0 ≤ P(E) ≤ 1 (d) 0 ≤ P(E) < 1 126  Mathematics–10

23. In DABC, ∠B = 90° and BD ^ AC. if AC = 9 cm and AD = 3 cm then BD is equal to 1 (a) 2 2 cm (b) 3 2 cm (c) 2 3 cm (d) 3 3 cm 24. The pair of linear equations 3x + 5y = 3 and 6x + ky = 8 do not have a solution if 1 1 (a) k = 5 (b) k = 10 (c) k ≠ 10 (d) k ≠ 5 1 25. If the circumference of a circle increases from 2p to 4p then its area ____ the original area. (a) half (b) double (c) three times (d) four times 26. Given that sin q= a , then tan q is equal to (a) b b2 b b a a a2 + (b) b2 − a2 (c) a2 − b2 (d) b2 − a2 27. If x = 2 sin2 q and y = 2 cos2 q + 1, then x + y is 1 (a) 3 (b) 2 (c) 1 (d) 1 2 22 28. If the difference between the circumference and the radius of a circle is 37 cm, p = 7 , the circumference (in cm) of the circle is 1 (a) 154 (b) 44 (c) 14 (d) 7 29. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) 1 (a) 100 (b) 1000 (c) 2520 (d) 5040 30. Three bells ring at intervals of 4, 7 and 14 minutes. All three rang at 6 AM. When will they ring together again? 1 (a) 6:07 AM (b) 6:14 AM (c) 6:28 AM (d) 6:25 AM 31. What is the age of father, if the sum of the ages of a father and his son in years is 65 and twice the difference of their ages in years is 50? 1 (a) 40 years (b) 45 years (c) 55 years (d) 65 years 32. What is the value of (tan q cosec q)2 – (sin q sec q)2? 1 (a) –1 (b) 0 (c) 1 (d) 2 33. The perimeter of two similar triangles are 26 cm and 39 cm. The ratio of their areas will be 1 (a) 2 : 3 (b) 6 : 9 (c) 4 : 6 (d) 4 : 9 34. There are 20 vehicles — cars and motorcycles in a parking area. If there are 56 wheels together, how many cars are there? 1 (a) 8 (b) 10 (c) 12 (d) 20 35. A man goes 15 m due west and then 8 m due north. How far is he from the starting point? 1 (a) 7 m (b) 10 m (c) 17 m (d) 23 m 36. What is the length of an altitude of an equilateral triangle of side 8 cm? 1 (a) 2 3 cm (b) 3 3 cm (c) 4 3 cm (d) 5 3 cm 37. If the letters of the word RAMANUJAN are put in a box and one letter is drawn at random. The probability that the letter is A is given by 1 (a) 3 (b) 1 (c) 3 (d) 1 5 2 7 3 38. Area of a sector of a circle is 1 to the area of circle. The degree measure of its minor arc is given by 1 6 (a) 90° (b) 60° (c) 45° (d) 30° 39. A vertical stick 20 m long casts a shadow 10 m long on the ground. At the same time a tower casts a shadow 50 m long. What is the height of the tower? 1 (a) 30 m (b) 50 m (c) 80 m (d) 100 m 40. What is the solution of the pair of linear equations 37x + 43y = 123, 43x + 37y = 117? 1 (a) x = 1, y = 1 (b) x = –1, y = 2 (c) x = –2, y = 1 (d) x = 1, y = 2 Sample Paper - 1  127

SECTION-C Case Study Based Questions Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted. Case Study-1 Pacific Ring of Fire The Pacific Ring of Fire is a major area in the basin of the Pacific Ocean where many earthquakes and volcanic eruptions occur. In a large horseshoe shape, it is associated with a nearly continuous series of oceanic trenches, volcanic arcs, and volcanic belts and plate movements. https://commons.wikimedia.org/wiki/File:Pacifick%C3%BD_ohniv%C3%BD_kruh.png Fault Lines Large faults within the Earth’s crust result from the action of plate tectonic forces, with the largest forming the boundaries between the plates. Energy release associated with rapid movement on active faults is the cause of most earthquakes. A normal fault A reverse fault A strike-slip fault An oblique fault https://commons.wikimedia.org/wiki/File:Faults6.png 128  Mathematics–10

Positions of some countries in the Pacific ring of fire is shown in the square grid below. Y X′ X Y′ Based on the given information, answer the Questions No. 41-45. 41. The distance between the point Country A and Country B is 1 (a) 4 units (b) 5 units (c) 6 units (d) 7 units 42. A relation between x and y such that the point (x, y) is equidistant from the Country C and Country D is given by 1 (a) x – y = 2 (b) x + y = 2 (c) 2x – y = 2 (d) 2x + y = 2 43. The fault line 3x + y – 9 = 0 divides the line joining the Country P(1, 3) and Country Q(2, 7) internally in the ratio 1 (a) 3 : 4 (b) 3 : 2 (c) 2 : 3 (d) 4 : 3 44. The distance of the Country M from the x-axis is 1 (a) 1 units (b) 2 units (c) 3 units (d) 5 units 45. What are the co-ordinates of the Country lying on the mid-point of Country A and Country D? 1 ( )(b) 2, 9 (a) (1, 3) 2 ( ) (c) 4, 5 ( )(d) 9 ,2 2 2 Case Study-2 Roller Coaster of Polynomials Polynomials are everywhere. They play a key role in the study of algebra, in analysis and on the whole many mathematical problems involving them. Since, polynomials are used to describe curves of various types engineers use polynomials to graph the curves of roller coasters. Sample Paper - 1  129

Based on the given information, answer the Questions No. 46-50. 46. If the Roller Coaster is represented by the following graph y = p(x), then name the type of the polynomial it traces. 1 (a) Linear (b) Quadratic (c) Cubic (d) Bi-quadratic 47. The Roller Coasters are represented by the following graphs y = p(x), Which Roller Coaster has more than three distinct zeroes? 1 (a) (b) 130  Mathematics–10

(c) (d) 48. If the Roller Coaster is represented by the cubic polynomial t(x) = px3 + qx2 + rx + s, then which of the following is always true? 1 (a) s ≠ 0 (b) r ≠ 0 (c) q ≠ 0 (d) p ≠ 0 49. If the path traced by the Roller Coaster is represented by the above graph y = p(x), the number of zeroes are given by 1 (a) 0 (b) 1 (c) 2 (d) 3 50. If the path traced by the Roller Coaster is represented by the above graph y = p(x), its zeroes will be  1 (a) –3, –6, –1 (b) 2, –6, –1 (c) –3, –1, 2 (d) 3, 1, –2 Sample Paper - 1  131

Marking Scheme 1. (d) P(perfect square) = 5 = 1 45 9 ( ) ( )Length θ 60° 22 2. (c) of the arc = 360° ( 2πr ) = 360° ×2× 7 × 21 = 22 cm 3. (a) tan q = sin θ = sin θ × sec θ = xy cos θ 4. (d) The lines are parallel hence no solution. 5. (b) P(even composite no.) = 2 = 1 6 3 6. (a) Let the cost of one chair = ` x Let the cost of one table = ` y 8x + 5y = 10500 5x + 3y = 6450 Solving the above equations Cost of each chair = x = ` 750 7. (c) cos q = 1 – cos2 q = sin2 q \\ sin2 q + sin4 q = cos q + cos2 q = 1 8. (a) Terminating 9. (c) 2­3 × 33 10. (c) 1st No. × 2nd No. = HCF × LCM 12960 = 18 × LCM LCM = 720 11. (c) AE = DE = a a b = x AC BC + y x = ay (a + b) 12. (d) (2 × 4 +1×1) , (2 × 6 +1× 3) = (3, 5) 33 13. (c) 3825 = 32 × 52 × 17 14. (d) AB2 = AD2 + BD2 AB = 5 cm AC2 = AB2 + CB2 AC = 13 cm cot q = CB = 12 AB 5 15. (a) x + y = 12 x – y = 8 Solving the above equations x = 10, y = 2 132

16. (d) AB2 = AC2 + AC2 = AC2 + BC2 Hence, angle C = 90° 17. (d) Let the zeroes be a and b a = –1 , a + b = − ( −7) Then, Hence, 1 b = 7 + 1 = 8 18. (a) P(same no. on each die) = 6 = 1 36 6 19. (b) (2, 6) =  (3p − 2) , (4 + 2q)  2 2  3p – 2 = 4, 4 + 2q = 12 p = 2, q = 4 Hence p + q = 6 20. (c) 147 = 49 = 49 120 40 23 × 5 Three decimal places. 21. (d) Perimeter of protractor = Circumference of semi-circle + 2 × radius = pr + 2r 22. (c) 0 ≤ P(E) ≤ 1 23. (b) CD = BD BD AD BD2 = CD × AD = 6 × 3 BD = 3 2 cm 24. (b) 3 = 5 ⇒ k = 10 6 k 25. (d) C1 = 2πr ⇒ 2π = 2πr   ⇒  r = 1 C2 2πR 4π 2πR R 2 ( ) ( )A1 πr 2 r 2 1 2 1 26. (d) πR 2 R 2 4 A2 = = = = sin q = a b H2 = P2 + B2 b2 = a2 + B2 ( )B = b2 − a2 = P = a B b2 − a2 ( )tanq 27. (a) x + y = 2 sin2 q + 2 cos2 q + 1 = 2(sin2 q + cos2 q) + 1 = 2 + 1 = 3 28. (b) 2pr – r = 37 { ( ) }r2×22 −1 = 37 7 7 r = 37 × 37 ⇒ r = 7 ( )Circumference = 2 × 22 × 7 = 44 cm 7 Marking Scheme  133

29. (c) 1 = 1 2 = 2 × 1 3 = 3 × 1 4 = 2 × 2 5 = 5 × 1 6 = 2 × 3 7 = 7 × 1 8 = 2 × 2 × 2 9 = 3 × 3 10 = 2 × 5 So, LCM of these numbers = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520 Hence, least number divisible by all the numbers from 1 to 10 is 2520 30. (c) LCM of 4, 7, 14 = 28 Bells will ring together again at 6:28 AM. 31. (b) Let age of father = x years Let age of son = y years x + y = 65 2(x – y)= 50 Solving the above equations Father’s Age = x = 45 years 32. (c) (tan q cosec q)2 – (sin q sec q)2 = tan2 q cosec2 q – sin2 q sec2 q =  sin 2 θ  × 1 − sin 2 θ × 1  cos2 θ  sin 2 cos2 θ θ ( )= 1 − sin2 θ = cos2 θ = 1 cos2 θ cos2 θ 2 26 2 ( )A1  P1  39 33. (d) =  P2  = A2 ( )A1 2 2 4 3 9 A2= = 34. (a) Let no. of cars = x Let no. of motorcycles = y x + y = 20 4x + 2y = 56 Solving the above equations No. of cars = x = 8 35. (c) H2 = P2 + B2 H2 = 152 + 82 H = 17 m ( )(altitude)2 = (side)2 − side 2 2 36. (c) = 82 – 42 = 64 – 16 = 48  \\  Altitude = 4 3 cm 134  Mathematics–10

37. (d) P = 3 = 1 9 3 38. (b) θ × πr 2 = 1 × πr2 360° 6 q = 60° 39. (d) Height of vertical stick Height of tower Shadow of vertical stick = Shadow of tower 20 Height of tower 10 = 50 Height of tower = 100 m 40. (d) 37x + 43y = 123 ...(1) ...(2) 43x + 37y = 117 ...(3) Adding (1) and (2) x + y = 3 ...(4) Subtracting (2) from (1) –x + y = 1 Adding (3) and (4), 2y = 4 y = 2 ⇒ x = 1 \\ Solution is x = 1 and y = 2 ( )32 + 42 41. (b) { }AB = (4 −1)2 + (0 − 4)2 = AB = 5 units 42. (a) (x – 7)2 + (y – 1)2 = (x – 3)2 + (y – 5)2 x2 + 49 – 14x + y2 + 1 – 2y = x2 + 9 – 6x + y2 + 25 – 10y Simplifying x – y = 2 43. (a) 3x + y – 9 = 0 Let R divide the line in ratio k : 1 ( ) R 2k +1, 7k + 3 k +1 k +1 ( ) ( ) 3 2k +1 + 7k + 3 − 9 = 0 k +1 k +1 4k – 3 = 0 k = 3 So, ratio = 3 : 4 4 44. (c) Distance of M from x-axis = (2 − 2)2 + (0 − 3)2 = 9 = 3 units 45. (b) ( ) ( )(1+3), (4 + 5) = 4,9 = 2, 9 46. (c) Cubic 2 2 22 2   47. (d) Four zeroes as the curve intersects the x-axis at 4 points. 48. (d) p ≠ 0 49. (d) Three zeroes as the curve intersects the x-axis at 3 points 50. (c) –3, –1, 2 Marking Scheme  135

Sample Paper - 2 Time Allowed: 90 Minutes Maximum Marks: 40 General Instructions: S ame as Sample Paper-1 SECTION-A Section A consists of 20 questions. Any 16 questions are to be attempted. 1. The (HCF × LCM) for the numbers 50 and 20 is  1 1 (a) 10 (b) 100 1 1 (c) 1000 (d) 50 2. The zeroes of the quadratic polynomial x2 + 13x + 40 are (a) –10, –4 (b) 8, 5 (c) – 8, – 5 (d) 20, 2 3. The distance between the points (cos θ, − sin θ ) and (−cos θ, sin θ) is (a) 3 (b) 4 (c) 2 (d) 5 4. The value of sin 45° + cos 45° is  1 (a) (b) 2 2 (c) 2 (d) 2 2 5. If sin2 A + cos2 A = 1, then which of the following is true? 1 (a) 0° ≤ A ≤ 90° (b) 0° < A ≤ 90° (c) 0° ≤ A < 90° (d) None of these 6. The value of sin 60° cos 30° + sin 30° cos 60° is 1 (a) 0 (b) 1 (c) 2 (d) 5 7. If an event cannot occur, then its probability is  1 (a) 1 3 (b) 4 1 (d) 0 (c) 2 8. The probability expressed as a percentage of a particular occurrence can never be 1 (a) less than 100 (b) less than 0 (c) greater than 1 (d) anything but a whole number 9. The diameter of circle whose area is equal to sum of area of two circles of diameter 16 cm and 12 cm is 1 (a) 56 cm (b) 42 cm (c) 28 cm (d) 20 cm 10. The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then the length of the side of the rhombus is 1 (a) 9 cm (b) 10 cm (c) 8 cm (d) 20 cm 136

1 1 11. If sin θ = , then the value of 2 cot2 θ + 2 is 3 (a) 12 (b) 16 (c) 18 (d) 14 12. A wire, bent in the form a square, encloses an area of 484 cm2. If same wire is bent so as to form a circle, then the area enclosed will be 1 (a) 484 cm2 (b) 616 cm2 (c) 644 cm2 (d) 252 cm2 13. The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is: 1 (a) 13 (b) 65 (c) 875 (d) 1,750 14. Graphically, the pair of equations 6x – 3y + 10 = 0 and 2x – y + 9 = 0 represents two lines which are 1 (a) intersecting at exactly one point (b) intersecting at exactly two points (c) coincident (d) parallel 15. The distance of the point P(2, 3) from the x-axis is: 1 (a) 2 (b) 3 (c) 1 (d) 5 16. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is: 1 (a) 2 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4 17. Area of the largest triangle that can be inscribed in a semi-circle of radius r units is: 1 (a) r2 sq. units (b) 1 r 2 sq. units (c) 2r2 sq. units (d) 2 r2 sq. units 2 18. Which of the following is not the graph of a quadratic polynomial? 1 (a) (b) (c) (d) 19. The pair of equations x + 2y + 5 = 0 and –3x – 6y + 1 = 0 have 1 1 (a) a unique solution (b) exactly two solutions (c) infinitely many solutions (d) no solution 20. The distance between the points A(0, 6) and B(0, –2) is (a) 6 (b) 8 (c) 4 (d) 2 Sample Paper - 2  137

SECTION-B Section B consists of 20 questions. Any 16 questions are to be attempted. 21. In the figure given here, ∠BAC = 90° and AD ^ BC. Then A 1 (a) BD × CD = BC2 (b) AB × AC = BC2 (c) BD× CD = AD2 (d) AB × AC = AD2 BD C 22. It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be 1 (a) 10 m (b) 15 m (c) 20 m (d) 24 m 23. If the zeroes of the quadratic polynomial ax2 + bx + c, a ≠ 0 are equal, then 1 (a) c and a have opposite signs (b) c and b have opposite signs (c) c and a have the same sign (d) c and b have the same sign 24. If a pair of linear equation is consistent, then the lines will be 1 (a) parallel (b) always coincident (c) intersecting or coincident (d) always intersecting 25. AOBC is a rectangle whose three vertices are A(0, 3), O(0,0) and B(5, 0). The length of its diagonal is: 1 (a) 5 (b) 3 (c) 34 (d) 4 26. The value of (tan 1° tan 2° tan 3° ... tan 89°) is 1 (a) 0 (b) 1 (c) 2 (d) 1 2 27. The area of the circle that can be inscribed in a square of side 6 cm is 1 (a) 36p cm2 (b) 18p cm2 (c) 12p cm2 (d) 9p cm2 1 28. The zeroes of the quadratic polynomial x2 + kx + k, k ≠ 0, (a) cannot both be positive (b) cannot both be negative (c) are always unequal (d) are always equal 29. The pair of equations y = 0 and y = –7 has 1 (a) one solution (b) two solutions (c) infinitely many solutions (d) no solution 30. The coordinates of the point which is equidistant from the three vertices of the DAOB as shown in the figure is 1 (a) (x, y) (b) (y, x) (c)  x , y  2 2 (d)  y , x  2 2 31. If cos 9a = sin a and 9a < 90°, then the value of tan 5a is 1 1 (a) 1 (b) 3 3 (c) 1 (d) 0 32. An event is very unlikely to happen. Its probability is closest to (a) 0.0001 (b) 0.001 (c) 0.01 (d) 0.1 138  Mathematics–10

33. The zeroes of the quadratic polynomial x2 + 99x + 127 are 1 (a) both positive. (b) both negative. (c) one positive and one negative. (d) both equal. 34. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, in years, of the son and the father are, respectively 1 (a) 4 and 24 (b) 5 and 30 (c) 6 and 36 (d) 3 and 24 35. The points (–4, 0), (4, 0) and (0, 3) are the vertices of a 1 (a) right triangle (b) isosceles triangle (c) equilateral triangle (d) scalene triangle 36. If cos (a + b) = 0, then sin(a – b) can be reduced to 1 (a) cos b (b) cos 2b (c) sin a (d) sin 2a 37. If the probability of an event is p, then the probability of its complementary event will be 1 (a) p – 1 (b) p (c) 1 – p (d) 1− 1 p 38. If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is –1, then the product of the other two zeroes is 1 (a) b – a + 1 (b) b – a – 1 (c) a – b + 1 (d) a – b – 1 39. If 2x + y = 23 and 4x – y = 19, then the values of (5y = 2x) and  y − 2 are 1  x (a) 30, 5 (b) 31, −5 (c) 32, 5 (d) None of these. 7 7 7 40. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is 1 (a) 5 (b) 12 (c) 11 (d) 7 + 5 SECTION-C Case Study Based Questions Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted. Case Study-1 Amar, Akbar and Anthony are playing a game. Amar climbs 5 stairs and gets down 2 stairs in one turn. Akbar goes up by 7 stairs and comes down by 2 stairs every time. Anthony goes 10 stairs up and 3 stairs down each time. Sample Paper - 2  139

Doing this they have to reach to the nearest point of 100th stairs and they will stop once they find it impossible to go forward. They can not cross 100th stair in anyway. 41. Who reaches the nearest point? 1 (a) Amar (b) Akbar (c) Anthony (d) All together reach to the nearest point. 42. How many times can they meet in between on same stair ? 1 (a) 3 (b) 4 (c) 5 (d) No, they cannot meet in between on same stair. 43. Who takes least number of steps to reach near hundred? 1 (a) Amar (b) Akbar (c) Anthony (d) All of them take equal number of steps. 44. What is the first stair where any two out of three will meet together? 1 (a) Amar and Akbar will meet for the first time on 15th stair. (b) Akbar and Anthony will meet for the first time on 35th stair. (c) Amar and Anthony will meet for the first time on 21th stair. (d) Amar and Akbar will meet for the first time on 21th stair. 45. What is the second stair where any two out of three will meet together? 1 (a) Amar and Akbar will meet on 21th stair. (b) Akbar and Anthony will meet on 35th stair. (c) Amar and Anthony will meet on 21th stair. (d) Amar and Anthony will meet on 35th stair. Case Study-2 For going to city B from city A, there is a route via city C such that AC ⊥ CB, AC = 2x km and CB = 2(x + 7) km. It is proposed to construct a 26 km highway which is directly connect the two cities A and B. In solving such problem authority of the cities have some questions in his mind. Give answer to his following questions: City C 2 x km 2(x + 7) km City A 26 km City B 46. Which theorem will you use to solve this problem? 1 (a) Pythagoras theorem (b) Basic proportionality theorem (c) Factor theorem (d) Fundamental Theorem of Arithmetic 47. What is the distance of AC? 1 (a) 15 km (b) 18 km (c) 10 km (d) 20 km 48. What is the distance of BC? 1 (a) 12 km (b) 10 km (c) 20 km (d) 24 km 49. Find how much distance will be saved in reaching city B from city A after the construction of the highway. 1 (a) 6 km (b) 8 km (c) 10 km (d) 12 km 50. If ∠A is supposed to be 30°, then ∠B = ? 1 (a) 30° (b) 45° (c) 60° (d) 90° 140  Mathematics–10

1. (c) 2. (c) ANSWERS 4. (b) 5. (a) 6. (b) 7. (d) 9. (d) 10. (b) 11. (c) 12. (b) 3. (c) 14. (d) 15. (b) 16. (c) 17. (a) 8. (b) 19. (d) 20. (b) 21. (c) 22. (a) 13. (a) 24. (c) 25. (c) 26. (b) 27. (d) 18. (d) 29. (d) 30. (a) 31. (c) 32. (a) 23. (c) 34. (c) 35. (b) 36. (b) 37. (c) 28. (a) 39. (b) 40. (b) 41. (a) 42. (d) 33. (b) 44. (a) 45. (c) 46. (a) 47. (c) 38. (a) 49. (b) 50. (c) 43. (c) 48. (d) Sample Paper - 2  141

Sample Paper - 3 Time: Allowed: 90 Minutes Maximum Marks: 40 General Instructions: Same as Sample Paper-1 SECTION-A Section A consists of 20 questions. Any 16 questions are to be attempted. 1. The values of x and y in the given figure respectively are 42 1 (a) 2, 5 (b) 5, 7 x 21 (c) 2, 7 3y (d) 3, 5 2. The LCM and HCF two numbers are 36 and 2 respectively. If one of the numbers is 18, then the other number is 1 (a) 1 (b) 2 (c) 3 (d) 4 3. The value of x satisfying both the equation 4x – 5 = y and 2x – y = 3, when y = –1 is 1 (a) 1 (b) –1 (c) 2 (d) –2 4. The value of sin 45°. cos 45° + sin 30° is 1 (a) 0 (b) 1 (c) 2 (d) 2 2 5. 3 tan2 30° + sec4 45° – tan2 60° equal to 1 (a) 0 (b) 1 (c) 2 (d) 3 1 6. The value of (1 – sin2 q) sec2 q is (a) –1 (b) 2 (c) 1 (d) 0 7. Which of the followings cannot be the probability of an event? 1 (a) 1 - (b) 0.1 (c) 3% (d) 17 3 16 8. In a single throw of two dice, the probability of getting an odd number on the first dice and 6 on the second dice is 1 111 1 (a) (b) (c) (d) 3 4 6 12 9. 5 − 3 − 2 is 1 (a) a rational number (b) a natural number (c) equal to zero (d) an irrational number 10. Two poles of heights 16 m and 7 m stand on a plane ground. If the distance between their feet is 12 m, the distance between their tops is 1 (a) 18 m (b) 15 m (c) 16 m (d) 20 m 11. The value of sin2 30° cos2 45° + 4 tan2 30° + 1 sin2 90° – 2cos2 90° is 1 45 46 2 47 49 (a) (b) (c) (d) 24 24 24 24 12. The diameter of a cycle wheel is 28 cm. The number of revolution it will make in moving 11 km is 1 (a) 12000 (b) 12200 (c) 12500 (d) 12400 142

13. If two positive integers a and b are written as a = x3y2 and b = xy3; x, y are prime numbers, then HCF(a, b) is 1 (a) xy (b) xy2 (c) x3y3 (d) x2y2 14. The pair of equations x = a and y = b graphically represents lines which are 1 (a) parallel (b) intersecting at (b, a) (c) coincident (d) intersecting at (a, b) 15. The distance of the point P(–6, 8) from the origin is 1 (a) 8 (b) 2 7 (c) 10 (d) 6 16. If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?1 (a) EF = DF (b) DE = FE (c) DE = DF (d) EF = DE PR PQ PQ RP QR PQ RP QR ( ) 17. If 4 tan q = 3, then 4sin θ − cos θ is equal to 1 4sin θ + cos θ (a) 2 (b) 1 (c) 1 (d) 3 3 2 3 4 18. If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is 1 (a) 4 (b) 2 (c) 1 (d) 3 19. One equation of a pair of dependent linear equations is –5x + 7y = 2. The second equation can be 1 (a) 10x + 14y + 4 = 0 (b) –10x – 14y + 4 = 0 (c) –10x + 14y + 4 = 0 (d) 10x – 14y = –4 20. The distance between points (0, 5) and (–5, 0) is 1 (a) 5 (b) 5 2 (c) 2 5 (d) 10 SECTION-B Section B consists of 20 questions. Any 16 questions are to be attempted. 21. If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?1 (a) EF = DF (b) DE = FE (c) DE = DF (d) EF = DE PR PQ PQ RP QR PQ RP QR 22. The area of the square that can be inscribed in a circle of radius 8 cm is 1 (a) 256 cm2 (b) 128 cm2 (c) 64 2 cm2 (d) 64 cm2 23. The decimal expansion of the rational number 14587 will terminate after 1 1250 (a) one decimal place (b) two decimal places (c) three decimal places (d) four decimal places 24. For what value of k, do the equations 3x – y + 8 = 0 and 6x – ky = –16 represent coincident lines? 1 (a) 1 (b) −1 (c) 2 (d) –2 2 2 ( ) a ,4 25. If P 3 is the mid-point of the line segment joining the points Q(–6, 5) and R(–2, 3), then the value of a is 1 1 (a) –4 (b) –12 1 (c) 12 (d) –6 26. In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE. Then, the two triangles are: (a) congruent but not similar (b) similar but not congruent (c) neither congruent nor similar (d) congruent as well as similar. 27. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is (a) 22 : 7 (b) 14 : 11 (c) 7 : 22 (d) 11 : 14 Sample Paper - 3  143

28. The decimal expansion of the sum of rational numbers 15 and 5 will terminate after 1 4 40 (a) one decimal place (b) two decimal places (c) three decimal places (d) four decimal places 29. The probability that a non-leap year selected at random will contain 53 Sundays is 1 (a) 1 (b) 2 (c) 3 (d) 5 7 7 7 7 30. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio 1 (a) 2 : 3 (b) 4 : 9 (c) 81 : 16 (d) 16 : 81 31. The value of 9 sec2 A – 9 tan2 A is 1 (a) 1 (b) 9 (c) 8 (d) 0 32. If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then 1 (a) R1 + R2 = R (b) R12 + R 2 = R2 (c) R1 + R2 < R (d) R12 + R 2 = R2 2 2 33. The number of polynomials having zeroes as –2 and 5 is 1 (a) 1 (b) 2 (c) 3 (d) more than 3 34. If x = a, y = b is the solution of the equations x –y = 2 and x + y = 4, then the values of a and b are, respectively 1 (a) 3 and 5 (b) 5 and 3 (c) 3 and 1 (d) –1 and 3 35. If DABC ~ DEDF and DABC is not similar to DDEF, then which of the following is not true? 1 (a) BC × EF = AC × FD (b) AB × EF = AC × DE (c) BC × DE = AB × EF (d) BC × DE = AB × FD 36. If sin A = 1 , then the value of cot A is 1 2 (a) 3 (b) 1 (c) 3 3 2 (d) 1 37. If a card is selected from a deck of 52 cards, then the probability of its being a red face card is 1 (a) 3 (b) 3 (c) 2 (d) 1 26 13 13 2 38. Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, the product of the other two zeroes is 1 (a) − c (b) c (c) 0 (d) − b a a a 39. The larger of two supplementary angles exceed the smaller by 18°, then the angles are 1 1 (a) 99°, 81° (b) 98°, 82° (c) 97°, 83° (d) None of these. 40. If in two triangles ABC and PQR, AB = BC = CA , then QR PR PQ (a) DPQR ~ DCAB (b) DPQR ~ DABC (c) DCBA ~ DPQR (d) DBCA ~ DPQR SECTION-C Case Study Based Questions Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted. Case Study-1 Due to heavy storm an electric wire got bent as shown in the figure. It followed a mathematical shape. Answer the following questions given here: 144  Mathematics–10

y 6 1 2 34 5 6 7 x 5 4 3 2 1 x' –6 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 y' 41. Name the shape in which the wire is bent. 1 1 (a) spiral (b) ellipse (c) linear (d) parabola 1 (d) 0 1 42. How many zeroes are there for the polynomial (shape of the wire). (d) –4, 2 1 (d) x2 + 2x + 3 (a) 2 (b) 3 (c) 1 (d) 0 43. The zeroes of the polynomial are (a) –1, 5 (b) –1, 3 (c) 3, 5 44. What will be the expression of the polynomial? (a) x2 + 2x – 3 (b) x2 – 2x + 3 (c) x2 – 2x – 3 45. What is the value of the polynomial if x = –1? (a) 6 (b) –18 (c) 18 Case Study-2 To conduct sports day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in figure. Niharika runs 1 th the distance AD on the 2nd line 1 4 and posts a green flag. Preet runs 5 th the distance Green Blue Red AD on the eighth line and posts a red flag. Rashmi is in between these two girls with blue flag. 46. Coordinates of Niharika are 1 (c) (–20, 8) (a) (2, 25) (b) (25, 2) (c) (–2, 25) (d) (–25, 2) 47. Coordinates of Preet are 1 (a) (20, 8) (b) (8, 20) (d) (–8, 20) Sample Paper - 3  145

48. What is the distance between two flags? 1 1 (a) 68 (b) 71 (c) 61 (d) 52 1 49. Position of Rashmi’s flag is (a) (2, 22.5) (b) (22.5, 2) (c) (–2, 22.5) (d) (–22.5, 2) 50. If each flower pot costs ` 50. How much they have to pay for 100 pots? (d) ` 6000 (a) ` 2000 (b) ` 5000 (c) ` 3000 ANSWERS 1. (c) 2. (d) 3. (a) 4. (b) 5. (c) 6. (c) 7. (d) 8. (d) 9. (d) 10. (b) 11. (c) 12. (c) 13. (b) 14. (d) 15. (c) 16. (b) 17. (b) 18. (b) 19. (d) 20. (b) 21. (d) 22. (b) 23. (d) 24. (c) 25. (b) 26. (b) 27. (b) 28. (c) 29. (a) 30. (d) 31. (b) 32. (b) 33. (d) 34. (c) 35. (c) 36. (a) 37. (a) 38. (b) 39. (a) 40. (a) 41. (d) 42. (a) 43. (b) 44. (c) 45. (d) 46. (a) 47. (b) 48. (c) 49. (a) 50. (b) 146  Mathematics–10