Arihant Maths (standard) class 10 term 2

CBSE Term II Mathematics X (Standard) 43 Subtracting Eq. (ii) from Eq. (i), we get Now, sum of AP series is [(m − 1) − (n − 1)]d = 1 − 1 Sn = n [2a + (n − 1) d] nm 2 ⇒ (m − n )d = m − n = 1 ⎛⎝⎜ x + 2 ⎟⎞⎠ ⎣⎢⎡2 × 1 + ⎜⎝⎛ x + 2 − 1⎠⎞⎟ 3⎤⎥⎦ mn 2 3 3 ⇒ d= 1 = x + 2 [2 + x − 1] mn 6 Put d = 1 in Eq (i), we get ⇒ 287 = x + 2 × (x + 1) mn 6 a + (m − 1) 1 = 1 ⇒ 1722 = x2 + 3x + 2 mn n ⇒ x2 + 3x − 1720 = 0 ⇒ a = 1 − 1 (m − 1) = 1 n mn mn ⇒ x2 + (43 − 40)x − 1720 = 0 [splitting middle term] ∴ amn = a + (mn − 1)d ⇒ x2 + 43x − 40x − 1720 = 0 = 1 + (mn − 1) × 1 mn mn ⇒ x(x + 43) − 40(x + 43) = 0 ⇒ (x − 40) (x + 43) = 0 = 1 [1 + mn − 1] = mn = 1 ⇒ x = 40, − 43 mn mn But x = − 43 is not possible, because it is an increasing AP. Hence, required value of x is 40. 44. Given first term of an AP, a = 54 Common difference, d = − 3 46. Given, equation is 1 + 5 + 9 + 13 + K + x = 1326 and nth term of an AP, Here, first term is a1 = 1 an = 0 last term is an = l = x ⇒ a + (n − 1)d = 0 Difference of two consecutive terms, ⇒ 54 + (n − 1)(−3) = 0 ⇒ 54 − 3n + 3 = 0 5 − 1 = 4 and 9 − 5 = 4, which is same. ⇒ 3n = 57 Thus, given series is an AP. ⇒ n = 57 Then, nth term of given AP is 3 an = a + (n − 1)d ⇒ n = 19 ∴ x = 1 + (n − 1)4 Now, sum of first 19 terms of given AP is ⇒ (n − 1) 4 = x − 1 ⇒ n −1= x −1 Sn = n [2a + (n − 1)d] 2 4 ⇒ n = x −1 +1 ∴ S19 = 19 [2 × 54 + (19 −1)( −3)] 2 4 ⇒ n=x+3 = 19 [108 − 54] 2 4 Now, sum of given AP is = 513 Sn = n [2 a + (n − 1)d] 45. Given equation is 1 + 4 + 7 + 10 + .... + x = 287 2 Consider series, 1 + 4 + 7 + 10 + .... + x ∴ 1326 = x+3 ⎡⎣⎢2 ×1 + x + 3 × 4⎥⎦⎤ 4×2 4 Here, a1 = 1, a2 = 4, a3 = 7 ⇒ 1326 × 8 = (x + 3) [2 + x + 3] Now, a2 − a1 = 4 − 1 = 3 ⇒ 10608 = (x + 3) (x + 5) ⇒ x2 + 8x + 15 − 10608 = 0 a3 − a2 = 7 − 4 = 3 It implies that common difference is constant say 3. So, it is ⇒ x2 + 8x − 10593 = 0 an AP series, whose first term is a = 1 and common difference d = 3. ⇒ x2 + (107 − 99)x − 10593 = 0 Here, last term of an AP is l = x [by splitting middle term] ⇒ x2 + 107x − 99x − 10593 = 0 Q l = a + (n − 1) d ⇒ x(x + 107) − 99(x + 107) = 0 ∴ x = 1 + (n − 1) × 3 ⇒ (x − 99) (x + 107) = 0 ⇒ x = 99, − 107 ⇒ x = 1 + 3n − 3 ⇒ x = 3n − 2 ⇒ n=x+2 3

44 CBSE Term II Mathematics X (Standard) 47. (i) Here, first term (a) = 1 48. Given AP sequence is − 4 , − 1, − 2 , .... , 4 1 333 and common difference (d) = (− 2) − 1 = − 3 Here, first term ( a) = − 4, 3 Q Sum of n terms of an AP, common difference (d ) = − 1 + 4 = 1 Sn = n [2 a + (n − 1) d] 33 2 and the last term (l ) = 4 1 = 13 ⇒ Sn = n [2 ×1 + (n − 1) × (− 3)] 33 2 ⇒ Sn = n (2 − 3n + 3) Q nth term of an AP, 2 l = an = a + (n − 1) d ⇒ Sn = n (5 − 3n ) ...(i) ⇒ 13 = − 4 + (n − 1) 1 2 33 3 We know that, if the last term (l) of an AP is known, then ⇒ 13 = − 4 + (n − 1) l = a + (n − 1) d ⇒ n − 1 = 17 ⇒ − 236 = 1 + (n − 1) (− 3) [Q l = − 236, given ] ⇒ n = 18 [even] ⇒ − 237 = − (n − 1) × 3 So, the two middle most terms are ⎝⎛⎜ n ⎠⎟⎞ th and ⎛⎝⎜ n + 1⎟⎠⎞ th. 2 2 ⇒ n − 1 = 79 ⇒ n = 80 i.e. ⎜⎛⎝ 128⎟⎠⎞ th and ⎜⎛⎝ 18 + 1⎟⎠⎞ th terms 2 Now, put the value of n in Eq. (i), we get Sn = 80 [5 − 3× 80] i.e. 9th and 10th terms. 2 a9 a 8d 4 8 ⎛⎝⎜ 13⎠⎞⎟ −4 + 8 4 ∴ = + = − 3 + = 3 = 3 = 40 (5 − 240) = 40 × (− 235) and a10 = a + 9d = −4 + 9 ⎝⎛⎜ 13⎟⎠⎞ = −4 + 9 = 5 3 3 3 = − 9400 Hence, the required sum is − 9400. So, sum of the two middle most terms Alternate Method = a 9 + a10 =4+ 5=9=3 Given, a = 1, d = − 3 and l = − 236 3 33 ∴ Sum of n terms of an AP, Sn = n [a + l] 49. Let the first term, common difference and the number of 2 terms in an AP are a, d and n, respectively. = 80 [1 + (− 236)] 2 [Q n = 80] We know that, the n th term of an AP, Tn = a + (n − 1) d ... (i) ∴ 4th term of an AP, = 40 × (− 235) = − 9400 T4 = a + (4 − 1) d = − 15 [given] (ii) Here, first term, a = 4 − 1 ⇒ a + 3d = − 15 ...(ii) n and 9th term of an AP, Common difference, T9 = a + (9 − 1) d = − 30 [given] d = ⎝⎜⎛ 4 − 2 ⎠⎞⎟ − ⎛⎜⎝ 4 − 1 ⎞⎠⎟ = −2 + 1 = −1 ⇒ a + 8d = − 30 ...(iii) n n n n n Now, subtract Eq. (ii) from Eq. (iii), we get Q Sum of n terms of an AP, a + 8d = − 30 Sn = n [2 a + (n − 1) d] a + 3d = − 15 2 −− + n ⎢⎣⎡2 ⎜⎝⎛ 4 − 1 ⎟⎠⎞ ⎝⎜⎛ −n1⎠⎞⎟ ⎤ 5d = − 15 2 n ⎦⎥ ⇒ Sn = + (n − 1 ) ⇒ d=−3 n ⎩⎨⎧8 2 1⎫ Put the value of d in Eq. (ii), we get 2 n n ⎬⎭ = − −1 + a + 3 (− 3) = − 15 n 1 ⇒ a − 9 = − 15 2 n = ⎜⎛⎝7 − ⎞⎟⎠ ⇒ a = − 15 + 9 ⇒ a=−6 = n × ⎛⎝⎜ 7 n− 1⎟⎞⎠ Q Sum of first n terms of an AP, 2 n Sn = n [2 a + (n − 1) d] = 7n − 1 2 2

CBSE Term II Mathematics X (Standard) 45 ∴ Sum of first 17 terms of an AP, On substituting d = 2 in Eq. (i), we get S17 = 17 [2 × (− 6) + (17 − 1) (− 3)] ⇒ 2a + 3 × 2 = 20 2 ⇒ 2a = 14 = 17 [− 12 + (16) (− 3)] a =7 2 Now, Sn = n [2a + (n − 1)d] = n [2(7) + (n − 1)2] = 17 (− 12 − 48) 2 2 2 = n [14 + 2n − 2] = n[6 + n ] = 6n + n 2 = 17 × (− 60) 2 2 Hence, the sum of first n terms is n 2 + 6n. = 17 × (− 30) 52. Let a and d be the first term and common difference of an AP. = − 510 Given that, a11 : a18 = 2 : 3 ⇒ a + 10d = 2 Hence, the required sum of first 17 terms of an AP is − 510. a + 17 d 3 50. Given first term of each sum of an AP is 1 and common ratio of each sum are 1, 2 and 3, respectively. ⇒ 3a + 30d = 2a + 34d ⇒ a = 4d ∴ S1 = n [2(1) + (n − 1)1] …(i) 2 Now, a5 = a + 4d = 4d + 4d = 8d [from Eq. (i)] = n [2 + n − 1] 2 and a21 = a + 20d = 4d + 20 d = 24d [from Eq. (i)] = n (n + 1) ∴ a5 : a21 = 8d : 24d = 1 : 3 2 Now, sum of the first five terms, ∴ S2 = n [2(1) + (n − 1)2 ] S 5 = 5 [2 a + (5 − 1) d] 2 2 = n [2 + 2n − 2] = 5 [2 (4d ) + 4d ] [from Eq. (i)] 2 2 = n2 = 5 (8d + 4d ) = 5 × 12 d = 30 d 22 n and S3 = 2 [2(1) + (n − 1)3] and sum of the first 21 terms, = n [2 + 3n − 3] S21 = 21 [2 a + (21 − 1)d] 2 2 = n (3n − 1) = 21 [2 (4d ) + 20d ] [from Eq. (i)] 2 2 LHS = S1 + S 3 = 21 (28 d ) = 294 d = n (n + 1) + n (3n − 1) 2 22 So, ratio of the sum of the first five terms to the sum of the first 21 terms = n [n + 1 + 3n − 1] S 5 : S21 = 30 d : 294 d = 5 : 49 2 53. Let the four consecutive number of an AP be = n × 4n a, a + d, a + 2d and a + 3d. 2 Since, sum of four consecutive number in AP is 32. = 2n 2 = 2S 2 Hence proved. ∴ a + a + d + a + 2d + a + 3d = 32 51. Given, S4 = 40 and S14 = 280 ⇒ 4a + 6d = 32 ⇒ 2a + 3d = 16 [divide by 2] Let a be the first term and d be the common difference of ⇒ a = 16 − 3d …(i) 4 [2 a given AP. Then, S4 = 40 ⇒ 2 + (4 − 1)d] = 40 2 ⎡⎣⎢Q Sn n {2a − 1)d}⎤⎥⎦ According to the question, 2 = + (n Product of first and last terms = 7 Product of two middle terms 15 ⇒ 2[2a + 3d] = 40 ⇒ 2a + 3d = 20 …(i) …(ii) a (a + 3d) = 7 and S14 = 280 (a + d) (a + 2d) 15 ⇒ 14[2a + (14 − 1)d] = 280 ⇒ 15a(a + 3d) = 7(a + d) (a + 2d) 2 ⇒ 15a2 + 45ad = 7a2 + 21ad + 14d2 ⇒ 2a + 13d = 40 ⇒ 8a2 + 24ad − 14d2 = 0 On subtracting Eq. (i) from Eq. (ii), we get 10d = 20 ⇒ d = 2 ⇒ 4a2 + 12ad − 7d2 = 0 [divide by 2]

46 CBSE Term II Mathematics X (Standard) ⇒ 4a2 + (14 − 2)ad − 7d2 = 0 55. Given AP is 20, 19 1 , 182 ,... . 33 [by splitting middle term] ⇒ 4a2 + 14ad − 2ad − 7d2 = 0 Here, a = 20 and d = 19 1 − 20 = 58 − 20 = 58 − 60 = −2 33 33 ⇒ 2a(2a + 7d) − d(2a + 7d) = 0 Let n terms of given AP be required to get sum 300. ⇒ (2a + 7d) (2a − d) = 0 ⇒ a = − 7d and a = d …(ii) We know that, Sn = n [2a + (n − 1)d] 2 22 n ⎢⎣⎡2(20) −32 ⎠⎞⎟ ⎤ Put a = − 7d in Eq. (i), we get ⇒ 300 = 2 + (n − 1)⎝⎜⎛ ⎦⎥ 2 − 7d = 16 − 3d [Qa = 20 and d = − 2 / 3] 2 2 ⇒ 600 = n ⎣⎡⎢40 − 2n + 2⎤ ⇒ −7d = 16 − 3d 3 3 ⎦⎥ ⇒ 4d = − 16 ⇒ d = − 4 ⇒ 600 = 1 [120n − 2n 2 + 2n ] Now, put a = d in Eq. (i), we get 3 2 ⇒ 600 × 3 = 122n − 2n 2 d = 16 − 3d ⇒ 1800 + 2n 2 − 122n = 0 22 ⇒ 2[n 2 − 61n + 900] = 0 ⇒ d = 16 − 3d ⇒ 4d = 16 ⇒ d = 4 ⇒ n 2 − 61n + 900 = 0 [divide by 2] For a = − 7d and d = − 4, then ⇒ n 2 − 36n − 25n + 900 = 0 2 ⇒ n (n − 36) − 25(n − 36) = 0 a = − 7 × (−4) = 14 ⇒ (n − 36)(n − 25) = 0 ⇒ n = 36 or 25 2 Since, a is positive and d is negative, so both values of n are possible. For a = d and d = 4, then a = 4 = 2 22 Hence, sum of 25 terms of given AP Therefore, the four consecutive numbers in an AP are = Sum of 36 terms of given AP = 300. 56. (i) Now, she takes ` 1 on day 1, ` 2 on day 2, ` 3 on day 3 14, [14 + (−4)], [14 + (2 × − 4)], [14 + (3 × − 4)] and so on till the end of the month, from this money. or 2, (2 + 4), (2 + 2 × 4), (2 + 3 × 4). Hence, the numbers are 14, 10, 6, 2 or 2, 6, 10, 14. i.e. 1 + 2 + 3 + 4 + ... + 31. 54. Given that, the AP is a, b, c. which form an AP in which terms are 31 and first term Here, first term = a, common difference = b − a (a) = 1, common difference (d) = 2 − 1 = 1 and last term, l = an = c ∴ Sum of first 31 terms = S31 Q an = l = a + (n − 1) d Sum of n terms, ⇒ c = a + (n − 1) (b − a) ⇒ (n − 1) = c − a Sn = n [2 a + (n − 1) d] 2 b−a ∴ S31 = 31 [2 × 1 + ( 31 − 1) × 1] ⇒ n= c−a +1 2 b−a = 31 (2 + 30) = 31 × 32 = 31 × 16 = 496 ⇒ n = c − a + b − a = c + b − 2a …(i) 22 b−a b−a So, Kanika takes ` 496 till the end of the month from this ∴ Sum of an AP, money. n (ii) Let her pocket money be ` x. 2 Sn = [2 a + (n − 1) d] Now, she spent ` 204 of her pocket money and found that at the end of the month she still has ` 100 with her. = (b + c − 2a) ⎡ + ⎧b + c − 2a − 1⎫⎬ (b − ⎤ 2 (b − a) ⎢2 a ⎨ b−a ⎭ a )⎥ Now, according to the condition, ⎣ ⎩ ⎦ (x − 496) − 204 = 100 ⇒ x − 700 = 100 = (b + c − 2a) ⎡⎣⎢2 a + c−a ⋅ (b − a )⎦⎥⎤ ∴ x = ` 800 2 (b − a) b−a Hence, ` 800 was her pocket money for the month. = ( b + c − 2a) (2a + c − a) (iii) Here, a = 1, d = 1, n = 13 2 (b − a) Now, a n = a + (n − 1) d ⇒ a13 = 1 + 12(1) = 1 + 12 = 13 = (b + c − 2a) ⋅(a + c) Hence proved. So, Kanika, saved ` 13 till January 13th, 2008. 2(b − a)

Chapter Test Multiple Choice Questions (iv) Total amount paid in 13th and 17th installment is 1. The list of numbers − 10, − 6, − 2, 2, ... is (a) ` 380 (b) ` 300 [NCERT Exemplar] (c) ` 360 (a) an AP with d = − 16 (b) an AP with d = 4 (d) ` 340 (c) an AP with d = − 4 (d) not an AP (v) If he increases the installment by ` 6 every 2. In an AP, if a = 3.5, d = 0 and n = 101, then an will be month, then the amount he will pay in (a) 0 (b) 3.5 53th installment is (c) 103.5 (d) 104.5 (a) ` 314 (b) ` 360 3. Is an sequence defined by an = 2n2 + 1 forms an (c) ` 412 (d) ` 416 AP? Short Answer Type Questions (a) Yes (b) Not 7. Two AP’s have the same common difference. (c) Cannot be determined (d) None of the above The first term of one AP is 2 and that of the other is 7. 4. The sum of first 20 terms of an AP in which The difference between their 10th terms is the same as the difference between their 21st a = 1 and 20th term = 58 is terms, which is the same as the difference between any two correspondi[nNgCEteRrTmEsx?eWmphlya?r] (a) 590 (b) 580 (c) 570 (d) 560 8. Determine the AP whose fifth term is 19 and 5. The 10th term of an AP is 52 and 16th term is 82, the difference of the eighth term from the thirteenth term is 20. then 32nd term of the AP is [NCERT Exemplar] (a) 152 (b) 159 9. Find the sum of all the 11 terms of an AP (c) 162 (d) 156 whose middle most term is 30. Case Based MCQs Long Answer Type Questions 6. Kartik starts repaying a loan as first installment 10. An AP consists of 37 terms. The sum of the of three middle most terms is 225 and the sum ` 100. He increases the installment by ` 5 every of the last three terms is 429. Find the AP. month. (i) AP formed from the given situation is 11. If sum of first 6 terms of an AP is 36 and that (a) 105, 110, 115, ...... (b) 100, 105, 110, ...... (c) 95, 100, 105, ...... (d) 110, 115, 120, ...... of the first 16 terms is 256, then find the sum of first 10 terms. (ii) The amount Kartik will pay in 30th 12. Which term of the AP : 121, 117, 113, ... is its installment is second negative term? (a) ` 265 (b) ` 235 (c) ` 255 (d) ` 245 13. The sum of the third and the seventh terms of (iii) If Kartik pays ` 795, then it is an AP is 6 and their product is 8. Find the sum (a) 140th installment of first sixteen terms of the AP. (b) 150th installment (c) 160th installment 14. Solve the equation −4 + (−1) + 2 + ..... + x = 437. (d) 170th installment Answers 1. (b) 2. (b) 3. (b) 4. (c) 5. (a) 6. (i) (b) (ii) (d) (iii) (a) (iv) (d) (v) (c) For Detailed Solutions Scan the code 7. (37) 8. 3, 7, 11, 15 9. 330 10. 3, 7, 11, 15 11. 100 12. 33rd term 13. 20 or 76 14. x = 50

48 CBSE Term II Mathematics X (Standard) CHAPTER 03 Circles In this Chapter... l Circle l Tangent of a Circle l Theorem Related to Tangent of a Circle A circle is a collection of all points in a plane which are at a Semi-circle constant distance i.e. radius from a fixed point i.e. centre. A diameter of a circle divides it into two equal parts or In the given figure, O is the centre of circle and OA is the radius of in two equal arcs. Each of these two arcs is called a the circle. Also, AB is the diameter of the circle. semi-circle. B A Circumference O The length of the complete circle is called the Two or more circles having the same centre are called concentric circumference of the circle. circles. Arc Some Important Terms Related to Circle A continuous piece of a circle is called an arc. In adjoining figure, P and Q are two points on a circle Chord which divide it into two parts, called the arcs. The larger part is called the major arc QRP and the smaller A line segment joining any two points on the circumference of the part is called the minor arc PMQ. circle is called a chord of the circle. If this chord passes through the centre, then this chord (or diamter) is the longest chord of the circle. R Major arc O P Minor arc Q AB M

CBSE Term II Mathematics X (Standard) 49 Segment Tangent to a Circle The region between a chord and either of its arcs is called a segment of the circular region or simply a segment of the A line which touches the circle at a point, is called tangent to circle. The segment formed by minor arc along with chord, is a circle. called minor segment and the segment formed by major arc, is called the major segment. In the figure, O is the centre of circle, AB is a tangent line and P is a point of contact. Major segment G There is only one tangent at a point of the circle. G A circle can have maximum two parallel tangents which can be drawn to the opposite sides of the centre. AB O Minor segment A B (Tangent line) P Sector The region between an arc and the two radii, joining the ends (Point of contact) of the arc to the centre, is called a sector. Length of a Tangent Major sector The length of the segment of the tangent, between the given O point (on the tangent) and the point of contact, is called the length of tangent from the given point. Minor O A sector B The sector formed by minor arc, is called minor sector and AB the sector formed by major arc, is called major sector. Segment of the tangent between Important Results Related to Circle Tangent line the given point B and point of contact A (i) The perpendicular drawn from the centre of a circle to In the above figure, AB is called the length of tangent. a chord bisects it and vice-versa. ∴ Length of tangent to the circle from an exterior point, (ii) Equal chords of a circle are equidistant from the centre. AB = (iii) The angle subtended by an arc (or corresponding (Distance of exterior point from centre)2 − (Radius)2 chord) at the centre of the circle is twice the angle subtended by the same arc at any point on the Number of Tangent from a Point on a Circle remaining part of the circle. (i) If point P lies outside the circle, then two tangents can C be drawn to the circle, i.e. PT1 and PT2 . θ T1 (Tangent) O 2θ B P A (iv) Equal chords of a circle subtend equal angles at T2 (Tangent) the centre. (ii) If point P lies on the circle, then there is one and only (v) The angle in a semi-circle is a right angle. one tangent to a circle passing through point P. (vi) Angles in the same segment of a circle are equal. P Tangent (vii) The sum of any pair of opposite angles of a cyclic quadrilateral is 180°. (viii) If two circles intersect at two points, then the line through the centres is the perpendicular bisector of the common chord.

50 CBSE Term II Mathematics X (Standard) (iii) If a point P lies inside the circle, then there is no Important Results Related to Tangent to a Circle tangent to a circle passing through a point lying inside the circle. (i) If two circles touch internally or externally, then point of contact lies on the straight line through the two centres. No Tangent O Oʹ P O P Oʹ P Theorems Related to Tangent of Circle (ii) A pair of tangents drawn at two points of a circle are either Theorem 1 The tangent at any point of a circle is parallel or they intersect each other at a point outside the perpendicular to the radius through the point of contact. circle. O (iii) If two tangents drawn to a circle are parallel to each other, then the line segment joining their point of contact is a APB diameter of the circle. Here, O is centre of circle and AB is tangent of circle at P (iv) If two tangents are drawn to a circle from an external point, and it is point of contact and OP is radius. then ∴ OP ⊥ AB. (a) They subtend equal angles at the centre, Theorem 2 A perpendicular drawn from the end point of i.e. ∠POA = ∠POB. radius is tangent to the circle. If OP ⊥ AB, then AB is tangent to circle. A O PO B (b) They are equally inclined to the segment joining the centre to that point, i.e. ∠APQ = ∠BPQ. APB (v) The opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Theorem 3 The lengths of two tangents drawn from an external point to a circle are equal. DC A O B A OP B Here, P is exterior point and PA and PB are tangents. ∴ PA = PB

CBSE Term II Mathematics X (Standard) 51 Solved Examples Example 1. Prove that a tangent to a circle is From Eqs. (i) and (ii), we get TP = TQ perpendicular to the radius through the point of Hence, T is the mid-point of the line segment PQ. contact. [CBSE 2020 (Standard)] Sol. Given A circle with centre O and a tangent AB at a point Example 3. In figure, PQ is tangent to the circle with P on the circle. To prove OP ⊥ AB centre O, at the point B. If ∠AOB =100°, then find Construction Take any point Q, other than P on the tangent ∠ABP. [CBSE 2020 (Standard)] AB and join OQ. O O A 100° Q B R P A P QB Sol. Given, ∠AOB = 100°, Proof Here, Q is a point on the tangent AB, other than the A point of contact P. So, Q lies outside the circle (if Q lies 100° O inside the circle, then AB becomes a secant and not a tangent to the circle). Let OQ intersects the circle at R . Then, OP = OR [radii of the circle] B PQ Now, OQ = OR + RQ ⇒ OQ > OR In ΔOAB, ⇒ OQ > OP or OP < OQ [Q OP = OR] OA = OB [radii of the circle] Thus, OP is shorter than any other segment joining O to any ⇒ ∠OBA = ∠OAB point of AB. Also, we know that the shortest distance between a point and a line is perpendicular distance from [angles opposite to equal sides are equal] …(i) the point to the line. In ΔOAB, So, OP is perpendicular to AB. ∠AOB + ∠OAB + ∠OBA = 180° i.e. OP ⊥ AB Hence proved. [by angle sum property of triangle] Example 2. In given figure, two circles touch each ⇒ 100° + ∠OBA + ∠OBA =180° [from Eq. (i)] other at the point C. Prove that the common ⇒ 2∠OBA =180° − 100° tangent to the circles at C, bisects the common ⇒ ∠OBA = 80° tangent at P and Q. 2 P TQ ⇒ ∠OBA = 40° …(ii) We know that, radius of circle is perpendicular to the tangent. A CB ∴ ∠OBP = 90° ⇒ ∠OBA + ∠ABP = 90° ⇒ 40° + ∠ABP = 90° ⇒ ∠ABP = 90° − 40° ⇒ ∠ABP = 50° Sol. We know that, tangents drawn from an external point are of Example 4. In below figure, PA is a tangent from equal length. Therefore, according to the given figure, TP = TC …(i) [Q point T is external] an external point P to a circle with centre O. and TQ = TC …(ii) [Q point T is external] If ∠POB = 115°, find ∠APO. [CBSE 2020 (Standard)] P TQ A AB PO 115° C B

52 CBSE Term II Mathematics X (Standard) Sol. Given, ∠POB = 115° ⇒ AP2 = OA 2 − OQ2 A [Q OP = OQ = radii of a circle] …(ii) P Now, in right angled ΔOQA, O OA 2 = OQ2 + AQ2 115° ⇒ AQ2 = OA 2 − OQ2 ...(iii) B From Eqs. (ii) and (iii), we get AP2 = AQ2 Since, AB is a straight line. ∴ ∠POB + ∠AOP = 180° ⇒ AP = AQ Hence proved. ⇒ 115° + ∠AOP = 180° ⇒ ∠AOP = 180° − 115° Example 6. Prove that the angle between the two ⇒ ∠AOP = 65° We know that, radius line is perpendicular to the tangent. tangents drawn from an external point to a circle is ∴ ∠PAO = 90° supplementary to the angle subtended by the line In ΔAOP, segment joining the points of contact at the centre. ∠PAO + ∠AOP + ∠APO = 180° Sol. Let PQ and PR be two tangents drawn from an external point [Q sum of all angles of a triangle is 180°] P to a circle with centre O. ∴ 90° + 65° + ∠APO = 180° ⇒ ∠APO = 180° − (65° + 90°) Q = 180° − 155° = 25° PO Example 5. Prove that the length of tangents drawn R from an external point to a circle are equal. To prove ∠QOR = 180° − ∠QPR [CBSE 2020 (Standard)] or ∠QOR + ∠QPR = 180° Proof In Δ OQP and ΔORP, Sol. Let AP and AQ are two tangents drawn from a point A to a circle with centre O. PQ = PR [Q tangents drawn from an P external point are equal in length] AO OQ = OR [radii of circle] OP = OP [common sides] Q ∴ ΔOQP ≅ ΔORP [by SSS congruence rule] Then, and ∠ QPO = ∠RPO [by CPCT] To prove AP = AQ ⇒ ∠POQ = ∠POR [by CPCT] and Construction Join OP, OQ and OA. ∠ QPR = 2 ∠ OPQ⎫ …(i) ⎬ Proof We know that, a tangent at any point of a circle is ∠ QOR = 2 ∠POQ ⎭ perpendicular to the radius through the point of contact. Now, in right angled ΔOQP, Here, AP is a tangent and OP is the radius of the circle ∠ QPO + ∠ QOP = 90° through P. ⇒ ∠ QOP = 90° − ∠ QPO ∴ OP ⊥ AP ⇒ 2 ∠QOP = 180° − 2 ∠QPO Similarly, OQ ⊥ AQ [multiplying both sides by 2] ⇒ ∠OPA = ∠OQA = 90° ...(i) ⇒ ∠ QOR = 180° − ∠ QPR [from Eq. (i)] First Method ⇒ ∠ QOR + ∠ QPR = 180° Hence proved. In ΔOPA and ΔOQA, we have Example 7. In figure, find the perimeter of ΔABC, OP = OQ [radii of a circle] if AP = 12 cm. ∠OPA = ∠OQA = 90° [from Eq. (i)] OA = OA [common sides] A So, ΔOPA ≅ ΔOQA [by RHS congruence rule] ∴ AP = AQ [by CPCT] D BC Second Method PQ In right angled ΔOPA, OA 2 = OP2 + AP2 [by Pythagoras theorem] ⇒ AP2 = OA 2 − OP2

CBSE Term II Mathematics X (Standard) 53 Sol. Given, AP = 12 cm ∴ AC = AR + RC ⇒ AQ = AP = 12 cm A ⇒ 11 = 4 + RC ⇒ RC =11 − 4 = 7 cm D BC Now, BC =BQ + QC PQ = 3 + 7 = 10 cm Hence, length of BC is 10 cm. Example 9. In the given figure, from an external point P, two tangents PQ and PR are drawn to a circle of radius 4 cm with centre O. If ∠QPR = 90°, then find the length of PQ. [Q tangents drawn from an external point 4 cm Q are equal in lengths] P Also, BD = BP [Q B is an external point] …(i) OR and CD = CQ [Q C is an external point] …(ii) Now, AP = AB + BP Sol. We know that, if pair of tangents are drawn from an external point P, then line joining from centre O to the point P, ⇒ 12 = AB + BD [from Eq. (i)] …(iii) bisects the angle P. and AQ = AC + CQ ⇒ 12 = AC + CD [from Eq. (i)] …(iv) Perimeter of ΔABC Q 4 cm = AB + BC + AC O = AB + BD + DC + AC 45° P 90° = 12 + 12 [from Eqs. (iii) and (iv)] = 24 cm R Hence, perimeter of a ΔABC is 24 cm. ∠QPR Example 8. In below figure, ΔABC is circumscribing a ∴ ∠OPQ = 2 circle, the length of BC is …… cm. = 90° = 45° 2 [CBSE 2020 (Standard)] A Also, radius of circle OQ is perpendicular to the tangent line QP. 3 cm 4 cm 11 cm Now, in right angled ΔOQP, PR tan 45° = OQ QP ⇒ 1= 4 QP BQ C ⇒ QP = 4 cm Hence, length of PQ is 4 cm. Sol. We know that, the tangents drawn from an external point to a circle are equal. Therefore, Example 10. In the given figure, if tangents PA and PB A from an external point P to a circle with centre O, 3 cm 4 cm are inclined to each other at an angle of 80°, then 4 cm find ∠AOB. [CBSE 2020 (Standard)] PR A 7 cm 80º P B 3 cm Q 7 cm C O B BP = BQ, [point B is an external] AP = AR [point A is an external] and CQ = CR [point C is an external]

54 CBSE Term II Mathematics X (Standard) Sol. Given, ∠APB = 80°. Example 12. In given figure, two tangents TP and TQ We know that, line drawn from centre of a circle to the are drawn to a circle with centre O from an external tangent is perpendicular. point T. Prove that ∠PTQ = 2∠OPQ. Since, OA⊥PA and OB⊥PB. …(i) [CBSE 2020 (Standard)] Then, ∠OAP = ∠OBP = 90° P A T O P 80º Q O Sol. Given, TP and TQ are two tangents of a circle with centre O B and points P and Q are point of contact. To prove ∠PTQ = 2∠OPQ Since, OAPB is a quadrilateral. Proof Let ∠PTQ = θ By using angle sum property of a quadrilateral, As we know that, the length of tangents drawn from an external point to a circle are equal. ∠AOB + ∠OBP + ∠APB + ∠OAP = 360° So, ΔTPQ is an isosceles triangle. ⇒ ∠AOB + 90° + 80° + 90° = 360° Therefore, according to the given figure, ⇒ ∠AOB = 360° − 260° ∠TPQ = ∠TQP = 1 (180° − θ) = 90°− θ ⇒ ∠AOB = 100° 22 As we know that, the tangents at any point of a circle is Example 11. In given figure, PA and PB are tangents to perpendicular to the radius through the point of contact. P the circle with centre O, such that ∠APB = 50°, then the measure of ∠OAB is ...... A P 50° Tq O O B Q A Sol. Given, ∠APB = 50° ∴ ∠OPT = 90° Now, ∠OPQ = ∠OPT − ∠TPQ = 90° − ⎜⎝⎛ 90° − 2θ⎞⎠⎟ = θ = ∠PTQ 2 2 P 50° O ⇒ ∠PTQ = 2∠OPQ Hence proved. B Example 13. In figure, a quadrilateral ABCD is drawn Since, P is an external point of a circle. to circumscribe a circle. Prove that Therefore, PA =PB AB + CD = BC + AD [Q tangents drawn from an external to a circle are equal] [CBSE 2020 (Standard)] ⇒ ∠PBA = ∠PAB A B [Q angles opposite to equal sides are equal] ...(i) In ΔAPB, ∠APB + ∠PBA + ∠PAB = 180° D C [Q sum of all angles of a triangle is 180°] ∴ 50° + 2∠PAB = 180° [from Eq. (i)] Sol. Given A quadrilateral ABCD is circumscribing a circle. ⇒ 2∠PAB = 130° To prove AB + CD = AD + BC ⇒ ∠PAB = 65° ...(ii) Proof Let P, Q, R and S be the point of contact. Also, radius OA is perpendicular to the tangent of a circle. AP Therefore, B ∠OAP = 90° S Q ⇒ ∠OAB + ∠PAB = 90° ⇒ ∠OAB + 65° = 90° [from Eq. (ii)] D R ⇒ ∠OAB = 90° − 65° = 25° C

CBSE Term II Mathematics X (Standard) 55 We know that, the length of tangents drawn from an external Now, ∠OPQ = 90° − 60° = 30° point to a circle are equal. [Q ∠OPT = 90°, as radius line OP is perpendicular to the tangent] ∴ AP = AS ⇒ ∠OQP = 30° [Q both are tangents to a circle from point A] …(i) [angles opposite to equal sides are equal] Similarly, BP = BQ, …(ii) In ΔOPQ, using angle sum property of a triangle, ∠POQ + ∠OPQ + ∠OQP = 180° CR = CQ …(iii) ⇒ ∠POQ + 30° + 30° = 180° ⇒ ∠POQ = 180° − 60° = 120° and DR = DS …(iv) ⇒ ∠PQʹ Q = 60° Adding Eqs. (i), (ii), (iii) and (iv), we get [angle subtended by an arc at centre is twice the angle subtended at remaining part of circle] (AP + BP) + (CR + DR ) = (AS + BQ) + (CQ + DS) ⇒ ∠PRQ = 180°−∠PQʹ Q = 120° ⇒ AB + CD = (AS + DS) + (BQ + CQ) [Q opposite angles are supplementary in a cyclic quadrilateral PQʹ QR] ⇒ AB + CD = AD + BC Hence proved. Example 16. In given figure, AB is a chord of circle Example 14. Prove that the tangents at the extremities with centre O, AOC is diameter and AT is tangent of any chord of a circle make equal angles with the at A. Prove that ∠BAT = ∠ACB. chord. [CBSE 2020 (Standard)] C Sol. Let AB be a chord of a circle having centre O. Let AP and OB BP be the tangents at A and B, which intersect at point P. To prove ∠PAC = ∠PBC Construction Join points C and P. Proof We know that, tangents drawn from an external point are equal. A OC P B AT ∴ In ΔPCA and ΔPCB, Sol. Given AB is a chord of a circle, AOC is a diameter of the PA =PB, [QP is an external point of a circle] circle having centre O and line AT is tangent at A. To prove ∠BAT = ∠ACB ∠APC = ∠BPC C [Q AP and BP are equally inclined to OP] OB and PC = PC [common sides] ∴ ΔPAC ~ ΔPBC [by SAS similarity rule] ⇒ ∠PAC = ∠PBC [by CPCT] Example 15. In given figure, PQ is a chord of a circle AT and PT is tangent at P such that ∠QPT = 60°, then Proof We know that, diameter of a circle subtends 90° to the the measure of ∠PRQ is ....... [CBSE 2020 (Standard)] semi-circle. ∴ ∠ABC = 90° Q Let ∠ACB = θ, O then ∠CAB = 180° −(90° + θ) 60° R [by using angle sum property of triangle] P ⇒ ∠CAB = 90° − θ ...(i) T We know that, radius of a circle is perpendicular to the Sol. Take a point Qʹ on circle and join PQʹ and QQʹ. tangent. ∴ ∠OAT = 90° Qʹ ⇒ ∠OAB + ∠BAT = 90° A Q ⇒ ∠CAB + ∠BAT = 90° [Q ∠OAT = ∠CAT] O ⇒ 90° − θ + ∠BAT = 90° 60º R PT ⇒ ∠BAT = θ ⇒ ∠BAT = ∠ACB Hence proved.

56 CBSE Term II Mathematics X (Standard) Chapter Practice PART 1 A Objective Questions O 115° G Multiple Choice Questions B P 1. If radii of two concentric circles are 4 cm and 5 cm, then length of each chord of one circle, which is (a) 25° (b) 20° (c) 30° (d) 65° tangent to the other circle, is [NCERT Exemplar] 8. In figure, AT is a tangent to the circle with centre O (a) 3 cm (b) 6 cm (c) 9 cm (d) 1 cm such that OT = 4 cm and ∠OTA = 30° . Then, AT is 2. The length of tangent from an external point P on a equal to [NCERT Exemplar] circle with centre O is always less than OP. [NCERT Exemplar] (a) True (b) False O (c) Can’t determined (d) None of these 4 cm 3. The length of the tangents to the circle from a point 30° AT at any distance of 5 cm from centre of the circle of radius 3 cm is (a) 4 cm (b) 2 cm (c) 2 3 cm (d) 4 3 cm (a) 2 cm (b) 4 cm (c) 8 cm (d) None of these 9. PQ is a tangent drawn from a point P to a circle 4. The length of the tangent drawn from a point 8 cm with centre O and QOR is a diameter of the circle away from the centre of circle of radius 6 cm is such that ∠POR = 135°, then ∠OPQ is (a) 7 cm (b) 2 7 cm (c) 10 cm (d) 5 cm (a) 60° (b) 45° (c) 30° (d) 90° 5. PQ is a tangent to a circle with centre O at the 10. A tangent PQ at a point P of a circle of radius 6 cm point P. If ΔOPQ is an isosceles triangle, then meets a line through the centre O at a point Q, so ∠OPQ is equal to that OQ = 14 cm, then length of PQ is (a) 30° (b) 45° (c) 60° (d) 90° (a) 4 10 cm (b) 6 10 cm 6. In figure, if O is the centre of a circle, PQ is a chord (c) 5 10 cm (d) 7 10 cm and the tangent PR at P makes an angle of 50° with 11. In figure, if PA and PB are tangents to the circle PQ, then ∠POQ is equal to [NCERT Exemplar] with centre O such that ∠ APB = 50°, then ∠OAB is PR equal to 50° [NCERT Exemplar] A O P 50° O Q (a) 100° (b) 80° (c) 90° (d) 75° 7. In the given figure, PA is a tangent from an external (a) 25° B (c) 40° point P to a circle with centre O. If ∠POB = 115°, (b) 30° then ∠APO is (d) 50°

CBSE Term II Mathematics X (Standard) 57 12. If angle between two tangents drawn from a point P (a) 41 cm (b) 41 cm (c) 40 cm (d) 40 cm to a circle of radius a and centre O is 90°, then OP = a 2. [NCERT Exemplar] 18. PA is a tangent to the circle with centre O. (a) True (b) False If BC = 3 cm, AC = 4 cm and ΔACB ~ΔPAO, then (c) Can’t say (d) Partially true or false OA is equal to [CBSE 2013] 13. In the given figure, find the value of x°. B A C O x° 30° B O AP (a) 130° (b) 75° (c) 120° (d) 60° (a) 2.7 cm (b) 5 cm (c) 5 cm (d) 5 cm 14. From the given figure, find the value of x° + y°. 2 P 19. In the given figure, if ∠ACB = 50°, then ∠ATO is A x° y° O QR CO T (a) 270° (b) 180° B (c) 90° (d) None of these (a) 30° 15. At one end A of a diameter AB of a circle of radius (b) 50° (c) 40° 5 cm, tangent XAY is drawn to the circle. The (d) Can’t be determined length of the chord CD parallel to XY and at a 20. In the adjoining figure, PQ is a chord of a circle with centre O and PT is a tangent at P such that distance 8 cm from A is [NCERT Exemplar] ∠QPT = 60°, then ∠PRQ = (a) 4 cm (b) 5 cm (c) 6 cm (d) 8 cm 16. In figure, AB is a chord of the circle and AOC is its Q diameter such that ∠ ACB = 50°. If AT is the tangent O to the circle at the point A, then ∠BAT is equal to [NCERT Exemplar] C R OB 60° AP T (a) 120° (b) 160° (c) 130° (d) 150° AT 21. In figure, if PQR is the tangent to a circle at Q, (a) 45° (b) 60° (c) 50° (d) 55° whose centre is O, AB is a chord parallel to PR and 17. In the adjoining figure, AD = 8 cm, AC = 6 cm and ∠BQR = 70°, then ∠AQB is equal to TB is the tangent at B to the circle with centre O. [NCERT Exemplar] If BT is 4 cm, then OT = [CBSE 2013] A DB A O CD P 70° R O (b) 40° Q B 4 cm T (a) 20° (c) 35° (d) 45°

58 CBSE Term II Mathematics X (Standard) 22. From an external point P, tangents PA and PB are 28. In figure, if ∠AOB = 125°, then ∠COD is equal to drawn to a circle with centre O. If CD is the [NCERT Exemplar] A tangent to the circle at a point E and PA = 14 cm, B then perimeter of ΔPCD is 125° (a) 14 cm (b) 21 cm (c) 28 cm (d) 35 cm O 23. Tangents AP and AQ are drawn to circle with centre O from an external point A, then ∠PAQ is equal to (a) 2∠OPQ (b) ∠OPQ (c) ∠OPQ (d) ∠OPQ D C 2 3 4 (b) 45° 24. In the given figure, two tangents AB andAC are (a) 62.5° (d) 55° drawn to a circle with centre O such that (c) 35° ∠BAC = 120°, then OA is equal to G Case Based MCQs O 29. A playground is in the shape of a triangle with right angle at B, AB = 3 m and BC = 4 m. A pit was dig (a) 2AB B C (d) 5AB inside it such that it touches the walls AC, BC and (b) 3AB 120° AB at P, Q and R, respectively such that AP = x m. A A (c) 4AB P 25. In the given figure, O is the centre of a circle, BOA r R is its diameter and the tangent at the point P meets O BA extended at T. If ∠PBO = 30°, then ∠PTA = [CBSE 2016] P B 30° T BQ C O A (a) 40° (b) 50° (c) 30° (d) 20° Based on the above information, answer the following questions. 26. In adjoining figure, PQ and PR are tangents to the circle with centre O and S is a point on the circle (i) The value of AR = such that ∠SQL = 50° and ∠SRM = 60°. Then, (a) 2x m (b) x / 2 m ∠QSR [NCERT Exemplar] (c) x m (d) 3x m L (ii) The value of BQ = (a) 2x m 50° Q (b) (3 − x) m SO P (c) (2 − x) m (d) 4x m (iii) The value of CQ = 60° R (a) (4 + x) m (b) (5 − x) m M (c) (1 + x) m (d) Both (b) and (c) (a) 40° (b) 50° (c) 60° (d) 70° (iv) Which of the following is correct? 27. In given figure, AB is diameter of a circle with (a) Quadrilateral AROP is a square centre O and AT is tangent. If ∠AOQ = 58° , then (b) Quadrilateral BROQ is a square ∠ATQ = [CBSE 2015] (c) Quadrilateral CQOP is a square B (d) None of the above (v) Radius of the pit is O (a) 1 m (b) 3 m 58° Q (c) 4 m (d) 5 m (a) 52° (b) 58° AT (d) 62° 30. A student draws two circles that touch each other (c) 61° externally at point K with centres A and B and radii 6 cm and 4 cm, respectively as shown in the figure.

CBSE Term II Mathematics X (Standard) 59 PX T 3 cm 5. If a number of circles touch a given line segment 6 cm A YQ PQ at a point A, then their centres lie on the K perpendicular bisector of PQ. Why or why not? 4 cm B [NCERT Exemplar] 8 cm 6. Out of the two concentric circles, the radius of S the outer circle is 5 cm and the chord AC of length 8 cm is a tangent to the inner circle. Find Based on the above information, answer the following the radius of the inner circle. questions. (i) The value of PA = 7. If a chord AB subtends an angle of 60° at the (a) 10 cm (b) 5 cm centre of a circle, then find the angle between the (c) 13 cm (d) Can’t be determined tangents at A and B. [NCERT Exemplar] (ii) The value of BQ = 8. From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one (a) 4 cm (b) 5 cm point E on the circle tangent is drawn, which (c) 6 cm (d) 18 cm intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the trianlge (iii) The value of PK = PCD. (a) 13 cm (b) 15 cm (c) 16 cm (d) 18 cm (iv) The value of QY = 9. Prove that the centre of a circle touching two (a) 2 cm (b) 5 cm intersecting lines lies on the angle bisector of the (c) 1 cm (d) 3 cm lines. [NCERT Exemplar] (v) If two circles touch externally, then the number of 10. If from an external point B of a circle with centre common tangents can be drawn is O, two tangents BC and BD are drawn, such that (a) 1 (b) 2 ∠DBC = 120°, prove that BC + BD = BO i.e. (c) 3 (d) None of these BO = 2 BC. PART 2 11. In figure, AB and CD are common tangents to Subjective Questions two circles of equal radii. Prove that AB = CD. G Short Answer Type Questions [NCERT Exemplar] 1. If PQ is a tangent to a circle with centre O and radius AB 6 cm such that ∠PQO = 60°, then find the length of a tangent PQ and a line OQ. CD 2. The tangent to the circumcircle of an isosceles ΔABC 12. In figure, common tangents AB and CD to two at A, in which AB = AC, is parallel to BC. circles intersect at E. Prove that AB = CD. [NCERT Exemplar] [NCERT Exemplar] A 3. If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in D figure. Prove that ∠BAT = ∠ ACB. [NCERT Exemplar] E C CB OB 13. If PA and PB are two tangents drawn from a point P to a circle with centre O touching it at A and B, prove that OP is perpendicular bisector of AB. AT [CBSE 2008] 4. Prove that a diameter AB of a circle bisects all 14. Tangents AP and AQ are drawn to circle with those chords, which are parallel to the tangent at centre O from an external point A. Prove that the point A. [NCERT Exemplar] ∠PAQ = 2∠OPQ. [CBSE 2013, 12,11, 09]

60 CBSE Term II Mathematics X (Standard) 15. In the given figure, ∠ADC = 90°, BC = 38 cm, 21. If a hexagon ABCDEF circumscribe a circle, prove that CD = 28 cm and BP = 25 cm, then find the radius of AB + CD + EF = BC + DE + FA [NCERT Exemplar] the circle. [CBSE 2011] 22. In the given figure, AD is a diameter of a circle with B centre O and AB is a tangent at A. C is a point on the circle such that DC produced intersects the Q tangent at B and ∠ABD = 50°. Find ∠CO[AC.BSE 2015] C D RO P DA OC S 16. ΔABC is a right angled triangle with ∠B = 90°, 50° BC = 3 cm and AB = 4 cm. A circle with centre O B and radius r cm has been inscribed in ΔABC. Find the radius of the incircle. A 17. The radii of two concentric circles are 13 cm and 23. Tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn parallel to the 8 cm. AB is a diameter of the bigger circle. BD is a tangent PQ. Find ∠RQS. [CBSE 2015] tangent to the smaller circle touching it at D. 24. PA and PB are the tangents to a circle, which Find the length of AD. [CBSE 2010] circumscribes an equilateral ΔABQ. If ∠PAB = 60°, 18. A circle is inscribed in a ΔABC having sides as shown in the figure, prove that QP bisects AB at AB = 8 cm, BC = 10 cm and CA = 12 cm, as shown in right angle. [CBSE 2015] figure. Find AD, BE and CF. [CBSE 2012] A A 60° Q P M FD C B B E 25. Two circles with centres O and Oʹ of radii 3 cm and G Long Answer Type Questions 4 cm, respectively intersect at two points P and Q, such that OP and Oʹ P are tangents to the two 19. Let s denotes the semi-perimeter of a ΔABC, in circles. Find the length of the common chord PQ. which BC = a, CA = b and AB = c. If a circle touches [NCERT Exemplar] the sides BC, CA, AB at D, E, F, respectively. Prove 26. If an isosceles ΔABC in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find the area of that BD = s − b. [NCERT Exemplar] the triangle. 20. AC and AD are tangents at C and D, respectively. 27. In a figure, the common tangents AB and CD of If ∠BCD = 44° , then find∠CAD, ∠ADC, ∠CBD two circles with centres O and Oʹ intersect at E. and ∠ACD. Prove that the points O, E and Oʹ are collinear. C [NCERT Exemplar] A A 44° D O O E Oʹ D CB B

CBSE Term II Mathematics X (Standard) 61 28. In figure, O is the centre of a circle of radius 5 cm, (ii) A circle of radius 3 cm is inscribed in a right angled ΔBAC such that BD = 9 cm and DC = 3 cm. T is a point such that OT = 13 and OT intersects the Find the length of AB. circle at E, if AB is the tangent to the circle at E, A find the length of AB. [NCERT Exemplar] P A OE T F RO E B B DC Q G Case Based Questions (a) 6 cm (b) 12 cm (c) 15 cm (d) 10 cm 29. Dheeraj loves geometry. So, he was curious to know more about the concepts of circles. His grand father (iii) In the given figure, what is the length of CD? is a mathematicians. So, he reached to his grand father to learn something interesting about tangents A 4 cm 2 cm B and circles. His grand father gave him knowledge on circles and tangents and ask him to solve the P following questions. 10 cm S 5 cm Q R C D (a) 11 cm (b) 9 cm (c) 7 cm (d) 13 cm (iv) If PA and PB are two tangents to a circle with centre O from an external point P such that ∠OPB = 50°, then find ∠BPA (a) 60° (b) 50° (c) 120° (d) 100° (i) In the given figure, AP, AQ and BC are tangents to (v) In the given figure, P is an external point from, the circle such that AB = 7 cm, BC = 4 cm and which tangents are drawn to two externally AC = 9 cm. Find AP touching circles. If PA = 11 cm, then find PC. A P B D A P C (a) 3.5 cm Q (c) 11 cm BC (a) 12 cm (b) 15 cm (c) 13 cm (d) 10 cm (b) 4 cm (d) Can’t be determined

62 CBSE Term II Mathematics X (Standard) SOLUTIONS Objective Questions 4. (b) Since, tangent to a circle is perpendicular to the radius through the point of contact. 1. (b) Let O be the centre of two concentric circles C1 and C2, ∴ ∠OTP = 90° whose radii are r1 = 4 cm and r2 = 5 cm . Now, we draw a chord AC of circle C2, which touches the circle C1 at B. T Also, join OB, which is perpendicular to AC. 6 cm O [Q tangent at any point of circle is perpendicular to radius through the point of contact] C2 P 8 cm C1 In ΔOTP, we have O OP2 = OT 2 + PT 2 AC ⇒ (8)2 = (6)2 + PT 2 B ⇒ PT 2 = 64 − 36 = 28 ⇒ PT = 28 = 2 7 cm Now, in right angled ΔOBC, by using Pythagoras theorem, 5. (b) Since, PQ is a tangent to a circle from a point P and OC 2 = BC 2 + BO 2 centre of circle is O. ∴ ΔOPQ is an isosceles triangle. [Q(hypotenuse) 2 = (base) 2 + (perpendicular) 2] Q ⇒ 5 2 = BC 2 + 4 2 ⇒ BC2 = 25 − 16 = 9 PO ⇒ BC = 3 cm ∴ Length of chord AC = 2 BC 2. (a) = 2 × 3 = 6 cm O ∠OQP = 90° P OP = QP ∴ ∠POQ = ∠OPQ In ΔOPQ, T ∠POQ + ∠OQP + ∠OPQ = 180° PT is a tangent drawn from external point P. Join OT. ⇒ 2∠OPQ = 180° − 90° [Q ∠POQ = ∠OPQ] Q OT ⊥ PT So, ΔOPT is a right angled triangle formed. ⇒ 2∠OPQ = 90° ⇒ ∠OPQ = 45° In right angled triangle, hypotenuse is always greater than any of the two sides of the triangle. 6. (a) Given, ∠ QPR = 50° ∴ OP > PT or PT < OP We know that, the tangent at any point of a circle is 3. (b) Given, OB = 5 cm and radius OA = 3 cm perpendicular to the radius through the point of contact. By Pythagoras theorem, in right angled ΔOAB, ∴ ∠OPR = 90° ⇒ ∠OPQ + ∠QPR = 90° [from figure] ⇒ ∠OPQ = 90° − 50° = 40° [Q ∠QPR = 50°] Now, OP = OQ = Radius of circle ∴ ∠OQP = ∠ OPQ = 40° O [since, angles opposite to equal sides are equal] 53 In Δ OPQ, ∠ O + ∠P + ∠ Q = 180° B [since, sum of all angles of a triangle = 180°] A ⇒ ∠ O = 180° − (40° + 40° ) [Q ∠P = 40° = ∠Q] OB2 = OA 2 + AB2 AB2 = (5)2 − (3)2 = 180° − 80° = 100° AB2 = 25 − 9 AB2 = 16 [Q OB = 5, AB = 3] 7. (a) Here, ∠OAP = 90° [Q tangent at any point of a circle is AB = 4 cm perpendicular to the radius] Now, ∠AOP + ∠BOP = 180° ⇒ ∠AOP + 115° = 180° ⇒ ∠AOP = (180° − 115°) = 65°

CBSE Term II Mathematics X (Standard) 63 And also, ∠OAP + ∠AOP + ∠APO = 180° ⇒ PQ2 = 160 [angle sum property of triangle] ⇒ PQ = 16 × 10 = 4 10 cm ⇒ 90°+ 65°+∠APO = 180° ⇒ 155° + ∠APO = 180° 11. (a) Given, PA and PB are tangent lines. ⇒ ∠APO = 180°−155° = 25° 8. (c) Join OA. ∴ PA = PB [since, the length of tangents drawn from an external point to a circle is equal] ⇒ ∠PBA = ∠PAB = θ [say] O In ΔPAB, ∠P + ∠ A + ∠B = 180° [since, sum of all angles of a triangle = 180°] 4 cm ⇒ 50° + θ + θ = 180° 30° AT ⇒ 2θ = 180° − 50° = 130° ⇒ θ = 65° We know that, the tangent at any point of a circle is Also, OA ⊥ PA perpendicular to the radius through the point of contact. [since, tangent at any point of a circle is perpendicular ∴ ∠OAT = 90° to the radius through the point of contact] In Δ OAT, cos 30° = AT ∴ ∠PAO = 90° OT ⇒ ∠PAB + ∠BAO = 90° ⇒ 3 = AT ⇒ 65° + ∠BAO = 90° 24 ⇒ ∠BAO = 90° − 65° = 25° ⇒ AT = 2 3 cm 12. (a) From point P, two tangents are drawn. 9. (b) Given, PQ is a tangent from point P, centre O and QOR Given, OT = a as diameter. Also, line OP bisects the ∠RPT. R ∴ ∠ TPO = ∠RPO = 45° Also, OT ⊥ PT 135° T O 45° a P 90° O QP In ΔPQO, ∠ROP = ∠OPQ + 90° R [Q exterior angle of a triangle is equal to the sum In right angled Δ OTP, of opposite angles] sin 45° = OT OP ∠OPQ = 135° − 90° = 45° ⇒ 1=a 2 OP 10. (a) Here, OP = 6 cm and OQ = 14 cm ⇒ OP = a 2 O 13. (c) Given, ∠OBA = 30° 6 cm 14 cm P Q In ΔABO, x° = ∠ABO + 90° We know that, tangent at any point of a circle is [Q external angle = sum of opposite internal angles] perpendicular to the radius through the point of contact. x° = 30° + 90° = 120° So, OP ⊥ PQ 14. (a) In ΔPOQ, Now, in right angled ΔOPQ, x° = ∠PQO + 90° OQ2 = OP2 + PQ2 [by Pythagoras theorem] [Q external angle = sum of opposite internal angles] = (180° − y°) + 90° ⇒ (14)2 = (6)2 + PQ2 = 270° − y° ⇒ PQ2 = 196 − 36 x° + y° = 270°

64 CBSE Term II Mathematics X (Standard) 15. (d) First, draw a circle of radius 5 cm having centre O. 18. (d) In ΔACB, A tangent XY is drawn at point A. ∠BCA = 90° [angle in a semi-circle] C ∴ AB2 = AC2 + BC2 [by Pythagoras theorem] B 5 cm ⇒ AB2 = 42 + 32 XE ⇒ AB2 = 16 + 9 cm 5 cm O 3 cm D ⇒ AB2 = 25 cm A 8 cm ⇒ AB = 5 cm ⇒ OA = 5 cm 2 Y 19. (c) ∠OAT = 90° [Q angle between radius and tangent] A chord CD is drawn, which is parallel to XY and at a Now, ∠BOA = 100° [angle subtended by an arc at distance of 8 cm from A. centre is twice the angle subtended at remaining part of circle] Also, AE = 8 cm. Join OC ⇒ ∠ATO = 180° − (∠TOA + ∠OAT) Now, in right angled ΔOEC, [angles property of a triangle] OC 2 = OE 2 + EC 2 = 180° − (50° + 90°) ⇒ EC 2 = OC 2 − OE 2 = 180° − 140° = 40° [by Pythagoras theorem] 20. (a) Take a point Qʹ on circle and join PQʹ and QQʹ. = 52 − 32 [Q OC = radius = 5 cm, OE = AE − AO = 8 − 5 = 3 cm] Qʹ Q = 25 − 9 = 16 O ⇒ EC = 4 cm 60º R Hence, length of chord CD = 2 CE AP T = 2 × 4 = 8 cm [since, perpendicular from centre to the Now, ∠OPQ = 90°−60° = 30° [Q ∠OPT = 90°] chord bisects the chord] 16. (c) In figure, AOC is a diameter of the circle. We know that, ⇒ ∠OQP = 30° [angles opposite to equal diameter subtends an angle 90° at the circle. sides are equal] So, ∠ ABC = 90° In ΔACB, ∠A + ∠B + ∠C = 180° ⇒ ∠POQ = 120° [since, sum of all angles of a triangle is 180°] [angle sum property of a triangle] ⇒ ∠PQʹ Q = 60° ⇒ ∠ A + 90° + 50° = 180° [angle subtended by an arc at centre is twice the angle subtended at remaining part of circle] ⇒ ∠A + 140 = 180 ⇒ ∠PRQ = 120° [Q opposite angles ⇒ ∠ A = 180° − 140° = 40° are ∠A or ∠OAB = 40° supplementary in a cyclic quadrilateral PQQʹR] Now, AT is the tangent to the circle at point A. So, OA is 21. (b) Given, AB ||PR perpendicular to AT. A DB ∴ ∠ OAT = 90° [from figure] ⇒ ∠ OAB + ∠BAT = 90° On putting ∠OAB = 40°, we get O ⇒ ∠BAT = 90° − 40° = 50° Hence, the value of ∠BAT is 50°. 70° P QR 17. (b) Clearly, ∠CAD = 90° [angle in a semi-circle] So, in ΔACD, CD 2 = AC2 + AD 2 = 36 + 64 = 100 ∴ ∠ ABQ = ∠BQR = 70° [alternate angles] [by Pythagoras theorem] Also, QD is perpendicular to AB and QD bisects AB. ⇒ CD = 10 cm In ΔQDA and ΔQDB, Therefore, OC = OD = OB = 5 cm [Q radius of a circle] Since, ∠OBT = 90° ∠QDA = ∠QDB [each 90°] [angle between radius and tangent] AD = BD So, in ΔOBT, OT 2 = OB2 + BT 2 QD = QD [common side] = 25 + 16 = 41 [by Pythagoras theorem] ∴ Δ ADQ ~ ΔBDQ ⇒ OT = 41 cm [by SAS similarity criterion]

CBSE Term II Mathematics X (Standard) 65 Then, ∠QAD = ∠ QBD [by CPCT] ...(i) ⇒ 1 = AB 2 OA Also, ∠ABQ = ∠BQR [alternate interior angle] ∴ ∠ABQ = 70° [Q ∠BQR = 70°] ⇒ OA = 2AB Hence, ∠QAB = 70° [from Eq. (i)] 25. (c) ∠OPB = 30° Now, in Δ ABQ, ∠ A + ∠B + ∠ Q = 180° [Q angles opposite to equal sides are equal] ⇒ ∠ Q = 180° − (70° + 70° ) = 40° and ∠OPT = 90° 22. (c) We have, PA = PB = 14 cm Now, ∠PTA = 180°−(∠OBP + ∠BPT ) A [angle sum property of a ΔBPT] C = 180° − (30° + 120°) 14 cm P EO [Q ∠BPT = 90° + 30° = 120°] = 180° − 150° D = 30° B 26. (d) ∠OQS = ∠OQL − ∠SQL [since, OQ ⊥ LP] = 90°−50° = 40° Also, CD is tangent at point E on the circle. Similarly, ∠ORS = 30° So, CA and CE are tangents to the circle from point C. Therefore, CA = CE, similarly DB = DE Now, ∠QSR = ∠OSR + ∠OSQ = ∠ORS + ∠OQS Now, perimeter of ΔPCD [Q angle opposite to equal sides are equal] = PC + CD + PD = PC + CE + ED + PD = PC + CA + PD + DB = 30° + 40° [Q CA = CE and DE = DB] = 70° = PA + PB = 14 + 14 = 28 cm 27. (c) ∠ABQ = 1 ∠AOQ = 1 ( 58° ) = 29° 23. (a) Here, AP = AQ 2 2 ⇒ ∠AQP = ∠APQ = x (say) [angle subtended by an arc at the centre is twice the angle [Q angles opposite to equal sides of a triangle are equal] subtended at remaining part of circle] P and ∠BAT = 90° [angle between radius and tangent] In ΔABT, we get, ∠ATQ = 180° − (29°+90°) = 61° [angle sum property of a triangle] OA 28. (d) We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Q i.e. ∠AOB + ∠COD = 180° In ΔAPQ, ∠PAQ = 180° − (∠APQ + ∠AQP) ⇒ ∠ COD = 180° − ∠AOB = 180° − (x + x) = 180° − 2x = 180° − 125° = 55° Q OP ⊥ AP ∴ ∠OPA = 90° 29. Here, in right angled ΔABC, AB = 3 m and BC = 4 m. ⇒ ∠OPQ + ∠APQ = 90° ∴By Pythagoras theorem, AC = (AB)2 + (BC)2 ⇒ ∠OPQ + x = 90° = (3)2 + (4)2 ⇒ ∠OPQ = 90° − x = 9 + 16 = 25 = 5 m ⇒ 2∠OPQ = 180°−2x ⇒ ∠PAQ = 2∠OPQ [multiplying by 2] Also, AP = x m (i) (c) AR = AP = x m … (i) 24. (a) In ΔOAB and ΔOAC, we have [since, length of tangents drawn from an ∠OBA = ∠OCA = 90° external point are equal] OA = OA [common] (ii) (b) BQ = BR = AB − AR = (3 − x) m [using Eq. (i)] and OB = OC [radii of circle] (iii) (d) CQ = CP = AC − AP = (5 − x) m So, by RHS congruence criterion, Also, CQ = BC − BQ = BC −BR ΔOBA ≅ ΔOCA = 4 − (3 − x) = 1 + x ⇒ ∠OAB = ∠OAC (iv) (b) Since, CQ = 5 − x = 1 + x = 1 × 120° = 60° 2 ⇒ 4 = 2x ⇒x = 2 In ΔOBA, we have ∴ AR = AP = 2 m, BR = BQ = 1 m cos 60° = AB OA and CP = CQ = 3 m

66 CBSE Term II Mathematics X (Standard) A ⇒ 3= 6 ⎡ sin 60° = 3⎤ 2 OQ ⎢Q ⎣ 2 ⎥ ⎦ P ⇒ OQ = 2 × 6 × 3 [rationalising] r 33 RO ⇒ OQ = 4 3 cm Hence, length of a tangent PQ is 2 3 cm and a line OQ BQ C is 4 3 cm. Also, OQ ⊥ BQ and OR ⊥ BR 2. Let EAF be tangent to the circumcircle of ΔABC. A ∴ BROQ is a square. EF (v) (a) Radius of the pit, OR = BR = 1 m 30. Here, AS = 6 cm, BT = 4 cm [Q radii of circles] (i) (c) Since, radius at point of contact is perpendicular to BC tangent. ∴ By Pythagoras theorem, we have PA = PS2 + AS2 To prove EAF ||BC = 82 + 62 ∠EAB = ∠ABC = 64 + 36 Here, AB = AC = 100 = 10 cm (ii) (b) Again, by Pythagoras theorem, we have ⇒ ∠ACB = ∠ABC …(i) BQ = TQ2 + BT 2 = 32 + 42 [angle between tangent and its chord equal to angle made by chord in the alternate segment] ∴ Also, ∠EAB = ∠BCA …(ii) = 9 + 16 = 25 = 5 cm From Eqs. (i) and (ii), we get (iii) (c) PK = PA + AK = 10 + 6 = 16 cm ∠EAB = ∠ ABC (iv) (c) QY = BQ − BY = 5 − 4 = 1 cm ⇒ EAF ||BC (v) (b) If two circles touch externally, then the number of common tangents can be drawn is 2. 3. Since, AC is a diameter line, so angle in semi-circle makes an angle 90°. Subjective Questions ∴ ∠ ABC = 90° 1. Given, PQ is a tangent, OP = 6 cm and ∠PQO = 60° In Δ ABC, ∠ CAB + ∠ ABC + ∠ ACB = 180° We know that, tangent at any point of a circle is perpendicular to the radius through the point of contact. [Q sum of all interior angles of any triangle is 180°] ∴ OP ⊥ PQ ⇒ ∠ CAB + ∠ ACB = 180° − 90° = 90° …(i) Since, diameter of a circle is perpendicular to the tangent. i.e. CA ⊥ AT ∴ ∠ CAT = 90° ⇒ ∠ CAB + ∠BAT = 90° …(ii) O From Eqs. (i) and (ii), we get ∠ CAB + ∠ ACB = ∠ CAB + ∠BAT 60° ⇒ ∠ ACB = ∠BAT Hence proved. Q 4. Given, AB is a diameter of the circle. P A tangent is drawn from point A. Draw a chord CD parallel to the tangent MAN. Now, in right angled ΔOPQ, ⎢⎡⎣Q tan θ = perpendicular ⎤ MC tan 60° = OP base ⎥⎦ PQ [Q tan 60° = 3] A O B ⇒ 3= 6 E PQ ⇒ PQ = 6 × 3 [rationalising] 33 ND ⇒ PQ = 2 3 cm and sin 60° = OP OQ So, CD is a chord of the circle and OA is a radius of the circle. ∠MAO = 90° ⎡ sin θ = perpendicular ⎤ ⎢⎣Q hypotenuse ⎥⎦ [tangent at any point of a circle is perpendicular to the radius through the point of contact]

CBSE Term II Mathematics X (Standard) 67 ∠ CEO = ∠MAO [corresponding angles] ⇒ DO2 = 52 − 42 = 25 − 16 = 9 ⇒ DO = 3 cm ∴ ∠CEO = 90° ∴ Radius of the inner circle OD = 3 cm 7. Since, a chord AB subtends an angle of 60° at the centre of a Thus, OE bisects CD, [perpendicular from centre of circle to circle. the chord bisects the chord] O Similarly, the diameter AB bisects all chords, which are parallel to the tangent at the point A. 60° AB 5. Given that, PQ is any line segment and S1, S2, S3, S4, ... circles are touch a line segment PQ at a point A. Let the centres of the circles S1, S2, S3, S4, ... be C1, C2, C3, C4,... respectively. C4 C3 S3S4 C S2 C2 S1 i.e. ∠ AOB = 60° As, OA = OB = Radius of the circle C1 ∴ ∠ OAB = ∠ OBA = 60° The tangent at points A and B is drawn, which intersects P Q at C. A We know, OA ⊥ AC and OB ⊥ BC. ∴ ∠ OAC = 90° and ∠OBC = 90° To prove Centres of these circles lie on the perpendicular ⇒ ∠ OAB + ∠BAC = 90° bisector of PQ. and ∠ OBA + ∠ABC = 90° ⇒ ∠BAC = 90° − 60° = 30° Now, joining each centre of the circles to the point A on the and ∠ ABC = 90° − 60° = 30° line segment PQ by a line segment, i.e. In Δ ABC, ∠BAC + ∠ CBA + ∠ ACB = 180° C1A, C2A, C3A, C4A,... so on. We know that, if we draw a line from the centre of a circle to [since, sum of all interior angles of a triangle is 180°] its tangent line, then the line is always perpendicular to the ⇒ ∠ ACB = 180° − (30° + 30°) = 120° tangent line. But it not bisect the line segment PQ. 8. Two tangents PA and PB are drawn to a circle with centre O from an external point P. So, C1A ⊥ PQ [for S1] C2A ⊥ PQ [for S2] AC C3A ⊥ PQ C4A ⊥ PQ [for S3] ... so on. [for S4] Since, each circle is passing through a point A. Therefore, all E P the line segments C1A, C2A, C3A, C4A,..., so on are coincident. O So, centre of each circle lies on the perpendicular line of PQ but they do not lie on the perpendicular bisector of PQ. Hence, a number of circles touch a given line segment PQ at D a point A, then their centres lie. B 6. Let C1 and C2 be the two circles having same centre O. AC is Perimeter of ΔPCD = PC + CD + PD a chord which touches the C1 at point D. = PC + CE + ED + PD C2 = PC + CA + DB + PD [Q CE = CA, DE = DB] C1 = PA + PB = 2PA = 2(10) [PA = PB tangents from O external point to a circle are equal] = 20 cm A DC Join OD. 9. Given Two tangents PQ and PR are drawn from an external point P to a circle with centre O. Also, OD ⊥ AC R ∴ AD = DC = 4 cm [perpendicular line OD bisects the chord] In right angled ΔAOD, OP OA2 = AD 2 + DO2 [by Pythagoras theorem, Q i.e. (hypotenuse)2 =(base)2 +(perpendicular)2]

68 CBSE Term II Mathematics X (Standard) To prove Centre of a circle touching two intersecting lines Also, ∠ OAB + ∠ OCD = 180° lies on the angle bisector of the lines. ∴ AB||CD In ∠RPQ. Similarly, BD is a straight line. Construction Join OR and OQ. and ∠ OʹBA = ∠ OʹDC = 90° In ΔPOR and ΔPOQ, Also, AC = BD [radii of two circles are equal] ∠PRO = ∠PQO = 90° In quadrilateral ABCD, [tangent at any point of a circle is perpendicular ∠ A = ∠B = ∠ C = ∠D = 90° to the radius through the point of contact] and AC = BD OR = OQ [radii of same circle] ABCD is a rectangle Since, OP is common. Hence, AB = CD ∴ ΔPRO ≅ ΔPQO [by RHS] [opposite sides of rectangle are equal] Hence, ∠RPO = ∠ QPO [by CPCT] 12. Given Common tangents AB and CD of two circles Thus, O lies on angle bisecter of PR and PQ. Hence proved. intersecting at E. 10. Two tangents BD and BC are drawn from an external point B. To prove AB = CD A C D B 120° O E D CB To prove BO = 2BC Proof EA = EC ...(i) Given, ∠DBC = 120° [the lengths of tangents drawn from an external point to a circle are equal] Join OC, OD and BO. Since, BC and BD are tangents. EB = ED ...(ii) ∴ OC ⊥ BC and OD ⊥ BD On adding Eqs. (i) and (ii), we get We know, OB is a angle bisector of ∠DBC. EA +EB = EC + ED ∴ ∠ OBC = ∠DBO = 60° ⇒ AB = CD Hence proved. In right angled Δ OBC, 13. Let OP intersect AB at a point C. cos 60° = BC Clearly, ∠APO = ∠BPO …(i) OB [Q O lies on bisector of ∠APB] 1 BC ⇒= A 2 OB ⇒ OB = 2 BC Also, BC = BD P CO [tangents drawn from external point to circle are equal] ∴ OB = BC + BC B ⇒ OB = BC + BD 11. Given AB and CD are tangents to two circles of equal radii. To prove AB = CD Now, in ΔACP and ΔBCP, A B AP = BP C1 O Oʹ C2 [Q length of tangents drawn from an external point to a circle are equal] PC = PC [common sides] and ∠APO = ∠BPO [from Eq. (i)] CD ∴ ΔACP ≅ ΔBCP [by SAS congruence rule] Construction Join Oʹ A, Oʹ C,OB and OD Then, AC = BC [by CPCT] Proof Now, ∠ OAB = 90° and ∠ACP = ∠BCP [by CPCT] [tangent at any point of a circle is perpendicular to radius = 1 × 180° = 90° through the point of contact] 2 Thus, AC is a straight line. [Q AB is a straight line] Hence, OP is perpendicular bisector of AB. Hence proved.

CBSE Term II Mathematics X (Standard) 69 14. P and ar (ΔOAC ) = 1 × OF × AC = 1 × r × 5 = 5r cm2 2 22 ∴ ar (ΔABC ) = ar (ΔOAB ) + ar (ΔOBC ) + ar (ΔOAC ) AO ⇒ 1 AB × BC = 4r + 3r + 5r ⇒ 1 × 3 × 4 = 12r 2 222 2 2 ⇒ r = 1 cm Q 17. Produce BD to meet the bigger circle at E. Join AE. AP = AQ Then, ∠AEB = 90° [Q angle in semi-circle] ⇒ ∠APQ = ∠AQP = x [say] A O [Q angles opposite to equal sides are equal] In ΔAPQ, ∠PAQ = 180° − (∠APQ + ∠AQP) [angle sum property of a triangle] BDE = 180° − (x + x) = 180° − 2x Q OP ⊥ AP Clearly, OD ⊥ BE [QBE is tangent to the smaller ∴ [Q radius is perpendicular to the tangent circle at D and OD is its radius] at the point of contact] BD = DE ∴ ∠OPA = 90° [QBE is a chord of the bigger circle and OD ⊥ BE] ⇒ ∠OPQ + ∠APQ = 90° Now, in ΔAEB, O and D are the mid-points of AB and BE, ⇒ ∠OPQ + x = 90° respectively. ⇒ ∠OPQ = 90° − x Therefore, by mid-point theorem, we have ∴ ∠PAQ = 2∠OPQ Hence proved. OD = 1 AE ⇒ AE =2 × OD =2 × 8 = 16 cm 2 15. CR = CQ = BC − BQ [Q OD = radius of smaller circle = 8 cm] = 38 − 25 = 13 cm ∴ RD = CD − CR = 28 − 13 = 15 cm In right angled ΔODB, [by Pythagoras theorem] Here, OR ⊥RD and OS ⊥ DA. OB2 = OD 2 + BD 2 [Q tangent is perpendicular to the radius ⇒ BD 2 = 169 − 64 = 105 through the point of contact] ⇒ BD = 105 cm = DE [QBD = DE] Also, ∠ADC = 90°, then fourth angle in quadrilateral ORDS will be 90°. Thus, ORDS will be a rectangle. Now, in right angled ΔAED, [by Pythagoras theorem] AD 2 = AE2 + ED 2 Q D is an external point of a circle. ⇒ AD = (16)2 + ( 105)2 ∴ DR = DS Also, opposites sides of rectangle are equal. = 256 + 105 = 361 = 19 cm ∴ RD = OR = OS = SD 18. We know that, tangents drawn from an exterior point to a Hence, quadrilateral DROS is a square. circle are equal in length. ∴ Radius = OR = RD = 15 cm ∴ AD = AF = x cm [say] 16. Let D, E and F are the points, where the incircle touches the BD = BE = y cm [say] sides AB, BC and CA, respectively. Join OA, OB and OC. CE = CF = z cm [say] A Given, AB = 8 cm ⇒ AD + BD = 8 cm ⇒ x+y=8 ...(i) F BC = 10 cm r r ⇒ BE + CE = 10 cm D O ⇒ y + z = 10 ...(ii) r and CA = 12 cm B C ⇒ CF + AF = 12 cm E In Δ ABC, AC2 = AB2 + BC2 [by Pythagoras theorem] ⇒ z + x = 12 ...(iii) = 42 + 32 = 16 + 9 = 25 On adding Eqs. (i), (ii) and (iii), we get ∴ AC = 5 cm 2(x + y + z) = 30 [taking positive square root, as length cannot be negative] ⇒ x + y + z = 15 ...(iv) Now, ar (ΔOAB ) = 1 × OD × AB = 1 × r × 4 = 4r cm2, On subtracting Eq. (ii) from Eq. (iv), we get 2 22 ar (ΔOBC ) = 1 × OE × BC = 1 × r × 3 = 3r cm2 x = 15 − 10 = 5 2 22 On subtracting Eq. (iii) from Eq. (iv), we get y = 15 − 12 = 3

70 CBSE Term II Mathematics X (Standard) On subtracting Eq. (i) from Eq. (iv), we get 21. Given, hexagon ABCDEF circumscribe a circle. A z = 15 − 8 = 7 PQ ∴ AD = x cm = 5 cm, FB BE = y cm = 3 cm and CF = z cm = 7 cm UR Hence, the length of AD, BE and CE are 5 cm, 3 cm and E C T S 7 cm, respectively. D 19. A circle is inscribed in the ΔABC, which touches the BC, CA and AB. Since, tangents drawn from an external point to a circle are equal in length A ∴ AQ = AP, BQ = BR, CR = CS, DS = DT, ET = EU, FP = FU FE So, AB + CD + EF = (AQ + QB) + (CS + SD ) + (EU + UF) = AP + BR + CR + DT + ET + FP BD C = (AP + FP) + (BR + CR ) + (DT + ET) ⇒ AB + CD + EF = AF + BC + DE Hence proved. Given, BC = a, CA = b and AB = c 22. ∠DAB = 90° By using the property, tangents are drawn from an external In ΔABD , ∠DAB + ∠ ABD + ∠ ADB = 180° point to the circle are equal in length. ⇒ ∠ ADB = 180° − 140° = 40° ∴ BD = BF = x [say] In ΔODC, OD = OC [radii of same circle] DC = CE = y [say] ⇒ ∠OCD = ∠CDO = 40° and AE = AF = z [say] [Q angles opposite to equal sides are equal] Now, BC + CA + AB = a + b + c ∴ ∠DOC + ∠OCD + ∠ CDO = 180° ⇒(BD + DC) + (CE + EA) + (AF + FB) = a + b + c [Q sum of all angles in a triangle is 180°] ⇒ (x + y) + (y + z) + (z + x) = a + b + c ⇒ ∠DOC = 100° ⇒ 2 (x + y + z) = 2s Since, AD is a straight line. [Q 2s = a + b + c = perimeter of ΔABC] ∴ ∠DOC + ∠ COA = 180° ⇒ ∠ COA = 80° ⇒ s = x + y + z ⇒ x = s − (y + z) 23. Let O be the centre of circle. ⇒ BD = s − b [Qb = AE + EC = z + y] Q Hence proved. 20. ∠OCA = 90° [angle between tangent and radius] SO 30° P Now, ∠OCA = ∠OCD + ∠ACD ⇒ ∠ACD = ∠OCA − ∠OCD A R ⇒ ∠ACD = 90° − 44° = 46° As, AC = AD Join OQ and OR. Then, [tangents drawn from an external point OQ ⊥ PQ and OR ⊥PR are equal in length] [Q tangent is perpendicular to the radius at the point of contact] So, ∠ADC = ∠ACD = 46° So, ∠ROQ + ∠RPQ = 180° [Q angles opposite to the equal sides are equal] [Q sum of all interior angles of quadrilateral is 360°] Also, ∠CAD + ∠ADC + ∠ACD = 180° ⇒ ∠ROQ = 150° [angle sum property of a ΔACD] But ∠RSQ = 1 ∠ROQ ⇒ ∠CAD = 180°−(46° + 46°) = 88° 2 = 1 × 150° = 75° Again, ∠COD = 180°− ∠CAD = 92° 2 Further, ∠OBD = ∠ODB [OB = OD radii of circle] Now, on extending QO to intersect RS at A, we get In ΔOBD, use exterior angle theorem ∠OQP = ∠QAS = 90° [alternate interior angle] [Q PQ||RS and ∠OQP = 90°] exterior angle ∠COD = ∠OBD + ∠ODB Therefore, from ΔQSA, = ∠OBD + ∠OBD ∠SQA = 180° − 90° − 75° = 15° ⇒ 2∠OBD = ∠COD [exterior angle theorem] ⇒ ∠CBD = 1 × 92°= 46° 2 Hence, ∠CAD = 88°, ∠ADC = 46°, ∠CBD = 46° and ∠ACD = 46°

CBSE Term II Mathematics X (Standard) 71 Also, ∠PQR + ∠PRQ + ∠QPR = 180° Let ON = x, then NOʹ = 5 − x ⇒ ∠PQR + ∠PQR + 30° = 180° [Q ∠PQR = ∠PRQ because PQ = PR] In right angled Δ OPN, (OP)2 = (ON)2 + (NP)2 [by Pythagoras theorem] ⇒ ∠PQR = 150° = 75° ⇒ (NP)2 = 32 − x2 = 9 − x2 …(i) 2 and in right angled ΔPNOʹ, ⇒ ∠AQR = ∠AQP − ∠PQR (POʹ )2 = (PN)2 + (NOʹ )2 [by Pythagoras theorem] = 90° − 75° = 15° ⇒ (4)2 = (PN)2 + (5 − x)2 So, ∠RQS = ∠SQA + ∠AQR ⇒ (PN)2 = 16 − (5 − x)2 …(ii) = 15° + 15° = 30° From Eqs. (i) and (ii), we get 9 − x2 = 16 − (5 − x)2 24. Clearly, ∠QAB = 60° and ∠QBA = 60° ⇒ 7 + x2 − (25 + x2 − 10x) = 0 ⇒ 10x = 18 [Q ΔABQ is an equilateral] So, ∠PAQ = ∠PAB + ∠QAB = 120° ∴ x = 1.8 Again, in right angled ΔOPN, Similarly, ∠PBQ = 120° OP2 = (ON)2 + (NP)2 [by Pythagoras theorem] [Q∠PAB = ∠PBA, as PA =PB] …(i) ⇒ 32 = (1.8)2 + (NP)2 ⇒ (NP)2 = 9 − 3.24 = 5.76 Now, in ΔPAQ and ΔPBQ, ∴ (NP) = 2.4 PA = PB ∴ Length of common chord, PQ = 2 PN = 2 × 2.4 = 4.8 cm [tangents drawn from external point] 26. In a circle, ΔABC is inscribed. ⇒ AQ = BQ [ΔABQ is an equilateral] Join OB, OC and OA. ⇒ ∠PAQ = ∠PBQ [each 120°, shown above] So, ΔPAQ ≅ ΔPBQ [by SAS similarity rule] ⇒ ∠APQ = ∠BPQ [by CPCT] …(ii) Conside ΔABO and ΔACO Let QP intersect AB at M. A 6 cm 1 2 Now, in ΔPAM and ΔPBM, 6 cm ∠APM = ∠BPM [from Eq. (ii)] BC M ⇒ PA = PB 99 O [tangents drawn from an external point] ⇒ PM = PM [common side] AB = AC [given] So, ΔPAM ≅ ΔPBM [by SAS congruence rule] ⇒ AM = BM BO = CO [radii of same circle] and ∠AMP = ∠BMP [by CPCT] ...(iii) But ∠AMP + ∠BMP = 180° AO = AO [common side] ⇒ ∠AMP + ∠AMP = 180° Hence Proved. ⇒ ∠AMP = 90° ∴ ΔABO ≅ Δ ACO [by SSS congruence rule] ⇒ ∠1 = ∠2 [CPCT] Now, in ΔABM and ΔACM, 25. Here, two circles are of radii OP = 3 cm and POʹ = 4 cm . AB = AC [given] These two circles intersect at P and Q. ∠1 = ∠2 [proved above] P AM = AM [common side] 3 cm 4 cm ∴ ΔAMB ≅ ΔAMC[by SAS congruence rule] O N Oʹ ⇒ ∠ AMB = ∠ AMC …(i) [CPCT] Also, ∠ AMB + ∠ AMC = 180° [linear pair] Q ⇒ ∠ AMB + ∠ AMB = 180° [from Eq. (i)] Here, OP and POʹ are two tangents drawn at point P. ⇒ ∠ AMB = 90° ∠ OPOʹ = 90° We know that a perpendicular from centre of circle bisects [tangent at any point of circle is perpendicular to radius the chord. So, OA is perpendicular bisector of BC. through the point of contact] Let AM = x, then OM = 9 − x [Q OA = radius = 9 cm] Join OOʹ and PN. In right angled ΔAMC, AC2 = AM2 + MC2 [by Pythagoras theorem] In right angled Δ OPOʹ, (OOʹ )2 = (OP)2 + (POʹ )2 [by Pythagoras theorem] i.e. (Hypotenuse)2 = (Base)2 + (Perpendicular)2 i.e. ( Hypotenuse)2 = (Base)2 + (Perpendicular)2 ⇒ MC2 = 62 − x2 …(i) = (3)2 + (4)2 = 25 and in right angled Δ OMC, ⇒ OOʹ = 5 cm OC2 = OM2 + MC2 [by Pythagoras theorem] Also, PN ⊥ OOʹ ⇒ MC2 = 92 − (9 − x)2 …(ii)

72 CBSE Term II Mathematics X (Standard) From Eqs. (i) and (ii), ⇒ ∠DEOʹ = 90° − 1 ∠DOʹ B …(v) 2 62 − x2 = 92 − (9 − x)2 ⇒ 36 − x2 = 81 − (81 + x2 − 18 x) [since, Oʹ E is the angle bisector of ∠DEB i.e. 1 ⇒ 36 = 18x ⇒ x = 2 2 ∠DEB = ∠DEOʹ] ∴ AM = x = 2 Similarly, ∠AEC = 180° − ∠AOC In right angled ΔABM, [by Pythagoras theorem] Divided by 2 on both sides, we get AB2 = BM2 + AM2 1 ∠AEC = 90° − 1 ∠AOC 22 62 = BM2 + 22 ⇒ ∠AEO = 90° − 1 ∠AOC …(vi) ⇒ BM2 = 36 − 4 = 32 ⇒ BM = 4 2 2 ∴ BC = 2 BM = 2 × 4 2 = 8 2 cm [since, OE is the angle bisector of ∠AEC i.e. 1 ∠AEC = ∠AEO] ∴ Area of ΔABC = 1 × Base × Height 2 2 Now, = 1 × BC × AM 2 ∠AED + ∠DEOʹ + ∠AEO = ∠AED + ⎝⎜⎛ 90° − 1 ∠DOʹ B⎞⎠⎟ 2 = 1 × 8 2 × 2 = 8 2 cm 2 2 + ⎛⎝⎜ 90° − 1 ∠AOC⎞⎟⎠ 2 Hence, the required area of ΔABC is 8 2 cm 2. 27. In the given figure, join AO, OC and Oʹ D , Oʹ B. = ∠AED + 180° − 1 ( ∠DOʹ B + ∠AOC) 2 Now, in ΔEOʹ D and ΔEOʹ B, = ∠AED + 180° − 1 (∠AED + ∠AED ) Oʹ D = Oʹ B [radius] 2 Oʹ E = Oʹ E [common side] [from Eqs. (iii) and (iv)] = ∠AED + 180° − 1 (2 × ∠AED ) ED = EB 2 [since, tangents drawn from an external point to the circle are equal in length] = ∠AED + 180° − ∠AED = 180° A ∴ ∠AEO + ∠AED + ∠DEOʹ = 180° D So, OEOʹ is straight line. O Oʹ Hence, O, E and Oʹ are collinear. Hence proved. E 28. Given, OT = 13 cm and OP = 5 cm B C Since, if we draw a line from the centre to the tangent of the circle, then it is always perpendicular to the tangent i.e. ∴ ΔEOʹ D ≅ ΔEOʹ B [by SSS similarity rule] OP ⊥ PT. ⇒ ∠Oʹ ED = ∠Oʹ EB [by CPCT] P A Oʹ E is the angle bisector of ∠DEB. …(i) Similarly, OE is the angle bisector of ∠AEC. Now, in quadrilateral DEBOʹ, OE T ∠Oʹ DE = ∠Oʹ BE = 90° [since, CED is a tangent to the circle and Oʹ D is the radius, B i.e. Oʹ D ⊥ CED] Q ⇒ ∠Oʹ DE + ∠Oʹ BE = 180° In right angled ΔOPT , ∴ ∠DEB + ∠DOʹ B = 180° OT 2 = OP2 + PT 2 [since, DEBOʹ is cyclic quadrilateral] …(ii) [by Pythagoras theorem, Since, AB is a straight line. (hypotenuse)2 = (base)2 + (perpendicular)2] ⇒ PT 2 = (13)2 − (5)2 ∴ ∠AED + ∠DEB = 180° ⇒ ∠AED + 180° − ∠DOʹ B = 180° [from Eq. (ii)] = 169 − 25 = 144 …(iii) ⇒ ∠AED = ∠DOʹ B ...(iv) ⇒ PT = 12 cm Similarly, ∠AED = ∠AOC Since, the length of pair of tangents from an external point T Again from Eq. (ii), is equal. ∠DEB = 180° − ∠DOʹ B ∴ QT = 12 cm Divided by 2 on both sides, we get Now, TA = PT − PA 1 ∠DEB = 90° − 1 ∠DOʹ B ⇒ TA = 12 − PA ...(i) 2 2 and TB = QT − QB

CBSE Term II Mathematics X (Standard) 73 ⇒ TB = 12 − QB ...(ii) Given, BD = FB = 9 cm, CD = CE = 3 cm In ΔABC, AB2 = AC2 + BC2 Again, using the property,length of pair of tangents from an external point is equal. ⇒ (AF + FB)2 = (AE + EC)2 + (BD + CD )2 ∴ PA = AE and QB = EB ...(iii) ⇒ (x + 9)2 = (x + 3)2 + 122 ∴ OT = 13 cm ⇒ x2 + 81 + 18x = x2 + 9 + 6x + 144 ∴ ET = OT − OE [Q OE = 5 cm = radius] ⇒ 18x + 81 = 6x + 9 + 144 ⇒ 12x = 72 ⇒ x = 6 cm ⇒ ET = 13 − 5 ⇒ ET = 8 cm ∴ AB = 6 + 9 = 15 cm (iii) (b) As we know that, tangents drawn from an external Since, AB is a tangent and OE is the radius. point are equal in length. Therefore, AP = AS = 4 cm ∴ DS = DR = 10 − 4 = 6 cm ∴ OE ⊥ AB [linear pair] And BP = BQ = 2 cm. So, CR = CQ = 5 − 2 = 3 cm So, CD =DR + CR = 6 + 3 = 9 cm ⇒ ∠OEA = 90° (iv) (d) Here ∠OAP = 90° ∴ ∠AET = 180° − ∠OEA A ⇒ ∠AET = 90° Now, in right angled ΔAET, (AT )2 = (AE)2 + (ET )2 [by Pythagoras theorem] ⇒ (PT − PA)2 = (AE)2 + (8)2 ⇒ (12 − PA)2 = (PA)2 + (8)2 [from Eq. (iii)] ⇒ 144 + (PA)2 − 24⋅PA = (PA)2 + 64 ⇒ 24⋅PA = 80 ⇒ PA = 10 cm P 50° O 3 ∴ AE = 10 cm [from Eq. (iii)] B 3 Similarly BE = 10 cm In ΔAOP and ΔBOP, 3 ∠OAP = ∠OBP [90° each] Hence, AB = AE + EB = 10 + 10 = 20 cm OA = OB [radii of circle] 333 PA = PB Hence,the required length AB is 20 cm . [tangents drawn from an external point are equal] 3 ∴ ΔAOP ~ ΔBOP [by SAS similarity] 29. (i) (d) We have, AP = AQ, BP = BD, CQ = CD … (i) ∴ ∠APO = ∠OPB [by CPCT] [Q tangents drawn from an external point = 50° are equal in length] ∴ ∠BPA = 50° + 50° = 100° Now, AB + BC + AC = 7 + 4 + 9 = 20 cm (v) (c) For bigger circle, PA = PB … (i) ⇒ AB + BD + CD + AC = 20 cm [Qtangents drawn from an external point are equal in length] … (ii) ⇒ AP + AQ = 20 cm ⇒ 2AP = 20 cm ⇒ AP = 10 cm Similarly, for smaller circle, PB = PC From Eqs. (i) and (ii), we get (ii) (c) Let AF = AE = x cm [Q tangents drawn from an external point to PA = PB = PC = 11 cm a circle are equal in length]

Chapter Test Multiple Choice Questions (iv) In PA and PB are two tangents, drawn to a circle 1. Two concentric circles are of radii 10 cm and with centre O from P such that ∠PBA = 60°, then 8 cm, then the length of the chord of the larger ∠OAB is circle, which touches the smaller circle is (a) 50° (b) 25° (c) 30° (d) 130° (a) 6 cm (b) 12 cm (v) In the adjoining figure, if PC is the tangent at A (c) 18 cm (d) 9 cm of the circle with ∠PAB = 62° and ∠AOB = 132°, then ∠ABC is 2. From a point P, which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the B pair of tangents PQ and PR to the circle is O drawn. Then, the area of the quadrilateral PQOR is [NCERT Exemplar] (a) 60 cm2 (b) 65 cm2 62° PA (c) 30 cm2 (d) 32.5 cm2 C 3. If two tangents inclined at an angle 60° are (a) 18° (b) 20° (c) 60° (d) Can’t be determined drawn to a circle of radius 3 cm, then the Short Answer Type Questions length of each tangent is [NCERT Exemplar] 5. Two tangents PQ and PR are drawn from an (a) 3 3 cm (b) 6 cm external point to a circle with centre O. Prove 2 (d) 3 3 cm that QORP is a cyclic quadrilateral. (c) 3 cm Case Based MCQs 6. In figure, AB and CD are common tangents to 4. For revision of chapter circles, a teacher two circles of unequal radii. Pro[vNeCtEhRaTt EAxBem=pClaDr]. planned a game with some questions written A on the paper, which are to be answered by the B students. For each correct answer, a student will get a prize. Some of the questions are given D below. C Answer the questions to check your Long Answer Type Questions knowledge. (i) In the given figure, x + y is O 7. A is a point at a distance 13 cm from the centre x (a) 60° O of a circle of radius 5 cm. AP and AQ are the (b) 90° C (c) 120° y tangents to the circle at P and Q. If a tangent (d) 145° A BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the (ii) In the given figure, PQ and PR are two tangents perimeter of the ΔABC. [NCERT Exemplar] to the circle, then ∠ROQ is 8. If a chord and a tangent intersect externally, S R then the product of the lengths of the segments of the chord is equal to the square of O the length of the tangent from the point of contact to the point of intersection. 20° P Answers Q 1. (b) 2. (a) 3. (d) (a) 30° (b) 60° (c) 105° (d) C 4. (i) (b) (ii) (d) (iii) (b) (iv) (c) (v) (b) 160° 45° (iii) In the adjoining figure, AB is a O 7. 24 cm chord of the circle and AOC is its diameter such that B ∠ACB = 45°, then ∠BAT is (a) 35° AT For Detailed Solutions (b) 45° Scan the code (c) 125° (d) 110°

CBSE Term II Mathematics X (Standard) 75 CHAPTER 04 Constructions In this Chapter... l Division of a Line Segment Internally in the Given Ratio l Construction of a Tangent to a Circle at a Point that lies on it l Construction of Tangent to a Circle from a Point Outside the Circle l Construction of Tangents to a Circle When Angle Constructions 1 Justification Division of a Line Segment Internally in the Given Ratio Since, Am C||ApB, so use the basic proportionality theorem in ΔABAp . To divide a line segment AB (say) internally in the given ratio m : n, where m and n are both positive integers, we use the Then, AAm = AC …(i) following steps A m A p CB Step I Draw the given line segment AB and any ray AX, By using construction, the ratio is …(ii) making an acute angle with the line segment AB. AAm = m This ray AX can be drawn above or below AB. AmAp (p − m) Step II Mark m + n = p points ∴From Eqs. (i) and (ii), (i.e. A 1 , A 2 , . . . , A m ,. . . , A p ) on the ray AX, such AC = m that AA 1 = A 1 A 2 = ... = A p − 1 A p CB (p − m) Step III Join BA p . Alternate Method Step IV Through the point Am , draw a line parallel to A pB (by making an angle equal to ∠AA pB at A m ) which To divide a line segment in the given ratio m : n, where intersects the line segment AB at point C. Thus, m and n are both positive integers, we can also use the point C divides the line segment AB internally in the following steps. ratio m : n, i.e. AC: CB = m : n. Step I Draw the given line segment AB (say) and any ray X AX making an acute angle with the line segment AB. A p(A m+n) Step II Draw another ray BY|| AX by making Am ∠ABY = ∠BAX. Step III Mark m points i.e. A1 , A2 ,. . . , Am on AX A1 A2 C B A and n points i.e. B1 , B2 , . . . ,Bn on BY such that AA1 = A1 A 2 = . . . = A m − 1 A m = BB1 = B1B2 = . . . = Bn − 1Bn

76 CBSE Term II Mathematics X (Standard) Step IV Join AmBn which intersects line segment AB at the So, in ΔABA5, by basic proportionality theorem, we get point C. Now, C is the required point which divides line AA3 = AC ...(i) segment AB internally in the ratio m : n. A 3A 5 CB By construction, we have X AA3 = 3 = 3 ...(ii) A 3A 5 (5 − 3) 2 Am Am – 1 On equating Eqs. (i) and (ii), we get A2 AC 3 A1 BC = 2 This shows that C divides AB internally in the ratio 3 : 2. A B C Construction 2 B1 B2 Construction of a Tangent to a Circle at a Point that lies on it Bn Bn – 1 We can construct a tangent to a circle at a point that lies on it Y by two cases which are given below Justification Case I By using the centre of circle To construct a tangent to a circle by using the centre, we use Step V Use the condition of similarity of two triangles in the following steps. ΔAAm C and ΔBBn C. Then, AA m = AC ...(i) Step I Take a point O as centre and draw a circle of BB n BC given radius. Step VI Write the ratio by using construction, ..(ii) Step II Take a point P on the circle, at which we want to AAm = m draw tangent. BBn n Step III Join OP, which is the radius of circle. Step VII Equating Eqs. (i) and (ii), we get AC = m BC n Step IV Take OP as base and construct ∠OPT = 90° at P. Example 1. Draw a line segment AB = 8 cm and divide it Step V Draw a ray PT and produce TP to Tʹ to get the required tangent TPTʹ. internally in the ratio 3 : 2 and also justify it. Sol. Steps of Construction (i) First, draw line segment, AB = 8 cm and draw a ray AX, which makes an acute angle with line segment AB. (ii) Mark m + n = 3 + 2 = 5 points i.e. A1, A2, A3, A4 and A5 O on the ray AX such that AA1 = A1A 2 = A 2A 3 = A 3A 4 = A 4A 5 Tʹ T P (iii) Join BA5. Example 2. Draw a circle of diameter AB = 5 cm with (iv) Through the point A3 (Q m = 3), draw a line A3C ||A5B centre O and then draw a tangent to the circle at (by making an angle equal to ∠AA5 B at A3), which point A or B. intersects the line segment AB at C. Sol. Given, diameter of circle = AB = 5 cm and centre is O. X ∴ Radius = OA = OB = 5 = 2. 5 cm 2 A5 A4 Steps of Construction A3 (i) Take a point O as centre and draw a circle of radius A2 2.5 cm. (ii) Draw diameter AOB. A1 (iii) Take OA as base and construct ∠OAT = 90° at A. A B C 8 cm Thus, point C divides the line segment AB internally in the ratio 3 : 2. Justification Since, A3C ||A5B.

CBSE Term II Mathematics X (Standard) 77 (iv) Produce TA to Tʹ to get the required tangent TATʹ. Construction 3 Similarly, we can draw a tangent at point B or any other point on the circle. Construction of Tangents to a Circle from a Point T Outside the Circle 2.5 cm If a point lies outside the circle, then there will be two B OA tangents to the circle from this point. Case I When centre of circle is known Tʹ If centre of circle is known, then to draw tangents from a given external point, we use the following steps Case II Without using the centre of circle To construct a tangent to a circle without using the Step I Draw a circle with centre O of given radius and take centre of circle, we use the following steps. a point P outside it. Step I Draw a circle of given radius and take a point P Step II Join OP and bisect it. Let its mid-point be M. Then, (at which we want to draw tangent) on the circle. MP = MO. Step II Draw any chord PQ through the given point P on the Step III On taking M as centre and MO or MP as radius, circle. draw a dotted circle, which intersects the given circle at points Q and Qʹ (say). Step III Take a point R in either the major arc or minor arc and join PR and QR. Q Step IV On taking PQ as base, construct ∠QPY equal to P O ∠PRQ and on the opposite side of R. M Step V Draw a ray PY and produce YP upto X to get the Qʹ required tangent YPX. R Step IV Join PQ and PQʹ. Thus, PQ and PQʹ are the required Q tangents drawn to the circle from the external point P. Here, we observe that PQ = PQʹ. X Y P Justification Join OQ. Then, ∠PQO = 90°, since it is constructed in the Example 3. Draw a circle of radius 6 cm. Take a point P semi-circle of dotted circle. It shows that OQ ⊥ PQ. Also, OQ on it. Without using the centre of the circle, draw a is radius of given circle, so PQ has to be a tangent of given tangent to the circle at point P. circle. Similarly, PQʹ is also a tangent to the given circle. Sol. Given, radius of circle = 6 cm Example 4. Draw a circle of radius 3.5 cm. From a point P, 6 cm from its centre, draw two tangents of the Steps of Construction circle. (i) Draw a circle of radius 6 cm and take a point P on the Sol. Given, a circle of radius 3.5 cm whose centre is O (say) and circle. a point P, 6 cm away from its centre. (ii) Draw a chord PQ through the point P on the circle. A (iii) Take a point R in the major arc and join PR and RQ. 6 cm P O MC (iv) On taking PQ as base, construct ∠QPY equal to ∠PRQ on the opposite side of the point R. 3.5 cm (v) Produce YP to X. Then, YPX is the required tangent at point P. R X Q B P Y

78 CBSE Term II Mathematics X (Standard) Steps of Construction Sol. Given, radius of circle = 2 cm and distance between point P (i) Draw a circle with O as centre and radius and centre = 6.5 cm OC = 3.5 cm. Take a point P such that OP = 6 cm. Steps of Construction (i) Draw a circle of radius 2 cm with centre O. (ii) Draw the bisector of OP which intersect OP at M. (ii) Take a point P outside it, such that its distance from (iii) Take M as centre and MO as radius, draw a dotted centre O is 6.5 cm. (iii) Consider O and P as centre and draw arcs of radius circle. Let this circle cuts the given circle at A and B. more than half of OP on both sides of OP which (iv) Join PA and PB. intersect each other at R and S. Join RS which bisects OP at M. Then, MP = MO. Hence, PA and PB are the required tangents. Case II When centre of circle is unknown (iv) Consider M as centre and MO as radius, draw a dotted circle which intersects given circle at Q and Qʹ. If centre of the circle is unknown, then to draw tangents of the circle, by using the following steps (v) Join PQ and PQʹ. Step I Firstly, draw the circle and then draw two Hence, we get the required tangents drawn from point non-parallel chords of the circle. P to the given circle. Step II Draw the perpendicular bisectors of both chords R which intersect each other at a point, say O. Then, this point O gives the centre of given circle. Now, Q we further use the steps given in case I to draw tangents. 1.1 cm M 2 cm O P 6.5 cm Alternate Method Qʹ If centre of circle is unknown, then we can draw tangents without finding centre of the circle. For this, we use the S following steps of construction. Construction 4 Step I Draw a circle of given radius and take a point P outside it. Construction of Tangents to a Circle When Angle between Them is Given D Sometimes, angle between two tangents (or pair of tangents) T1 is given and we have to draw these tangents. Then, we use the following steps of construction. CK B Step I First, draw the given circle with centre O and radius PA r cm. T2 Step II Through P, draw a line (i.e. secant) intersecting the A O Q given circle at points A and B, respectively and r produce it to C in opposite direction of AB such that α AP = CP. P α Step III Now, bisect the segment CB at K. Then, take K as R centre and KB (or KC) as radius, draw a semi-circle. Step II Draw any diameter say AOQ of this circle. Step IV At point P, draw PD ⊥ CB which cuts the Step III Make given angle α at centre O with OQ (say) as semi-circle at D. base which intersect the circle at point R (say) or draw the radius OR meets the circle at R such that Step V Take P as centre and PD as radius draw arcs to ∠QOR = α. intersect the given circle at points T1 and T2 . Step IV Now, draw perpendiculars to OA at A and to OR at Step VI Join PT1 and PT2 which are the required tangents. R, which intersect the tangents each other at a point say P. Example 5. Draw a circle of radius 2 cm with centre O and take a point P outside the circle such that OP = 6.5 cm. From P, draw two tangents to the circle.

CBSE Term II Mathematics X (Standard) 79 Then, AP and RP are the required pair of tangents to given (iii) Draw the radius OR meets the circle at R such that ∠QOR = 30°. circle, inclined at an angle α, i.e. angle between pair of tangents is α. Justification P By construction, ∠OAP = 90° and OA is radius. E 30° O 4 cm N 30° Q So, PA is a tangent to the circle. D R Similarly, PR is a tangent to the circle. Also, ∠AOR = 180° − ∠QOR [Q AOQ is a straight line] = 180° − α Now, in quadrilateral AORP, (iv) Draw PD ⊥ PQ and RE ⊥ OR, which intersect each other at point N. Then, NP and NR are the required ∠APR + ∠PAO + ∠AOR + ∠PRO = 360° tangents to the given circle inclined to each other at an angle of 30°. ⇒ ∠APR + 90° +180° − α + 90° = 360° Justification ⇒ ∠APR = α By construction, ∠OPN = 90° and OP is radius. Example 6. Draw a pair of tangents to a circle of radius ∴ PN is a tangent to the circle. 4 cm which are inclined to each other at an angle of Similarly, NR is a tangent to the circle. 30°. Now, ∠POR = 180° − 30°= 150° Sol. Given, a circle of radius 4 cm. We have to construct a pair of [QPOQ is a straight line and ∠QOR = 45°] tangents, which are inclined to each other at an angle of 30°. In quadrilateral OPNR, Steps of Construction ∠OPN = 90° , ∠POR = 150° and ∠ORN = 90° (i) Draw a circle with O as centre and radius 4 cm. ∴ ∠PNR = 360° − (90° + 150° + 90°) = 30° (ii) Draw any diameter POQ of this circle.

80 CBSE Term II Mathematics X (Standard) Chapter Practice PART 1 6. To divide a line segment AB in the ratio 4 : 7, a ray Objective Questions AX is drawn first such that ∠ BAX is an acute angle and then points A1 , A 2 , A 3 ,. . . are located at equal distances on the ray AX and the point B is joined to G Multiple Choice Questions (a) A12 (b) A11 (c) A10 (d) A9 1. To divide a line segment AB in ratio m : n 7. To divide a line segment AB in the ratio 5 : 6, draw a (m and n are positive integers), draw a ray AX to that ∠BAX is an acute angle and the mark point on ray AX such that ∠ BAX is an acute angle, then draw ray AX at equal distances such that the minimum a ray BY parallel to AX and the points number of these points is A1 , A 2 , A 3 ,. . . and B1 , B 2 , B 3 ,. . . are located to equal distances on ray AX and BY, respectively. (a) greater of m and n (b) m + n Then, the points joined are (c) m + n − 1 (d) mn (a) A5 and B6 (b) A6 and B5 2. To divide a line segment AB in the ratio 5 : 7, first (c) A4 and B5 (d) A5 and B4 a ray AX is drawn, so that ∠ BAX is an acute angle and then at equal distances points are marked on 8. To divide a line segment AB in the ratio 6 : 7, a ray the ray AX such that the minimum number of these points is AX is drawn first such that ∠BAX is an acute angle and then points A1 , A 2 , A 3, … are located equal distances on the ray AX and the point B is joined (a) 8 (b) 10 (c) 11 (d) 12 with 3. To divide a line segment AB in the ratio 3 : 5 first a (a) A12 (b) A13 (c)A10 (d)A11 ray AX is drawn so that ∠BAX is an acute angle 9. In the given figure, find the ratio, when P divides and then at equal distances points are marked on AB internally. the ray AX such that the minimum number of A3 X these points is A2 A1 (a) 8 (b) 9 (c) 10 (d) 11 A B 4. To divide a line segment AB in the ratio 4 : 5, first Y P B1 B4 B3 B2 a ray AX is drawn making ∠BAX an acute angle and then points A1 , A 2 , A 3 , . . at equal distances (a) 3 : 2 (b) 2 : 3 (c) 4 : 3 (d) 3 : 4 are marked on the ray AX and the point B is joined to 10. From the following ratios, a line segment cannot be divided into A ratio. (a) A4 (b) A 5 (c) A9 (d) A7 (a) A → 5 : 1 (b) A → 1 : 1 5 55 5. The ratio of division of the line segment AB by the point P from A in the following figure is [CBSE 2012] (c) A → 2 : 5 (d) A → 1 : 1 52 5 A5 A4 11. To draw a pair of tangents to a circle, which are A3 inclined to each other at an angle of 60°, it is A2 A1 required to draw tangents at end points of those two A PB radii of the circle, the angle between them should be (a) 2 : 3 (b) 3 : 2 (a) 135° (b) 90° (c) 60° (d) 120° (c) 3 : 5 (d) 2 : 5

CBSE Term II Mathematics X (Standard) 81 12. A pair of tangents can be constructed from a 7. Draw a circle of radius 3.5 cm. Take a point P outside point P to a circle of radius 3.5 cm, situated at a distance of 3 cm from the centre. the circle at a distance of 7 cm from the centre of the (a) True circle and construct a pair of tangents to the circle (b) False (c) Can’t determined from that point. [CBSE 2020 (Standard)] (d) None of the above 8. Draw a line segment AB of length 9 cm. Taking A as 13. A pair of tangents can be constructed to a circle inclined at an angle of 170°. centre, draw a circle of radius 5 cm and taking B as (a) True centre, draw another circle of radius 3 cm. Construct (b) False (c) Can’t determined tangents to each circle from the centre of the other (d) None of the above circle. [CBSE 2020 (Standard)] 9. Draw a circle with the help of circular solid ring. Construct a pair of tangents from a point P outside the circle. Also, justify the construction. PART 2 G Long Answer Type Questions Subjective Questions 10. Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is 60°. G Short Answer Type Questions Also, justify the construction. Measure the distance between the centre of the circle and the point of 1. Draw a line segment of length 7 cm. Find a point intersection of tangents. P on it, which divides it in the ratio 3 : 5. 11. Construct a tangent to a circle of radius 1.8 cm from a 2. Draw a circle of diameter AB = 6 cm with centre point on the concentric circle of radius 2.8 cm and O and then draw a tangent to the circle at point A measure its length. Also, verify the measurement by or B. actual calculation. 3. Draw a circle of radius 5 cm. Take a point P on it. 12. Draw a circle of radius 2.8 cm. From an external point Without using the centre of the circle, draw a P, draw tangents to the circle without using the centre tangent to the circle at point P. of the circle. 4. Draw a circle of radius 6 cm and draw a tangent 13. Draw a pair of tangents to a circle of radius 3 cm, to this circle, making an angle of 30° with a line which are inclined to each other at an angle of 45°. passing through the centre. 14. Let ABC be a right angled triangle, in which 5. Draw a circle of radius 4 cm. From a point 6 cm AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is the away from its centre, construct a pair of tangents perpendicular from B on AC. The circle through to the circle and measure their lengths. B, C and D is drawn. Construct the tangents from A to this circle. Also, justify the construction. [CBSE 2019] 15. Draw a circle of radius 3 cm. Take two points P and Q 6. Draw a circle of radius 1cm. From a point P, 2.2 cm apart from the centre of the circle, draw on one of its extended diameter each at a distance of tangents to the circle. 7 cm from its centre. Draw tangents to the circle from these two points P and Q. [NCERT]

82 CBSE Term II Mathematics X (Standard) SOLUTIONS Objective Questions 11. (d) The angle between them should be 120° because in that case the figure formed by the intersection point of pair of 1. (b) To divide a line segment in the ratio m : n, the maximum tangent, the two end points of those two radii (at which tangents are drawn) and the centre of the circle is a number of the points to mark are m + n. quadrilateral. 2. (d) We know that, to divide a line segment AB in the ratio P m : n, first draw a ray AX, which makes an acute angle ∠BAX, then marked m + n points at equal distance. O 60° θR Here, m = 5, n = 7 So, minimum number of these points = m + n = 5 + 7 = 12. Q 3. (a) Minimum number of points = 3 + 5 = 8 From figure, it is quadrilateral 4. (c) Here, 4 + 5 = 9 points are located at equal distances on ∠POQ + ∠PRQ = 180° [Q sum of opposite angles are 180°] the ray AX, so B is joined to last point A 9. 60° + θ = 180° 5. (b) The ratio of division of the line segment AB by the point ∴ θ = 120 Hence, the required angle between them is 120°. P from A is AP : BP = 3 : 2 . 12. (b) False, since, the radius of the circle is 3.5 cm 6. (b) Here, minimum 4 + 7 = 11 points are located at equal i.e. r = 3. 5 cm and a point P is situated at a distance of 3 cm from the centre i.e. d = 3 cm distances on the ray AX and then B is joined to last point We see that, r > d is A11. i.e. a point P lies inside the circle. So, no tangent can be 7. (a) Given, a line segment AB and we have to divide it in the drawn to a circle from a point lying inside it. ratio 5 : 6. 13. True Y B B 6 5B 4B 3B 2B A1 C B A 1A 2A 3A 4A X 5 Steps of Construction 170° 1. Draw a ray AX, making an acute ∠BAX. If the angle between the pair of tangents is always greater than 0 or less than 180°, then we can construct a pair of 2. Draw a ray BY parallel to AX by making ∠ABY equal tangents to a circle. to ∠BAX. Hence, we can draw a pair of tangents to a circle inclined at an angle of 170°. 3. Now, locate the points A1, A 2, A 3, A 4 and A 5 ( m = 5) on AX and B1, B2, B3, B4, B5 and B6 ( n = 6) such that Subjective Questions all the points are at equal distance from each other. 1. Steps of Construction 4. Join B6A5, which intersect AB at a point C. 1. Draw a line segment AB = 7 cm. Then, AC : BC = 5 : 6 2. Draw a ray AX, making an acute ∠BAX. 3. Along AX, mark 3 + 5 = 8 points 8. (b) A6+ 7 i.e. A13 is joined to the point B. i.e. A 1 , A 2 , A 3 , A 4 , A 5 , A 6 , A 7 and A 8 such that AA 1 = A 1 A 2 = A 2 A 3 = A 3 A 4 9. (d) From given figure, it is clear that there are three points at = A 4A 5 = A 5A 6 = A 6A 7 = A 7A 8 equal distances on AX and four points at equal distances on 4. Join A 8B. BY. Here, P divides AB on joining A3B4. So, P divides AB 5. From A 3 , draw A 3 C|| A 8 B, meeting AB at C. internally in the ratio 3 : 4. [by making an angle equal to ∠BA 8 A at A 3 ] Then, C is the point on AB, which divides it in the 10. (c) Since, (b) 1 : 1 = 1 : 1 ratio 3 : 5. (a) 5 : 1 = 5 : 1 55 5 (c) 2 : 5 = 2 2 : 5 (d) 1 : 1 = 1: 5 52 5 Since, (a), (b) and (d) are the ratio of both integers. So, it is possible to divide a line segment into these points. Hence, option (c) is correct.

CBSE Term II Mathematics X (Standard) 83 Thus, AC : CB = 3 : 5 4. Steps of Construction (i) Draw a circle with centre O and radius 6 cm. X A8 P 30° Q B 60°6 cm A A7 O A6 A5 (ii) Draw a radius OA and produce it to B. A4 (iii) Construct an ∠AOP equal to the complement of 30° A3 A2 i.e. equal to 60°. A1 (iv) Draw a perpendicular to OP at P, which intersects OB at AC B point Q. Hence, PQ is the required tangent such that ∠OQP = 30°. 7 cm 5. Given, a point Mʹ is at a distance of 6 cm from the centre of a circle of radius 4 cm. 2. Given, diameter of circle = AB = 6 cm and centre is O. Steps of Construction ∴ Radius = OA = OB = 6 = 3 cm (i) Draw a circle of radius 4 cm. Let centre of this circle is 2 O. Steps of Construction (ii) Join OMʹ and bisect it. Let M be mid-point of OMʹ. (i) Take a point O as centre and draw a circle of radius 3 cm. P (ii) Draw diameter AOB. Mʹ M O 4 cm (iii) Take OA as base and construct ∠OAT = 90° at A. (iv) Draw a ray AT and produce TA to Tʹ to get the required 6 cm tangent TATʹ. Q Similarly, we can draw a tangent at point B or any other point on the circle. (iii) Taking M as centre and MO as radius, draw a circle to intersect circle (0, 4) at two points, P and Q. T (iv) Join PMʹ and QMʹ. PMʹ and QMʹ are the required 3 cm tangents from Mʹ to circle C (0, 4). BA The measure length of the tangents are 4.48 cm. O 6. Given, radius of circle = 1 cm and distance between point P and centre = 2.2 cm. Tʹ Steps of Construction (i) Draw a circle of radius 1 cm with centre O. 3. Given, radius of circle = 5 cm (ii) Take a point P outside it such that its distance from Steps of Construction centre O is 2.2 cm. (i) Draw a circle of radius 5 cm and take a point P on the circle. (iii) Take O and P as centre and draw arcs of radius more (ii) Draw a chord PQ through the point P on the circle. than half of OP on both sides of OP, which intersect each other at R and S. Join RS, which bisects OP at M. (iii) Take a point R in the major arc and join PR and RQ. Then, MP = MO. (iv) On taking PQ as base, construct ∠QPY equal to ∠PRQ on the opposite side of the point R. (v) Draw a ray PY and produce YP to X. Then, YPX is the required tangent at point P. R X Q P Y

84 CBSE Term II Mathematics X (Standard) R M N Q 3 cm 5 cm B A O 9 cm Q P 1.1 cm M 1 cm O 2.2 cm P Qʹ 9. Steps of Construction S (i) First, draw a circle with the help of given circular solid (iv) On taking M as centre and MO as radius, draw a dotted ring and then draw two non-parallel chords AB and CD. circle, which intersects given circle at Q and Qʹ. TA (v) Join PQ and PQʹ. Thus, we get the required tangents PM D drawn from point P to the given circle. OB 7. Steps of Construction Tʹ C (i) Draw a circle with O as centre and radius OC = 3.5 cm. Take a point P such that OP = 7 cm. (ii) Draw perpendicular bisectors of AB and CD, which (ii) Draw the bisector of OP, which intersects OP at M. intersect each other at point O. Then, O is the centre of the circle. (iii) On taking M as centre and MO as radius, draw a dotted circle. Let this circle cuts the given circle at A and B. (iii) Now, take a point P outside the circle and join OP. (iv) Join PA and PB. (iv) Draw bisector of OP. Let its mid-point be M. Thus, PA and PB are the required tangents. (v) On taking M as centre and MP as radius, draw a dotted A circle which intersect the given circle at T and Tʹ. 7 cm P (vi) Join PT and PTʹ. O 3.5 cm M C Then, PT and PTʹ are the required pair of tangents drawn to the circle from P. B Justification 8. Given, a line segment AB = 9 cm, two circles with centres A Join OT. and B of radii 5 cm and 3 cm, respectively. Then, ∠PTO = 90° [angle in semi-circle of dotted circle] We have to construct two tangents to each circle from the centre of the other circle. This shows that OT ⊥ PT. Steps of Construction Also, OT is radius of given circle, so PT has to be a tangent of (i) Draw a line segment AB = 9 cm. given circle. Similarly, PTʹ is also a tangent of given circle. (ii) Draw a circle with centre A and radius 5 cm and another 10. Steps of Construction circle with centre B and radius 3 cm. (i) Take a point O on the plane of the paper and draw a circle (iii) Now, bisect AB. Let O be the mid-point of AB. with centre O and radius OA = 4 cm. (iv) Take O as centre and AO as radius and draw a dotted (ii) At O construct radii OA and OB such that ∠AOB equal to circle, which intersects the two given circles at N, Q, M 120° i.e. supplement of the angle between the tangents. and P. (v) Join AN, AQ, BM and BP. These are the required tangents (iii) Draw perpendiculars to OA and OB at A and B, to each circle from the centre of the other circle. respectively. Suppose these perpendiculars intersect at P. Then, PA and PB are required tangents. P 60° A 4 cm 120° O B

CBSE Term II Mathematics X (Standard) 85 The distance between the centre of the circle and the point D of intersection of tangents is 8 cm. M Justification In quadrilateral OAPB, we have C OA B P ∠OAP = ∠OBP = 90° and ∠AOB = 120° N ∴ ∠OAP + ∠OBP + ∠AOB + ∠APB = 360° ⇒ 90° + 90° + 120° + ∠APB = 360° (iii) Now, bisect BC and take its mid-point as O. Draw a ∴ ∠APB = 360° − (90° + 90° + 120° ) semi-circle with centre O and radius OB (or OC ). = 360° − 300° = 60° (iv) Draw PD ⊥ BC, which intersects the semi-circle at D. 11. Given, two concentric circles of radii 2.8 cm and 1.8 cm with (v) With centre P and radius PD draw two arcs, which common centre say, O. intersects the given circle at points M and N. A (vi) Join PM and PN. Thus, PM and PN are the required tangents to the given circle. C1 C2 P MO 13. Given, a circle of radius 3 cm. We have to construct a pair of tangents, which are inclined to each other at an angle of 45°. B Steps of Construction Steps of Construction (i) Draw a circle with O as centre and radius 3 cm. (i) Draw two circles with common centre O and radii (ii) Draw any diameter POQ of this circle. 2.8 cm and 1.8 cm, respectively. (ii) Take a point P on the outer circle and join OP. (iii) Draw the radius OR meets the circle at R such that (iii) Draw bisector of OP. Let mid-point of OP be M. ∠QOR = 45°. (iv) Taking M as centre and PM as radius, draw a dotted circle, P O 3 cm which intersects the inner circle at two points say A and B. 45° Q (v) Join AP and BP. Then, AP and BP are required tangents. 45° EN R On measuring the lengths, we get PA = PB = 2.14 cm Calculation D Join OA. Then, OA = 1.8 cm [radius of inner circle C1] (iv) Draw PD ⊥ PQ and RE ⊥ OR, which intersects each other at point N. OP = 2.8 cm [radius of outer circle C2] Then, NP and NR are the required tangents to the given and ∠PAO = 90° circle inclined to each other at an angle of 45°. Justification [Q angle in semi-circle of constructed circle] By construction, ∠OPN = 90° and OP is radius. So, in ΔPAO, by Pythagoras theorem, ∴ PN is a tangent to the circle. Similarly, NR is a tangent to the circle. OP2 = OA 2 + AP2 Now, ∠POR = 180° − 45°= 135° ⇒ (2.8)2 = (1.8)2 + AP2 ⇒ 7.84 = 3.24 + AP2 [QPOQ is a straight line and ∠QOR = 45°] ⇒ AP2 = 7.84 − 3.24 = 4.6 In quadrilateral OPNR, ⇒ AP = 2.14 cm ∠OPN = 90° , ∠POR = 135° [taking positive square root, as length cannot be negative] and ∠ORN = 90° ∴ ∠PNR = 360° − (90° + 135° + 90°) = 45° ⇒ PA = PB = 2.14 cm 14. Given, ABC is a right angled triangle, in which AB = 6 cm, Hence, the length of tangents is 2.14 cm. BC = 8 cm, ∠B = 90° and BD is perpendicular to AC. 12. Given, a circle of radius 2.8 cm and we have to draw Then, ∠ADB = ∠CDB = 90° tangents without using the centre. Steps of Construction (i) First, draw a circle of radius 2.8 cm and take a point P outside the circle. (ii) Through P, draw a secant PAB, which intersects the circle at A and B and extend it to C in opposite direction of AB such that PC = PA.

86 CBSE Term II Mathematics X (Standard) Steps of Construction We have to construct the tangents to the circle from the (i) Draw the line segments AB = 6 cm and BC = 8 cm given points P and Q. perpendicular to each other. Join AC. Thus, ΔABC is the given right angled triangle. M Pʹ (ii) Draw perpendicular bisector of BC, which meets BC at O. 3 cm F Q O 7 cm (iii) With O as centre and OB as radius draw a circle, which intersects AC at D, then ∠BDC = 90°. Thus, PE BD is perpendicular to AC. 7 cm (iv) With A as centre and AB as radius draw an arc, N Qʹ cutting the circle at M. Steps of Construction (v) Join AM. Thus, AB and AM are required tangents. (i) Draw a circle of radius 3 cm with centre at O. X (ii) Produce its diameter on both sides and take points P and Q A on it such that OP = OQ = 7 cm D M (iii) Draw bisector of OP and OQ, which intersect OP and OQ at E and F, respectively. 6 cm (iv) Now, take E as centre and OE as radius, draw a dotted circle B C which intersects the given circle at two points M, N. Again, O take F as centre and OF as radius, draw another dotted circle which intersects the given circle at two points Pʹ and Qʹ. 8 cm (v) Join PM, PN, QPʹ and QQʹ. These are the required tangents Justification from P and Q to the given circle. Since ΔABC is right angled triangle with ∠ABC = 90° ∴ BO ⊥ AB. Justification Also, BO is the radius of circle. So, AB has to be tangent of the circle. Similarly, AM is also a tangent to the circle. Join OM and ON. The ∠OMP is the angle that lies in the semi-circle of the dotted circle with centre E. Therefore, 15. Given, two points P and Q on the extended diameter of a ∠OMP = 90° ⇒ OM ⊥PM circle with radius 3 cm such that OP = OQ = 7 cm Since, OM is radius of the circle. So, MP has to be a tangent to the circle. Similarly, PN, QPʹ and QQʹ are also tangents to the given circle.

Chapter Test Multiple Choice Questions these two radii of the circle, the angle between two radii is 1. To divide a line segment AB in the ratio 2 : 5, (a) 105° (b) 70° first a ray AX is drawn, so that ∠BAX is an acute (c) 125° angle and then at equal distances, then the (d) 135° number of points located on the ray AX is Short Answer Type Questions [CBSE 2011] 5. Draw a line segment AB = 6.5 cm and divide it (a) 7 (b) 10 (c) 2 (d) 5 internally in the ratio 3 : 5. 2. To divide a line segment AB in the ratio 7 : 5, 6. Draw two tangents at the end points of the first a ray AX is drawn, so that ∠BAX is acute diameter of a circle of radius 3.5 cm. Are these angle and then at equal distance points are tangents parallel? marked. Then, the minimum number of these Long Answer Type Questions points is 7. Draw two concentric circles of radii 3 cm and (a) 5 (b) 35 5 cm. Taking a point on outer circle construct the pair of tangents to the other. Measure the (c) 7 (d) 12 length of a tangent and verify it by actual calculation. 3. By geometrical construction, it is possible to 8. Draw a line segment AB of length 7 cm. Taking divide a line segment in the ratio 3 : 1 . 3 A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius (a) True (b) False 2 cm. Construct tangents to each circle from the centre of the other circle. (c) Can’t determined (d) None of these 4. To draw a pair of tangents to a circle, which are inclined to each other at an angle of 55°, it is required to draw tangents at the end points of Answers 7. 4 cm For Detailed Solutions Scan the code 1. (a) 2. (d) 3. (a) 4. (c) 6. These are parallel.

88 CBSE Term II Mathematics X (Standard) CHAPTER 05 Applications of Trigonometry In this Chapter... l Line of Sight & Horizontal Line l Angle of Elevation l Angle of Depression Line of Sight Some Important Points The line of sight is the line drawn from the eye of an observer to (i) The angle of elevation of a point P as seen from a the point where the object is viewed by the observer. point O is always equal to the angle of depression of O as seen from P. Horizontal Line Angle of P The line which goes parallel from eye to ground, is called depression A horizontal line. Line of sight of Angle of Elevation Angle The angle of elevation of an object viewed, is the angle formed by O elevation the line of sight with the horizontal, when it is above the horizontal level, i.e. the case when we raise our head to look at the object. Horizontal line P(object) (ii) The angles of elevation and depression are always acute angles. Line of sight Angle of (iii) If the observer moves towards the perpendicular line (tower/building), then angle of elevation O elevation A increases and if the observer moves away from the Horizontal line perpendicular line (tower/building), then angle of elevation decreases. Eye (iv) If the height of tower is doubled and the distance Angle of Depression Horizontal line between the observer and foot of the tower is also OA doubled, then the angle of elevation remains same. The angle of depression of an object viewed, is the angle formed by the line Angle of (v) If the angle of elevation of Sun, above a tower of sight with the horizontal, when it is depression decreases, then the length of shadow of a tower below the horizontal level, i.e. the case increases and vice-versa. when we lower our head to look at the Line of sight object. P(object)

CBSE Term II Mathematics X (Standard) 89 Solved Examples Example 1. In figure, a tightly stretched rope of length In right angled ΔACB, tan 30° = AB 20 m is tied from the top of a vertical pole to the BC ground. Find the height of the pole, if the angle made by the rope with the ground is 30°. ⇒ 1 =h 3 30 [CBSE 2020 (Standard)] ⇒ h = 30 × 3 B 33 20 m [multiply numerator and denominator by 3] 30º = 30 × 3 AC 3 Sol. Let AB = h be the height of the pole. = 10 3 m Given, length of rope, BC = 20 m B Hence, height of the tower is 10 3 m. Example 3. The ratio of the length of a vertical rod and the length of its shadow is 1 : 3. Find the angle of hm 20 m elevation of the Sun at that moment? A 30º [CBSE 2020 (Standard)] C Sol. Let AB be the vertical rod and BC be its shadow and θ be the angle of elevation of the Sun. A In right angled ΔACB, Vertical rod sin 30° = Perpendicular Hypotenuse q BC ∴ 1= h 2 20 Shadow ⇒ h = 20 = 10 m We have, AB : BC =1 : 3 2 Hence, height of the pole is 10 m. Example 2. In figure, the angle of elevation of the top Let AB = x, then BC = x 3 of a tower from a point C on the ground, which is In ΔABC, tan θ = AB BC 30 m away from the foot of the tower, is 30°. Find the height of the tower. [CBSE 2020 (Standard)] tan θ = x 1 = A ⇒ x3 3 ⇒ tan θ = tan 30° ⇒ θ = 30° 30° C Example 4. A vertical tower stands on a horizontal B 30 m Sol. Let height of a tower be AB = h m plane and is surmounted by a vertical flag-staff of height 6 m. At a point on the plane, the angle of A elevation of the bottom and top of the flag-staff are 30° and 45°, respectively. Find the height of the tower. (take, 3 = 1.73) [CBSE 2020 (Standard)] Sol. Let BC = h be the height of the tower, CD = 6 m be the h height of the flag-staff and A is any point on the ground. 30° Consider, AB = x m. B 30 m C Given, the angle of elevation from point A to the points C and D are ∠CAB = 30° and ∠DAB = 45°.

90 CBSE Term II Mathematics X (Standard) D Also, let AB = x m. 6m In right angled ΔABD, C tan 60° = Perpendicular = BD Base AB ⇒ 3 = BC + CD [Q tan 60° = 3] h x ⇒ 3 = h + 1.6 x A 30° 45° B xm 3x = h + 1. 6 In right angled ΔABC, ⇒ h = 3x − 1.6 …(i) tan 30° = BC AB In right angled ΔCBA, ⇒ 1 =h tan 45° = BC 3x AB ⇒ x= 3hm …(i) ⇒ 1 = h [Q tan 45° = 1] x Now, in right angled ΔABD, ⇒ x=h tan 45° = BD AB On putting x = h in Eq. (i), we get ⇒ 1= 6+ h h = 3 h − 1.6 x [QBD = BC + CD = 6 + h] ⇒ h ( 3 − 1) = 1.6 ⇒ x=6+ h …(ii) ⇒ h = 1.6 × 3 + 1 [rationalising] ⇒ 3h = 6 + h [from Eq. (i)] ( 3 − 1) 3 + 1 ⇒ h( 3 − 1) = 6 = 1 . 6( 3 + 1) [Q ( a + b)(a − b) = a 2 − b2] ( 3 )2 − (1)2 ⇒ h = 6 × 3+1 (rationalisation) ( 3 −1) 3 + 1 = 1 .6 ( 3 + 1) 2 = 6( 3 + 1) [Q(a − b)(a + b) = a2 − b2] ( 3 )2 − (1)2 = 0.8 (1.73 + 1) = 6(1.73 + 1) = 6 × 2.73 = 0.8 (2.73)= 2.184 m 3 −1 2 Hence, the height of the pedestal is 2.184 m. = 3 × 2.73 = 8.19 m Example 6. From a point on the ground, the angles of Hence, height of tower is 8.19 m. elevation of the bottom and the top of a transmission Example 5. A statue 1.6 m tall, stands on the top of a tower fixed at the top of a 20 m high building are 45° and 60°, respectively. Find the height of the pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the tower. (use 3 = 1.73 ) same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. Sol. Let AB = 20 m be the height of the building and BC = h m be the height of transmission tower. The angles of elevation (use 3 = 1.73) from a ground point D to the points B and C are Sol. Let BC = h m be the height of the pedestal and CD = 1.6 m ∠ADB = 45° and ∠ADC = 60° C be the length of the statue, which is standing on the pedestal. hm D B C 20 m h 45° 60° A D A xm B In right angled ΔADB, tan 45° = AB Again, let point A be a fixed point on the ground such that AD …(i) the angles of elevation of the top of the statue and bottom of the statue (i.e. top of the pedestal) are ⇒ 1 = 20 AD ∠DAB = 60° and ∠CAB = 45°. ⇒ AD = 20 m

CBSE Term II Mathematics X (Standard) 91 Now, in right angled ΔADC, [Q from Eq. (i)] ⇒ h = 25 3 m tan 60° = AC AD Hence, height of the pole is 25 3 m and distances of the point from the poles are 25 m and 75 m. ⇒ 3 = AB + BC 20 Example 8. The angle of elevation of the top of a ⇒ 20 3 = 20 + h building from the foot of a tower is 30° and the angle of elevation of the top of a tower from the foot ⇒ h = 20( 3 −1) = 20(1.73 −1) of the building is 60°. If the tower is 50 m high, then find the height of the building. = 20 × 0.73 = 14.60 m Hence, height of the transmission tower is 14.6 m. Sol. Let AB = 50 m, CD = h be the height of the tower and building. Then, Example 7. Two poles of equal heights are standing B opposite to each other on either side of the road, which is 100 m wide. From a point between them D on the road, the angles of elevation of the top of the 50 m poles are 60° and 30°, respectively. Find the height hm of the poles and the distance of the point from the poles. [CBSE 2020 (Standard)] A 30º 60º C Sol. Let AB = 100 m be the width of the road. On both sides of ∠CAD = 30° and ∠ACB = 60° the road, poles AE = BD = h m are standing. Let C be any point on AB such that from point C, angles of elevation are In right angled ΔACD , tan 30° = CD ⇒ 1 = h ∠BCD = 60° and ∠ACE = 30° AC 3 AC ED ⇒ AC = 3 h …(i) h hm In right angled ΔCAB, 30º 60º tan 60° = AB A (100 – x) m C x m AC 100 m B ⇒ 3 = 50 AC Let BC = x m , then AC = AB − BC = (100 − x) m ⇒ 3 = 50 [from Eq. (i)] 3h In right angled ΔCAE, tan 30° = Perpendicular ⇒ h = 50 = 16.67 m Base 3 ∴ 1 = AE ⇒ 1 = h Hence, the height of the building is 16.67 m. 3 AC 3 (100 − x) Example 9. From the top of a 7 m high building the ⇒ h = (100 − x) 3 angle of elevation of the top of a tower is 60° and ⇒ h 3 = 100 − x …(i) the angle of depression of its foot is 45°. Determine …(ii) and in right angled ΔBCD, the height of the tower. [CBSE 2020 (Standard)] tan 60° = BD Sol. Let AB = 7 m be the height of the building and EC be the BC height of tower. ⇒ 3=h A is the point from where elevation of tower is 60° and the x angle of depression of its foot is 45°. ⇒ h = 3x Here, EC = DE + CD Also, CD = AB = 7 m Put h = 3x in Eq. (i), we get and BC = AD 3x × 3 = 100 − x E ⇒ 3x + x = 100 ⇒ 4x = 100 A 60° D ⇒ x = 100 = 25 m 45° 4 7m ∴ BC = 25 m and AC = 100 − x = 100 − 25 = 75 m Put x = 25 in Eq. (i), we get ⇒ h 3 = 100 − 25 ⇒ h = 75 × 3 = 75 3 B 45° C 33 3

92 CBSE Term II Mathematics X (Standard) Now, in right angled ΔABC, ⇒ 3 = CʹB + BM PM tan 45° = AB BC ⇒ 3 = 10 + h + 10 [from Eq. (i)] 3h 7 ⇒ 1 = BC ⇒ 3h = 20 + h ⇒ 2h = 20 ⇒ BC = 7 m ⇒ h = 10 m Now, the height of the cloud from the surface of lake Also, in right angled ΔADE, =BC tan 60° = DE = BM + h AD = 10 + 10 = 20 m ⇒ 3 = DE [Q AD = BC] Example 11. From a point on a bridge across a river, 7 the angles of depression of the banks on opposite ⇒ DE = 7 3 m sides of the river are 30° and 45°, respectively. If the bridge is at a height of 30 m from sea level, then ∴ Height of the tower, EC = DE + CD find the width of the river. (use 3 = 1 . 73) = (7 3 + 7) m Sol. Let A be a point on the bridge and points B and D are on the = 7( 3 + 1) m opposite side of the banks. Then, angles of depression from point A to the opposite banks are Hence, height of the tower is 7( 3 + 1) m. ∠EAB = 30° and ∠FAD = 45° Example 10. If the angle of elevation of a cloud from a ⇒ ∠CBA = 30° and ∠CDA = 45° point 10 m above a lake is 30° and the angle of [Q alternate angles are equal] depression of its reflection in the lake is 60°, find the height of the cloud from the surface of lake. E A F 30º 45º Bridge [CBSE 2020 (Standard)] 30 m Sol. Let AB be the surface of the lake and P be the point of observation such that AP = 10 m. Let C be the position of the cloud and Cʹ be the reflection in the lake, then CB = CʹB. C h 30º 45º B CD M (River bank) 10 m (Opposite B river bank) P 30° In right angled ΔACB, 10 m 60° AC tan 30° = BC A ⇒ 1 = 30 3 BC (10 +h)m ⇒ BC = 30 3 m Cʹ and in right angled ΔACD, tan 45° = AC Let CM = h, then CB = 10 + h CD ⇒ CʹB = 10 + h In right angled ΔCMP, ⇒ 1 = AC 30 tan 30° = CM PM ⇒ AC = 30 m Hence, width of the river is ⇒ 1=h 3 PM BD = BC + CD = 30 3 + 30 ⇒ PM = 3h ...(i) = 30 × 1.73 + 30 In right angled ΔPMCʹ, = 51.9 + 30 tan 60° = CʹM = 81.9 PM Hence, width of the river is 81.9 m.