Properties of Triangle Exercise # 4

EXERCISE # 4 a2 = 4x2 sin2 A/2  a = 2x sin A/2  Old IIT-JEE Questions Q.1 Sol.  a = 2x sin 60° = 3x Thus required ratio is In a triangle ABC, A = 45º, b = 10 cm and c = 10 2 cm, then [REE 2000] (A) a = 10 cm (B) a = 10 2 cm (C) C = 90º (D) B = 45º [A, C, D] a 3x 3   a  b  c x  x  3x 2  3 Q.3 In any  ABC having sides a, b, c opposite to angles A, B, C respectively [IIT Scr. 2005] A BC (A) a sin   = (b  c) cos 2 2   A BC (B) a cos = (b  c) sin 2 2 BC A (C) a cos = (b  c) sin 2 2 BC A (D) a sin = (b  c) cos 2 2 Sol. [A] A sin B  sin C A bc  cos = cos sin A a 2 2 BC BC 2 cos sin 2 2 cos A = sin B  C = A A 2 2 2 sin cos 2 2 L.H.S. option A is correct. (B) Clearly B is wrong. A a (C)  cos bc 2 sin A A = cos sin B  sin C 2 A A 2 cos sin A sin A 2 2 = cos = BC BC 2 2 cos B  C 2 sin cos 2 2 2 Clearly (C) and (D) are wrong. Q.4 Internal angle bisector of A of triangle ABC, meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at P and the side AB at Q. If a, b, c represent the sides of ABC then [IIT-2006] 2bc A (A) AD = cos bc 2 4bc A (B) PQ = sin bc 2 (C) The triangle APQ is isosceles (D) AP is HM of b and c Sol. [A, B, C, D] b = 10, c = 10 2 , A = 45°  cos A =  1 = b2  c2  a 2 2bc 100 200 – a 2 2 2.10.10 2  a = 100  a = 10 Again from sine rule 2 a b c = = sin A sin B sin C  10 2 10 10 2 = = 1 sin B sin C  sin B = 1  B = 45° 2 and sin C = 1  C = 90° Option A, C, D are correct. Q.2 If the angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the longest side to the perimeter is [IIT 2003] (A) 3 : (2  3 ) (B) 1 : 6 (C) 1 : 2 + 3 (D) 2 : 3 Sol.[A] Here, ratio of angles are 4 : 1 : 1  4x + x + x = 180°  x = 30°  A = 120°, B = C = 30° Thus the ratio of longest side to perimeter C 30° a b 120° 30° A c B a = abc Let b=c=x  a2 = b2 + c2 – 2bc cos A  a2 = 2x2 – 2x2 cos A = 2x2(1 – cos A) Q. 6 A/2 A/2 A P B D Sol. [4] By geometry in APQ, we have AP = AQ, APQ is isosceles Now ar (ABC) = ar (ABD) + ar(ADC) 1 1 A 1  bc sin A = c.(AD) sin + b. (AD) 2 2 2 2 A sin 2 A 2bc cos 2  AD = bc A Also AD = AP cos 2 2 bc  AP = = H.M. of b and c. bc Again PQ = 2DP= 2.AD tan A/2 A sin 2 bc A 2  PQ = 2. cos . bc 2 cos A 2 A 4 bc  PQ = sin bc 2 A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle, then [IIT-2008] 1 1 2 (A) + < PS ST QS SR (B) 1 1 + > PS ST 2 QS SR 4 1 1 (C) + < PS ST QR 4 1 1 (D) + > PS ST QR Sol. A C Q Q.5 Let ABC and ABC be two non-congruent triangles with sides AB = 4, AC = AC = 2 2 and angle B = 30°. The absolute value of the difference between the areas of these triangles, is............ [IIT- 2009] [B, D] By using A.M  G.M. inequality, we get the answers 4  2 B  2  30° C C D AD = 4 sin 30° = 2 difference of areas = ACC  2 = 2 2 sin  = 45° side CC = 2 (2 2 cos 45°) = 4 1 ACC = . 4. 2 = 4 2 Q.7  6 and let a, b and c denote the lengths of the sides opposite to A, B and C respectively. The value(s) of x for which a = x2 + x +1, b = x2 – 1 and c = 2x + 1 is (are) [IIT-2010] (A)  2  3 (B) 1+ 3 Let ABC be a triangle such that  ACB =   (C) 2  3 Sol. (D) 4 3 In a triangle all sides are positive. A c b a B C c>0  x2 – 1 > 0  x < – 1 or x > 1 b>0  2x + 1 > 0 1 x>– 2 from (1) & (2) x 1 Now, a – b  x2 + x + 1 – 2x – 1  x2 – x  x(x – 1) as, x > 1 (from (3)) x(x – 1) > 0 a>b Also a – b = x + 2 > 0  a is the greatest side. .....(1) .....(2) .....(3)  A is the largest angle. ON = 2 cos b2  c2  a 2 cos A = 2bc = = ( x 2  1) 2  (2x  1) 2  ( x 2  x  1) 2 2( x  1)(2x  1) 2     OM + ON = 2  cos  cos  k 2k    3 1 3  = 2 cos cos 2 4k 4k 3 1 3  = cos cos 4 4k 4k 1 3  3 1 × = cos cos 4k 4k 2 2 2 comparing,   cos = cos 4k 12   =  k=3 4k 12 [k] = 3 2 2( x 2  1)(2x  1) = = ( x 2  1) 2  (3x  2  x 2 )(x – x 2 ) 2( x 2  1)(2x  1) ( x 2  1) 2  ( x )(x 2  3x  2)(x – 1) 2( x 2  1)(2x  1) = ( x 2  1) 2  ( x )(x  1)(x  2)(x  1) 2(2x  1) = ( x 2  1)  ( x )(x  2) 2(2x  1) = (2x  1) 1 x 2  1  x 2  2x = =– 2(2x  1) 2 2(2x  1) Q.9 1 2 A = 120º  B = 30º B = C = 30º  b=c  x2 – 1 = 2x + 1  x2 – 2x – 2 = 0 cos A = –  x=1+ 3 x=1– 3 & Q.8 Two parallel chords of a circle of radius 2 are at a distance 3 + 1 apart. If the chords subtend at 2  the center, angles of and , where k > 0, k k then the value of [k] is [Note : [k] denotes the largest integer less than or equal to k] [IIT-2010] (A) Sol.  k A In ONQ, = 2 M B 2  3   45  (C)   (D)    4   4  45 4 [C] b c Q 2 O 2 (B) P ....(1) k 2 3 4 Q N 7 2 5 , where a, b and c are the lengths of the 2 sides of the triangle opposite to the angles at P, Q 2 sin P  sin 2P and R respectively. Then equals 2 sin P  sin 2P [IIT-2012] Sol.[3] In OMB P Let PQR be a triangle of area  with a = 2, b = and c = Neglect using (3)  OM = 2 cos 2k .....(2) (1) + (2) ( x  1)  (2x  1  x 2  x  1)(2x  1  x 2  x  1) 2  k R a a2   b  7 / 2  s = 4 c  5 / 2  2 sin P  2 sin P cos P 1  cos P = 2 sin P  2 sin P cos P 1  cos P 1 3 4.2. . = 2 2 6 = tan2 P 2 2  =     1  3     (s  b)(s  c)  3 2 2 =    =  s(s  a )  32 4(2)  9   3  =    =  16.6   4  2 EXERCISE # 4 Q.No. Ans. 6. 4 7. (B) 8. 3 9. (C) 1 A,C,D 2 A 3 A 4 A,B,C,D 5 B,D 2