1) there is a regular polygon with r/R = 1/2 2) there is a regular polygon with r/R = 1/√2 3) there is a regular polygon with r/R = 2/3 4)... View Article
1) c + a 2) a + b + c 3) a + b 4) b + c Solution: Given ∠C = π/2 Let r is the inradius and R is the circumradius of the ∆ABC. The triangle... View Article
1) a2b2c2 2) 2 a2b2c2 3) 4 a2b2c3/R2 4) a2b2c2/8R3 Solution: Let p1 be the altitude AD, p2 be the altitude BE and p3 be the altitude CF.... View Article
1) 4√3 2) 12 - 7√3 3) 12 + 7√3 4) None of these Solution: Given the triangle is isosceles with one angle 1200. Let A = 1200 Then b = c B =... View Article
1) 15/4 2) 11/5 3) 16/7 4) 16/3 Solution: Given a : b : c = 4 : 5 : 6 Let R be the radius of the circumcircle and r be the radius of the... View Article
1) tan π/n : π/n 2) cos π/n : π/n 3) sin π/n : π/n 4) cot π/n : π/n Solution: Let r be the radius of the circle and a be the side of... View Article
1) 16 2) 18 3) 24 4) 22 Solution: Let s - a = 2k - 2 s - b = 2k s - c = 2k + 2, k ∈ I, k > 1 Adding we get, s = 6k So, a = 4k + 2 b =... View Article
1) HP 2) Arithmetico - Geometric Progression 3) AP 4) GP Solution: Consider ∆BDA, cos (90 - B) = AD/c AD = c sin B Similarly, BE = a sin... View Article
1) √2 2) (√3 + 1)/2 3) (√3 - 1)/2 4) 1 Solution: Given a = 3, b = 5 and c = 4 Here b2 = a2 + c2 So Pythagoras theorem is satisfied. Thus B... View Article
1) tan h-1 (sin θ) 2) tan h-1 (∞) 3) (1 / 2) tan h-1 (sin θ) 4) None of these Solution: (3) (1 / 2) tan h-1 (sin θ) tan-1 [cos θ + i sin θ]... View Article