Consider an obtuse angled triangle A B C in which the difference between the largest and the smallest angle is π 2 and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1 . (There are two questions based on PARAGRAPH   "   1   " , the question given below is one of them) Then the inradius of the triangle A B C is
A. A
Let angle A be obtuse angle and let sides be a - d ,   a ,   a + d So, plotting the diagram we get, Now let sides be, a − d , a , a + d Also given angle A − C = π 2 And circum radius R = 1 Now by sine rule we get, a + d sin ⁡ A = a sin ⁡ B = a − d sin ⁡ C = 2 ∵ A = π 2 + C ⇒ sin ⁡ A = sin ⁡ π 2 + C ⇒ sin ⁡ A = cos ⁡ C ⇒ a + d 2 = 1 − sin 2 ⁡ C ⇒ a + d 2 2 = 1 - a - d 2 2 ⇒ 2 a 2 + d 2 4 = 1 ⇒ a 2 + d 2 = 2         . . . ( 1 ) Now using cosine rule we get, cos ⁡ B = ( a − d ) 2 + ( a + d ) 2 − a 2 2 a 2 − d 2 ⇒ 1 - sin 2 B = 2 a 2 + d 2 - a 2 2 a 2 - d 2 ⇒ 1 - a 2 4 = 4 - a 2 2 a 2 - d 2    ∵ a 2 + d 2 = 2 ⇒ a 2 - d 2 2 = 4 - a 2         . . . ( 2 ) From ( 1 )   &   ( 2 ) a 2 = 7 4 , d 2 = 1 4 Area of triangle Δ = a a 2 - d 2 4 α = 7 2 × 6 4 × 4 Now using the formula of inradius we get, r = area of triangle semi perimeter = Δ S ⇒ r = 7 2 × 6 16 3 2 × 7 2 = 4 16 = 1 4 = 0 . 25
Related: Mathematics — Properties of Triangles · All PYQ Banks