Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C respectively. Suppose \(\mathrm{a}=6, \mathrm{~b}=10\) and the area of the triangle is \(15 \sqrt{3}\). If \(\angle \mathrm{ACB}\) is obtuse and if r denotes the radius of the incircle of the triangle, then \(\mathrm{r}^{2}\) is equal to

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Properties of Triangle

Consider a triangle ABC and let a, b and c denote the lengths of the sides opposite to vertices A, B and C respectively. Suppose

a = 6, b = 10

and the area of the triangle is

153

. If

∠ ACB

is obtuse and if r denotes the radius of the incircle of the triangle, then

r^{2}

is equal to

Answer: 3

Given

a = 6

b = 10

, and area

Δ = 153

.

Using the area formula:

Δ = \frac{1}{2} ab sin C ⟹ 153 = \frac{1}{2} \cdot 6 \cdot 10 \cdot sin C

sin C = \frac{30 3}{60} = \frac{3}{2}

Since

∠ C

is obtuse,

C = 12 0^{\circ}

.

Using the cosine rule:

c^{2} = a^{2} + b^{2} - 2 ab cos C = 36 + 100 - 2 \cdot 6 \cdot 10 \cdot cos 12 0^{\circ}

= 136 - 120 \cdot (- \frac{1}{2}) = 136 + 60 = 196

c = 14

The semi-perimeter is:

s = \frac{a + b + c}{2} = \frac{6 + 10 + 14}{2} = 15

The inradius is:

r = \frac{Δ}{s} = \frac{15 3}{15} = 3

Therefore:

r^{2} = (3)^{2} = 3

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Integer answer