Question 21

# In a triangle, the two longest sides are 13 cm and 12 cm. The angles of the triangle are in arithmetic progression. The radius of the circle inscribed in this triangle is :

Solution

Let x be the other side.

It is given that the angles are in AP, so the angles are 60-x, 60, 60+x

cos 60=$$\ \frac{\ x^2+169-144}{2\times\ x\times\ 13}$$

$$x^2-13x+25=0$$

x=10.6, 2.35

Inradii (r) = Area /Semi perimeter

when x= 10.6

$$\ \frac{\ \ \frac{\ 1}{2}\cdot12\cdot10.6*sin 60}{17.8}$$ = 3.09

when x= 2.34

$$\ \frac{\ \ \frac{\ 1}{2}\cdot12\cdot2.34*sin 60}{13.67}$$ = 0.9

x=1cm(approx)