The sides of a triangle are 21, 20 and 13 cm. The given triangle is divided into two triangles by the perpendicular on the longest side from the opposite vertex. What is value of 30% area of the smaller triangle formed?

Let the lengths of BD, CD and AD be $$x,\ y\ \&\ h$$, respectively.

=> $$ BD+CD = x+y=21$$.....(1) 

As AD is altitude, using Pythagoras' theorem, we get

$$13^2=x^2+h^2$$ .....(2)

$$20^2=y^2+h^2$$.....(3)

Eq (3) - (2) gives 

$$20^2-13^2=y^2-x^2$$ 

$$231=\left(y-x\right)\ \left(y+x\right)$$

 $$\left(y-x\right)=\frac{231}{x+y}=\frac{231}{21}=\ 11$$......(4)

Adding Eq(1) & (4), we get $$2y=32$$  => $$y=16 cm$$ 

we get $$x+16=21$$ => $$x=5 cm$$

From Eq(2) => $$13^2=5^2+h^2$$ 

$$h^2=13^2-5^2$$ => $$h=12 cm$$

The area of the smaller triangle ABD = $$\ \frac{\ BD\times\ AD}{2}$$

= $$\ \frac{\ 5\times\ 12}{2}$$ = $$30 cm^2$$

30% of area of smaller triangle = $$30\%\ \times\ 30=9\ cm^2$$

Option (C) is the answer