Question

The area of a right angle triangle is $600$ sq.cm .If the base of the triangle exceeds the altitude by $10$ cm ,find the base?

Answer 1

Given base of triangle exceeds altitude by $10$ cm

Let the altitude be $x$ cm

Therefore, base $=(x+10)$ cm

Area of triangle $=\frac{1}{2}\times$ base $\times$ altitude

$\Rightarrow 600=\frac{1}{2}\times (x+10)\times x$

$\Rightarrow 1200=x^2+10x$

$\Rightarrow x^2+10x-1200=0$

$\Rightarrow x^2+40x-30x-1200=0$

$\Rightarrow x(x+40)-30(x+40)=0$

$\Rightarrow (x+40)(x-30)=0$

$\therefore x=-40,x=30$

But $x$ cannot be negative.

Hence, $x=30$cm

That is altitude$=30$cm

$\therefore$ Base$=40$cm