Question

The area of a right angled triangle is $600{cm}^2$. If the base of the triangle exceeds the altitude by $10cm$, find the dimensions of the triangle.

Answer 1

Let the altitude $BC$ of right-angled triangle $ABC$ be $x$ cm.

Then Base $=BC=(x+10)$cm

Area $=\frac{1}{2}(Base\times Height)$

Area $=\frac{1}{2}(x+10)x{cm}^2$

$\Rightarrow \frac{1}{2}x(x+10)=600$ ......($\because Area=600{cm}^2$)

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

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

$\therefore$ $Hypotenus{e}^2$ $={30}^2+{40}^2=2500$

$\therefore$ Hypotenuse$=50cm$

Hence, Base $=(30+10)cm=40cm$ , Altitude $=30cm$ and Hypotenuse$=50cm$