Question

The base of a triangle is 4 cm longer than its altitude. If the area of the triangle is 48 sq. cm, then find its base and altitude.

Answer 1

Let the altitude of the triangle be x cm.

By the given condition, the base of the triangle is (x + 4) cm.

Now, the area of the triangle = $\frac{1}{2}\left(base\right)\times \left(height\right)$

By the given condition $\frac{1}{2}(x+4)\left(x\right)=48$

$\to x^2+4x-96=0\to (x+12)(x-8)=0$

$\to x=-12$ or $8$

But x = 12 - is not possible (since the length should be positive)

Therefore, x = 8 and hence, $x+4=12$.

Thus, the altitude of the triangle is 8 cm and the base of the triangle is 12 cm.