Question

The altitude of a triangle is two third the length of its corresponding base.If the altitude is increased by 4 cm and base is decreased by 2 cm, the area of triangle remains the same.Find the base and the altitude of the triangle

Answer 1

Let the base $=x,$ then altitude $=\frac{2x}{3}$
area of triangle $=\frac{1}{2}\times$ base$\times$ height $=\frac{1}{2}\times x\times \frac{2x}{3}=\frac{x^2}{3}$

Now altitude $=(\frac{2x}{3}+4),$ base $=(x-2)$
area of the triangle $=\frac{1}{2}\times (x-2)\times (\frac{2x}{3}+4)$

by question,
$\frac{x^2}{3}=\frac{1}{2}\times (x-2)\times (\frac{2x}{3}+4)$

$\frac{x^2}{3}=\frac{1}{2}\times (x-2)\times \left(\frac{2x+12}{3}\right)$

$\frac{x^2}{3}=\frac{1}{2}\times (x-2)\times \frac{2(x+6)}{3}$

$\frac{x^2}{3}=\frac{1}{3}\times (x-2)\times (x+6)$

$3x^2=3(x^2+4x-12)$

$-12x=-36$

$x=3$
base $=3cm$
altitude $=\frac{2x}{3}=\frac{6}{3}=2cm$