Question

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

Answer 1

Let the base be $=x$
Altitude $=\frac{2}{3}x$
Area $=\frac{1}{2}\times x\times \frac{2}{3}x=\frac{x^2}{3}$
New altitude $=\frac{2}{3}x+4$
New base $=x-2$
New area $=\frac{1}{2}\times (x-2)\times (\frac{2}{3}x-4)$

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

As per the given condition,

$x^2=(x-2)(x-6)$

$x^2=x^2-2x-6x+12$

$8x=12$

Base $x=1.5cm$

$Altitude=1cm$

Hence, this is the answer.