Question

An altitude of a triangle is $\frac{5}{3}$ times the length of its corresponding base. If the altitude is increased by $4cm$ and the base decreased by $2cm$, the area of the triangle remains the same. Find the altitude and the base of the triangle.

Answer 1

$\Rightarrow$ Let base of the triangle be $xcm$

$\Rightarrow$ Then, corresponding altitude $=\frac{5}{3}xcm$

$\Rightarrow$ Area of the triangle $=\frac{1}{2}\times base\times height$

$=\frac{1}{2}\times x\times \frac{5x}{3}$

$=\frac{5x^2}{6}c{m}^2$ ------- ( 1 )

$\Rightarrow$ Base of the new triangle $=(x-2)cm$

$\Rightarrow$ Corresponding altitude $=\frac{5x}{3}+4=\frac{5x+12}{3}cm$

$\Rightarrow$ Area of new triangle $=\frac{1}{2}\times (x-2)\times \frac{5x+12}{3}=\frac{(x-2)(5x+12)}{6}$ ------ ( 2 )

Area of the original triangle = Area of the new triangle

$\therefore$ $\frac{5x^2}{6}=\frac{(x-2)(5x+12)}{6}$

$\therefore$ $5x^2=(x-2)(5x+12)$

$\therefore$ $5x^2=5x^2+12x-10x-24$

$\therefore$ $2x=24$

$\therefore$ $x=12cm$

$\therefore$ Base of an original triangle is $12cm$ and altitude $=\frac{5}{3}\times 12=20cm$