# An altitude of a triangle is five-thirds thelength of its corresponding base.

Question:

An altitude of a triangle is five-thirds thelength of its corresponding base. If the altitude be increased by 4 cm and the base decreased by 2 cm, the area of the triangle remains the same. Find the base and the altitude of the triangle.

Solution:

Let the length of the base of the triangle be $\mathrm{x} \mathrm{cm}$.

Then, its altitude will be $\frac{5}{3} x \mathrm{~cm}$.

Area of the triangle $=\frac{1}{2}(x)\left(\frac{5}{3} x\right)=\frac{5}{6} x^{2}$

$\therefore \frac{1}{2}(x-2)\left(\frac{5}{3} x+4\right)=\frac{5}{6} x^{2}$

$\Rightarrow\left(\frac{x-2}{2}\right)\left(\frac{5 x+12}{3}\right)=\frac{5 x^{2}}{6}$

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

$\Rightarrow \frac{5 x^{2}+12 x-10 x-24}{6}=\frac{5 x^{2}}{6}$

$\Rightarrow 5 x^{2}+2 x-24=5 x^{2} \quad$ (cancelling the denominators from both the sides since they are same)

$\Rightarrow 5 x^{2}-5 x^{2}+2 x=24$

$\Rightarrow 2 x=24$

$\Rightarrow x=\frac{24}{2}=12 m$

Therefore, the base of the triangle is $12 \mathrm{~m}$.

Altitude of the triangle $=\frac{5}{3} x=\frac{5}{3}(12)=20 \mathrm{~m}$