Question

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

• A. The base of the triangle is $12cm$ and altitude is $20cm$.
• B. The base of the triangle is $4cm$ and altitude is $34cm$.
• C. The base of the triangle is $16cm$ and altitude is $12cm$.
• D. The base of the triangle is $8cm$ and altitude is $32cm$.

Answer 1

The correct option is A The base of the triangle is $12cm$ and altitude is $20cm$.

Let the base of the triangle be $x$ cm.

Then, the altitude of the triangle $=\frac{5x}{3}$

So, area of the triangle $=\frac{1}{2}\times base\times altitude$

$=\frac{1}{2}\times x\times \frac{5}{3}x=\frac{5}{6}{x}^2$ ...(1)

On increasing the altitude by $4cm$ and the decreasing base by $2cm$, the area remains the same.

Therefore, $\frac{1}{2}\times (x-2)\times (\frac{5x}{3}+4)=\frac{5}{6}{x}^2$ ...[using (1)]

$⟹\frac{1}{2}\times (x-2)\left(\frac{5x+12}{3}\right)=\frac{5}{6}{x}^2$

$⟹(x-2)(5x+12)=5x^2$

$⟹5x^2-10x+12x-24=5x^2$

$⟹2x-24=0$

$⟹2x=24$ or $x=12$.

Now, altitude of the triangle $=\frac{5x}{3}=\frac{5\times 12}{3}=20cm$

Hence, the base of the triangle is $12cm$ and altitude is $20cm$.