Question

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

• A. $3\frac{5}{7}$
• B. $2\frac{5}{7}$
• C. $2\frac{6}{7}$
• D. $3\frac{6}{7}$

Answer 1

Answer: (C)

$2\frac{6}{7}$

$\text{Let the base of the triangle be}xcm.$

$\text{Then, the corresponding altitude}\text{of the triangle would be}\frac{3x}{5}cm.$

$\text{If altitude is increased by 4 cm,}\text{The new altitude}=\frac{3x}{5}+4cm$

$\text{New base}=x-2cm$

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

$\text{Since the area remains the same,}\frac{1}{2}\times x\times \frac{3x}{5}=\frac{1}{2}\times (\frac{3x}{5}+4)\times (x-2)$
$\Rightarrow \frac{3x^2}{5}=\frac{3x^2}{5}+4x-\frac{6x}{5}-8$
$\Rightarrow 4x-\frac{6x}{5}=8\Rightarrow \frac{20x-6x}{5}=8$
$\Rightarrow 14x=40$
$\Rightarrow x=\frac{40}{14}=\frac{20}{7}\Rightarrow x=2\frac{6}{7}\text{cm}$
$\text{Hence, base of the triangle is}2\frac{6}{7}cm\text{long.}$