Question

The altitude of a triangle is three-fifth the length of its 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 of the triangle in cm. [Area of triangle = $\frac{1}{2}\times \text{base}\times \text{height}$]

• A. 20/14
• B. 30/14
• C. 40/14
• D. 50/14

Answer 1

The correct option is C

40/14

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

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

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

New base = $x-2$

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

Since the area remains the same, $\frac{1}{2}\times x\times \frac{3x}{5}=\frac{1}{2}\times (\frac{3x}{5}+4)\times (x-2)$

$\frac{3x^2}{5}=\frac{3x^2}{5}+4x-\frac{6x}{5}-8$
$\frac{20x-6x}{5}=8$
$20x-6x=40$
$14x=40$
$x=\frac{40}{14}=\frac{20}{7}$

The base of the triangle is $\frac{20}{7}$ cm long.