The altitude of a triangle is three-fifth the length of its corresponding base.

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.

Options:

A . 357

B . 257

C . 267

D . 367

Answer: Option C
:
C

Let the base of the triangle be x c m .

Then, the corresponding altitude of the triangle would be \frac{3 x}{5} c m .

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

New base = x - 2 c m

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

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

\Rightarrow \frac{3 x^{2}}{5} = \frac{3 x^{2}}{5} + 4 x - \frac{6 x}{5} - 8

\Rightarrow 4 x - \frac{6 x}{5} = 8 \Rightarrow \frac{20 x - 6 x}{5} = 8

\Rightarrow 14 x = 40

\Rightarrow x = \frac{40}{14} = \frac{20}{7} \Rightarrow x = 2 \frac{6}{7} cm

Hence, base of the triangle is 2 \frac{6}{7} c m long.

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