Question

The altitude of a triangle is two-thirds the length of its corresponding base. If the altitude is increased
by 4 cm and base is decreased by 2 cm, the area of the triangle remains the same. Find the base and the
altitude of the triangle​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

Let the base of right angle triangle be '3x' cm

Now,

According to statement,

The altitude of a triangle is two-thirds the length of its corresponding base.

This implies, altitude of right angle triangle be '2x' cm.

We know,

• Area of triangle is given by

$\rm :\implies\:Area_{(triangle)} = \dfrac{1}{2} \times base \times height$

$\rm :\implies\:Area_{(triangle)} = \dfrac{1}{2} \times 3x \times 2x$

$\boxed{ \pink{ \rm \: Area_{(triangle)} \: = \: {3x}^{2} }} \: - - - (i)$

According to next condition,

• The altitude is increased by 4 cm

$\rm :\implies\:altitude \: = (2 \: x \: + \: 4 )\: = 2(x + 2) \: cm$

• The base is decreased by 2 cm

$\rm :\implies\:base \: = ( \: 3x - 2 \: ) \: cm$

So,

• Area of triangle is given by

$\rm :\implies\:Area_{(triangle)} = \dfrac{1}{2} \times base \times height$

$\rm :\implies\:Area_{(triangle)} = \dfrac{1}{2} \times 2(x + 2) \times (3x - 2)$

$\boxed{ \pink{\rm :\implies\:Area_{(triangle)} = (x + 2)(3x - 2)}} - (ii)$

On equating equation (i) and equation (ii), we get

$\rm :\implies\: {3x}^{2} = (x + 2)(3x - 2)$

$\rm :\implies\: {3x}^{2} = {3x}^{2} - 2x + 6x - 4$

$\rm :\implies\:4x - 4 = 0$

$\rm :\implies\:4x = 4$

$\rm :\implies\:x = 1$

Hence,

• Base of triangle = 3x = 3 × 1 = 3 cm

• Altitude of triangle = 2x = 2 × 1 = 2 cm