Question

the altitude of triangle is three-fifth of the length of the corresponding base. if the altitude is decreased by, 4 CM and the corresponding base is increased by 10 CM, the area of the triangle remains same. find the base and the altitude of the triangle.?

Answer 1

Hope this will help you

Answer 2

Base of triangle is 20 cm and altitude is 12 cm.

Step-by-step explanation:

1. First consider old triangle, which base and altitude are

  Let

  Base = X  (cm)     ...a)

  $\mathbf{Altitude= \frac{3}{5}X=\frac{3X}{5}}$     ...b)      (Given)

  Then

 $\mathbf{Area of old triangle= \frac{1}{2}\times Base\times Altitude=\frac{1}{2}\times X\times \frac{3X}{5}}$

Which can be write as

$\mathbf{Area of old triangle=\frac{1}{2}\times \frac{3X^{2}}{5}}$    ...1)

2. Now consider new triangle, which base and altitude are

  Base = (X+10)  (cm)       (Given)

  $\mathbf{Altitude=\left ( \frac{3X}{5}-4 \right )}$       (Given)

  Then

 $\mathbf{Area of new triangle= \frac{1}{2}\times Base\times Altitude=\frac{1}{2}\times (X+10)\times \left ( \frac{3X}{5}-4 \right )}$  

Which can be write as

$\mathbf{Area of new triangle=\frac{1}{2}\times (X+10)\times \left ( \frac{3X}{5}-4 \right )}$        ...2)

3. It is given that

Area of new triangle = Area of old triangle     ...3)

So form equation 1) and equation 2)

$\mathbf{\frac{1}{2}\times (X+10)\times \left ( \frac{3X}{5}-4 \right )=\frac{1}{2}\times \frac{3X^{2}}{5}}$

$\mathbf{(X+10)\times \left ( \frac{3X}{5}-4 \right )=\frac{3X^{2}}{5}}$

$\mathbf{\frac{3X^{2}}{5}-4X+10\times \frac{3X}{5}-40=\frac{3X^{2}}{5}}$

$\mathbf{-4X+6X-40=0}$

So on solving

X = 20

4. So from equation a) and equation b)

    Base = X =20 (cm)     ...a)

  $\mathbf{Altitude=\frac{3X}{5}=12(cm)}$