Question

The area of a right angled triangle is $20c{m}^2$ and one of the sides containing the right angle is $4$ cm. Then the altitude on hypotenuse is

• A. $8$ cm
• B. $10$ cm
• C. $\frac{10}{\sqrt{41}}$ cm
• D. $\frac{20}{\sqrt{29}}$ cm

Answer 1

The correct option is D $\frac{20}{\sqrt{29}}$ cm

Let length of the other perpendicular side be $a$ cm and altitude on the hypotenuse is $p$ cm.
$\Rightarrow \frac{1}{2}\times 4\times a=20$
$\Rightarrow a=10$ cm
$\therefore$ Hypotenuse $=\sqrt{a^2+4^2}=\sqrt{{10}^2+4^2}=2\sqrt{29}$
Area of the triangle $=\frac{1}{2}\times 2\sqrt{29}\times p=20$ sq. cm
$\therefore p=\frac{20}{\sqrt{29}}$