Question

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

• A. 8 cm
• B. 10 cm
• C. $\frac{10}{\sqrt{4}1}$ cm
• D. $\frac{20}{\sqrt{2}9}$ cm

Answer 1

The correct option is D $\frac{20}{\sqrt{2}9}$ cm

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