Question

The area of a right angle triangle is $20c{m}^2$ and one of the sides containing the right triangle is 4 cm Then the altitude on the 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 $\frac{20}{\sqrt{29}}cm$

Given the area of right angle triangle is 20 sq cm and one side is 4 is 4 cm

we know that area of right angle triangle=$\frac{1}{2}hight\times base$

$\therefore 20=\frac{1}{2}\times h\times 4\Rightarrow 20=2h\Rightarrow h=10$ cm

Then hypotenuse =$\sqrt{(10{)}^2+(4{)}^2}=\sqrt{100+16}=\sqrt{116}$

Let the altitude of hypotenuse is x

Then area of triangle=$\frac{1}{2}\times \sqrt{116}\times x=\sqrt{29}x$

But area of triangle =20 sqcm

$\therefore \sqrt{29}x=20\Rightarrow x=\frac{20}{\sqrt{29}}$ cm

So altitude of hypotenuse=$\frac{20}{\sqrt{29}}$cm