# The area of a right triangle is 180 m². The height of the right triangle is 40 m. What is the length of the hypotenuse of the right triangle?

May 27, 2018

Assuming the height is one of the legs, the other leg is $9$ and the hypotenuse is $41$.

#### Explanation:

It's not clear if the height is one of the legs or the altitude to the hypotenuse.

Either way we have $A = \frac{1}{2} b h$ or $b = \frac{2 A}{h} = 2 \cdot \frac{180}{40} = 9$ m.

If $h = 40$ is the height to the hypotenuse, the hypotenuse is $9$ m and we're done. Let's assume $h = 40$ is one of the legs so $b = 9$ is the other leg so

${c}^{2} = {h}^{2} + {b}^{2} = {40}^{2} + {9}^{2} = 1681$

$c = 41$

May 27, 2018

$41 m$

#### Explanation:

$\text{In the right} \triangle A B C$

$\therefore \angle B = {90}^{\circ}$ $\text{given}$

$\therefore B C = 40 m = \text{height given} = o p p o s i t e$

$\therefore \text{Area of a triangle} = \frac{1}{2} b a s e \times h = 180 {m}^{2}$

:."1/2basexx40=180

$\therefore \frac{1}{2} b a s e = \frac{180}{40}$

$\therefore \frac{1}{2} b a s e = 4.5 m$

$\therefore b a s e = 9 m = A B = a \mathrm{dj} a c e n t$

$\therefore \frac{o p p o s i t e}{a \mathrm{dj} a c e n t} = \frac{B C}{A B} = \frac{40}{9} = \tan \angle A = 4.444444444$

$\therefore \angle A = {77}^{\circ} 19 ' 11 ' '$

$\therefore \text{AC= hypotenuse} , \frac{A C}{B C} = \sec {77}^{\circ} 19 ' 11 ' ' = 4.555555556$

$\therefore A C = \sec {77}^{\circ} 19 ' 11 ' ' \times 9 = 41 m = \text{ hypotenuse}$