Question

What is the area of a triangle for one of the legs being 3in and the hypotenuse being 9in

Answer 1

The area of a triangle for one of the legs being 3 inches and the hypotenuse being 9 inches is 12.727 square inches

Solution:

Given that to find area of a triangle for one of the legs being 3 inches and the hypotenuse being 9 inches

From given information,

Let "c" = hypotenuse = 9 inches

Let "a" = length of one of the leg of triangle = 3 inches

To find: area of triangle

The area of triangle when hypotenuse and length of one side of triangle is given:

$A = \frac{1}{2} a \sqrt{c^2 - a^2}$

Where, "c" is the length of hypotenuse

"a" is the length of one side of triangle

Substituting the given values we get,

$A = \frac{1}{2} \times 3 \times \sqrt{9^2 - 3^2}$

$A =\frac{1}{2} \times 3 \times \sqrt{81-9}\\\\A =\frac{1}{2} \times 3 \times \sqrt{72}\\\\A =\frac{1}{2} \times 3 \times 8.48528\\\\A = \frac{1}{2} \times 25.45584\\\\A = 12.727$

Thus area of triangle is 12.727 square inches