The hypotenuse and a leg of a particular right triangle are \(\sqrt97\) inches and 4 inches, respectively. The area of this triangle is what common fraction of a square foot?
+128475 The other leg = sqrt [ (√97)^2 - 4^2] = sqrt [ 97 - 16] = sqrt 81 = 9
The area is =
(1/2) (product of the leg lengths) =
(1/2) (4 * 9) =
(1/2) (36) =
18
Let x = length of the 3rd side of the triangle. By Pythagorean's Theorem,
x^2 + 4^2 = (√97)^2
x^2 + 16 = 97
x^2 = 81
x = 9 or x = -9
A negative length makes no sense so discard x = -9, leaving just
x = 9
The nonhypotenuse sides can be used as the base and height of the triangle so its area =
base * height / 2 =
9 * 4 / 2 =
18 in^2
