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?

Guest May 15, 2020

#1**+1 **

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

CPhill May 15, 2020

#4**0 **

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

Guest May 15, 2020