+0

0
42
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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?

May 15, 2020

#1
+111360
+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

May 15, 2020
#2
0

But it's aking for a fraction

May 15, 2020
#3
+111360
+1

Oops....I forgot we needed a sq ft answer

I gave my answer in sq in

1 sq ft   =144 in^2

So

18in ^2   /144 in^2   =    1/8  sq ft

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

May 15, 2020