Question

a right triangle has a hypotensue of length 10 ft and one leg o length 7 ft. how long is the other leg? round the answer to the nearest tenth of a foot.

Answer 1

x = 7.1 to the nearest tenth.

Step-by-step explanation:

Hypotenuse = 10;

Opposite = 7; and

Adjacent = x

Pythagoras rule states that square Hypotenuse side is equal to square Opposite side Plus square Adjacent side, so we solve;

(hyp)² = (opp)² + (adj)². Plug the in:

(10)² = (7)² + (x)²

Evaluate

100 = 49 + x²

subtract 49 from both sides

100 - 49 = 49 - 49 + x²

51 = x²

square root both sides to get rid of the square

√51 = √x², the square cancels the square root

√51 = 7.14

7.14 = x

or x = 7.14 to the nearest tenth.

x = 7.1

Check

(10)² = (7)² + (7.14)²

100 = 49 + 50.97 ( approximately = 51)

100 = 49 + 51

100 = 100.

Hope this helps!

Answer 2

7.1

Explanation:

By using Pythagorean theorem, a^2+b^2=c^2, we can find the missing length of the leg. a^2+7^2=10^2. 7 is the leg length given and 10 is the hypotenuse. Then we solve from there.

a^2+49=100

a^2=51

(sqrt)a^2=(sqrt)51

a=7.14142.....

rounded to the nearest tenth is 7.1