Question

directions: completely read and work through each problem below. show all work to receive full credit. use the pythagorean theorem to solve for x. round your answer to the nearest tenth. $a^2 + b^2 = c^2$ 1. right triangle with hypotenuse 14 mi, one leg 5 mi, the other leg x 2. right triangle with legs 9 mi and 13 mi, hypotenuse x 3. the diagonal length of a square field is 52 meters. find the perimeter of the square.

Explanation:

Step1: Solve for x (Problem 1)

Apply Pythagorean theorem: $x^2 + 5^2 = 14^2$
$x^2 = 14^2 - 5^2 = 196 - 25 = 171$
$x = \sqrt{171} \approx 13.1$

Step2: Solve for x (Problem 2)

Apply Pythagorean theorem: $9^2 + 13^2 = x^2$
$x^2 = 81 + 169 = 250$
$x = \sqrt{250} \approx 15.8$

Step3: Find square side length

Let side = $s$. Diagonal: $s\sqrt{2}=52$
$s = \frac{52}{\sqrt{2}} = 26\sqrt{2} \approx 36.77$

Step4: Calculate square perimeter

Perimeter = $4s$
$4 \times 36.77 \approx 147.1$

Answer:

1. $x \approx 13.1$ mi
2. $x \approx 15.8$ mi
3. $\approx 147.1$ meters