Question

Plz Help... Find x in the given triangle.

Answer 1

Answer:

$\huge\boxed{x=2\sqrt{286}}$

Step-by-step explanation:

Use the Pythagorean theorem:

$leg^2+leg^2=hypotenuse^2$

We have:

$leg=x\\leg=15\\hypotenuse=37$

Substitute and solve for x:

$x^2+15^2=37^2\\\\x^2+225=1269\qquad|\text{subtract 225 from both sides}\\\\x^2+225-225=1269-225\\\\x^2=1144\to x=\sqrt{1144}\\\\x=\sqrt{4\cdot286}\\\\x=\sqrt4\cdot\sqrt{286}\\\\x=2\sqrt{286}$