### Home > CCG > Chapter 10 > Lesson 10.1.5 > Problem10-59

10-59.

In the figure below, $\overleftrightarrow{ E X }$ is tangent to $⊙O$ at point $X$. $\overline{OE} = 20$ cm and $\overline{XE} = 15$ cm. Homework Help ✎

1. What is the area of the circle?

According to the definition of the tangent line, what is the measure of $∠EXO$?
How can you use this to find the length of $\overline{XO}$?

A tangent line is perpendicular to a radius, so the measure of $∠EXO = 90º$, so the triangle is a right triangle. So we can use the Pythagorean Theorem.

$A = πr²$
$A = π(13.23)²$
$A = 175π\text{ cm}²$
$A = 549.78\text{ cm}²$

$(\overline{XE})² + (\overline{XO})² = (\overline{OE})²$
$15² + (\overline{XO})² = 20²$
$225 + \overline{XO^2} = 400$
$\overline{XO^2} = 175$
$\overline{XO} = 13.23$

1. What is the area of the sector bounded by $\overline{OX}$ and $\overline{ON}$?

Find the measure of $∠XON$. Use this to find the area of the sector.

$A ≈ 74.21\text{ cm}²$

1. Find the area of the region bounded by $\overline{XE}$, $\overline{NE}$, and $\overarc{ N X }$.

Find the area of the triangle formed by $E, X,\text{ and }O$. Subtract the area of the sector from the area of the triangle.