CPM Homework Help : INT2 Problem 9-110

A regular pentagon with divided into 5 equal triangles about the center point, C. Each side length of the pentagon is 10. What is the area of the shaded region for the regular pentagon at right if the length of each side of the pentagon is $10$ units? Assume that point $C$ is the center of the pentagon.

Hint:

A regular pentagon is made up of $5$ isosceles triangles with angle measures $72°$ , $54°$ , and $54°$ .

Step 1:

First find the area of one of the five triangles within the pentagon, like the one shown at right. Triangle, with dashed segment, from top vertex, perpendicular to bottom side, creating 2 right triangles. Bottom side of large triangle, labeled, 10, Dashed segment labeled, y. Angles opposite the long leg, in each right triangle, labeled 54 degrees. Angles opposite the short leg, in each right triangle, labeled 36 degrees.

Step 2:

$\text{tan}(54°)=\frac{y}{5}$

$y = 5\tan(54°)$

$y ≈ 6.88$

Step 3:

Area $= .5$ (base)(height) $= .5(10)(6.88) ≈ 34.41$

Answer:

The area of the shaded region is equal to three of these triangles, so $3(34.41) ≈ 103.23$ units