Question

o is the center of the regular pentagon below. find its area. round to the nearest tenth if necessary.

Explanation:

Step1: Divide the pentagon into 5 triangles

A regular pentagon can be divided into 5 congruent isosceles triangles, each with a vertex angle at the center \( O \). The central angle for each triangle is \( \frac{360^\circ}{5} = 72^\circ \), and the two equal sides (radii) are 8 units.

Step2: Find the area of one triangle

The area of a triangle with two sides \( a, b \) and included angle \( \theta \) is given by \( \frac{1}{2}ab\sin\theta \). Here, \( a = b = 8 \) and \( \theta = 72^\circ \). So the area of one triangle is \( \frac{1}{2} \times 8 \times 8 \times \sin(72^\circ) \).

Calculating \( \sin(72^\circ) \approx 0.9511 \), we get:
\( \frac{1}{2} \times 8 \times 8 \times 0.9511 = 4 \times 8 \times 0.9511 = 32 \times 0.9511 \approx 30.4352 \)

Step3: Find the area of the pentagon

Since there are 5 such triangles, the area of the pentagon is \( 5 \times \) area of one triangle.
\( 5 \times 30.4352 \approx 152.176 \)

Rounding to the nearest tenth, we get approximately 152.2.

Answer:

The area of the regular pentagon is approximately \(\boxed{152.2}\) (square units).