Question

What is the area of a regular pentagon with a side of 12 in.? Round the answer to the nearest tenth.

Answer 1

@tylermcmullen23

Answer 2

a=1/4sqrt5(5+2sqrt5)a^2

Answer 3

The sum of the measures of the interior angles is:

180(n - 2) = 180(5 - 2) = 180 * 3 = 540

Each interior angle measures: 540/5 = \(108^o\)

Answer 4

\[\frac{ 1 }{ 4 }\sqrt{5(5+2\sqrt{5})a^2}\]

Answer 5

so just replace "a" with 12 and solve

Answer 6

BTW The options are
A. 495.5 in^2
B. 311.8 in^2
C. 247.7 in^2
D. 124.7 in^2

Answer 7

Thank you @tylermcmullen23

Answer 8

\(\tan 54^o = \dfrac{h}{6} \)

\(h = 6\tan 54^o\)

\(A_{pentagon} = 10 \times A_{small ~right ~triangle}\)

\(A_{pentagon} = 10 \times \dfrac{bh}{2} \)

\(A_{pentagon} = 10 \times \dfrac{6 ~in. \times 6 \tan 54^o~in.}{2} \)

\(A_{pentagon} = 247.7 ~in.^2\)