The Pentagon

Geometry Level 3

Find the length of a side of a regular pentagon that is inscribed in a circle of radius 2.

10 − 2 5 \sqrt{10-2\sqrt{5}} none of the other three choices 1 + 5 1+\sqrt{5} 5 − 5 5-\sqrt{5}

Jaydee Lucero by Jaydee Lucero , 24, Philippines

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3 solutions

By law of cosines, we have

x 2 = 2 2 + 2 2 − 2 ( 2 ) ( 2 ) cos ⁡ 72 x^2=2^2+2^2-2(2)(2)\cos~72

x 2 = 8 − 8 ( 5 − 1 4 ) x^2=8-8\left(\dfrac{\sqrt{5}-1}{4}\right)

x 2 = 8 − 2 ( 5 − 1 ) x^2=8-2(\sqrt{5}-1)

x 2 = 8 − 2 5 + 2 x^2=8-2\sqrt{5}+2

x 2 = 10 − 2 5 x^2=10-2\sqrt{5}

x = 10 − 2 5 \boxed{x=\sqrt{10-2\sqrt{5}}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

You've drawn the circle wrong, the pentagon is supposed to be inscribed in the circle. That then gives you the length of 2 for the two legs of the triangle.

Now you've got a isosceles triangle with two base angles of 54°, a top angle of 72° and a height of 2. That would give you 1/2x=2/tan 54° so x=4/tan 54°=2,9.

S Broekhuis - 4 months, 1 week ago
Sanjeet Raria
Aug 25, 2014

Draw two radii passing through two consecutive vertices of the regular polygon. Now the angle between them is 360/5=72°… making a room for cosine formula, which would result in easy solution.

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Not a four level problem!!

Sanjeet Raria - 6 years, 9 months ago

Well, that's an easier solution. :D

That will show that sin ⁡ 3 6 ∘ = 5 − 5 8 \sin 36^\circ = \sqrt{\dfrac{5-\sqrt {5}}{8}} and cos ⁡ 3 6 ∘ = 1 + 5 4 \cos 36^\circ = \dfrac{1+\sqrt{5}}{4} . Can you prove that? :)

Jaydee Lucero - 6 years, 9 months ago
S Broekhuis
Feb 1, 2021

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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