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If the pentagon is a regular pentagon, then each of the central angles (formed by joinning the center with the vertices) is72⁰ (1/5 of 360⁰). So that 5 congruent isosceles triangle are formed, with base angles of 54⁰. If we draw the height, of these triangles (from the center to the sides of the pentagon), 10 congruent right triangles are formed. Let denote with:

b the sides of the pentagon, and

h the height. In one of the right triangles we have: tan 54⁰ = h/(b/2)

tan 54⁰ = 2h/b

1.38 = 2h/b

2h = 1.38b

h = (1.38/2)b

h = .69b So that, h/b = .69b/b = .69

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βˆ™ 3y ago

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Q: What is the ratio of the height to the sides of a pentagon?
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