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In the figure above, a regular octagon with a side of 2 inches is surr

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In the figure above, a regular octagon with a side of 2 inches is surr  [#permalink]

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New post 01 Jul 2018, 18:27
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In the figure above, a regular octagon with a side of 2 inches is surrounded by eight identical right triangles whose legs are 3 and 4 inches. What is the largest possible perimeter of the shape formed?

A) 24
B) 32
C) 40
D) 48
E) 64

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Re: In the figure above, a regular octagon with a side of 2 inches is surr  [#permalink]

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New post 01 Jul 2018, 18:46
IMO E. 64
Part of hypotenuse is 5+2-3=4
8 sides of length 4 (part of hypotenuse)
8 sides of length 4 (leg)
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Re: In the figure above, a regular octagon with a side of 2 inches is surr  [#permalink]

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New post 01 Jul 2018, 19:37
Option E, IMO

Perimeter (exposed lengths) for one triangle = 4 (from one side) + {5 - (3-2)} (from hypotenuse) = 4+ 4 = 8
Total perimeter = 8*8 (8 triangles) = 64 inches
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Re: In the figure above, a regular octagon with a side of 2 inches is surr  [#permalink]

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New post 02 Jul 2018, 06:51
We can use the Pythagorean triplet 3-4-5: the triangles' hypotenuses are 5.
i) The part of the hypotenuse which is part of the perimeter is 5 minus [part of the triangle's leg of 3]: 5 —1 = 4
ii) The leg of triangle which is part of perimeter is 4.
Since there are 8 triangles, the perimeter is 8 [leg of 4 + part of hypotenuse which is included in perimeter] = 8 (4+4) = 64
Answer (E) 64 .
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Re: In the figure above, a regular octagon with a side of 2 inches is surr &nbs [#permalink] 02 Jul 2018, 06:51
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