In the figure above, a regular octagon with a side of 2 inches is surrounded by eight identical right triangles whose

M7MBA wrote: ↑

Fri Dec 11, 2020 2:37 am

gmatbusters (3).jpg

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

Answer: E

Solution:

The perimeter of the shape formed consists of two types of sides: the 4 inch leg of a right triangle and a portion of the hypotenuse of a right triangle. If we can determine the length of the latter type of a side, we can determine the perimeter of the shape. Let this unknown length be x.

Notice that the hypotenuse of each right triangle consists of a line segment of length x and a portion of a 3 inch leg. Notice also that the part of each 3 inch leg which does not overlap with a hypotenuse is a side of the regular octagon, which has length 2. Thus, the part of a 3 inch leg which overlaps with a hypotenuse has length 3 - 2 = 1. It follows that x equals 5 - 1 = 4.

Since there are 8 right triangles, there are 8 legs of length 4 and 8 sides with length x = 4. It follows that the perimeter of the shape is 8 x 4 + 8 x 4 = 64.

Answer: E