A graphic designer is creating a logo for a company. The logo is shown in the figure above. The logo

Question

A graphic designer is creating a logo for a company. The logo is shown in the figure above. The logo is in the shape of a trapezoid and consists of three congruent equilateral triangles. If the perimeter of the logo is 20 centimeters, what is the combined area of the shaded regions, in square centimeters, of the logo?

A.

$2 times the square root of 3$

B.

$4 times the square root of 3$

C.

$8 times the square root of 3$

D.

16

Accepted Answer

Choice C is correct. It&rsquo;s given that the logo is in the shape of a trapezoid that consists of three congruent equilateral triangles, and that the perimeter of the trapezoid is 20 centimeters (cm). Since the perimeter of the trapezoid is the sum of the lengths of 5 of the sides of the triangles, the length of each side of an equilateral triangle is . Dividing up one equilateral triangle into two right triangles yields a pair of congruent 30&deg;-60&deg;-90&deg; triangles. The shorter leg of each right triangle is half the length of the side of an equilateral triangle, or 2 cm. Using the Pythagorean Theorem, , the height of the equilateral triangle can be found. Substituting and and solving for b yields cm. The area of one equilateral triangle is , where and . Therefore, the area of one equilateral triangle is . The shaded area consists of two such triangles, so its area is . Alternate approach: The area of a trapezoid can be found by evaluating the expression , where is the length of one base, is the length of the other base, and h is the height of the trapezoid. Substituting , , and yields the expression , or , which gives an area of for the trapezoid. Since two-thirds of the trapezoid is shaded, the area of the shaded region is . Choice A is incorrect. This is the height of the trapezoid. Choice B is incorrect. This is the area of one of the equilateral triangles, not two. Choice D is incorrect and may result from using a height of 4 for each triangle rather than the height of .

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