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Frank needs to find the area enclosed by the figure. The figure is made by attaching semicircles to each side of a 44-m-by-44-m square. Frank says the area is 1,103.52 m^2. Find the area enclosed by the figure. Use 3.14 for π. What error might Frank have made?

A. 1,103.52 m^2; Frank made no error.
B. 1,237.78 m^2; Frank underestimated the area.
C. 1,372.04 m^2; Frank overestimated the area.
D. 1,506.30 m^2; Frank calculated the area of the square only.

User ENeF
by
8.8k points

1 Answer

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The correct statement for the area enclosed by the figure is:

B. 1,237.78 m^2; Frank underestimated the area.

How to find the area enclosed by the figure?

Side length of the square = 44m

The semicircle is attached to each side of the square. So the diameter of the semicircle is the length of the square.

Radius of the semicircle = 44/2 = 22m

Area of the square = 44 * 44 = 1936 square meters

Area of 1 semicircle = π(22 * 22) /2

= (3.14) (484) /2

= 1384.74/2

= 759.88 square meters

Area of 4 semicircles = 4 * 759.88

= 3,039.52 square meters

Total area = 1936 + 3,039.52

= 4,975.52 square meters

The area enclosed by the figure is 4,975.52 square meters.

User Wsda
by
7.6k points
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