9514 1404 393
Answer:
(d) A = 14.28 in², P = 14.28 in
Explanation:
The figure is wholly contained within a 4" square, which has an area of (4 in)² = 16 in², and a perimeter of 4(4 in) = 16 in. Since the figure is smaller in area and has a shorter perimeter (the top corners are rounded, not square), both answer values must be less than 16.
The only reasonable choice is the last choice: 14.28 in², 14.28 in.
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If you want to figure this out in detail, you have the area of a rectangle that is 2 in by 4 in, and the area of a semicircle of radius 2 in. The total area is ...
A = LW +1/2πr²
A = (2 in)(4 in) + 1/2(3.14)(2 in)² = 8 in² +6.28 in²
A = 14.28 in²
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The perimeter is half that of a 4" square, plus half that of a 4" circle.
P = 1/2(4(4 in) +π(4 in)) = (2 in)(4 +π) = 2(7.14) in
P = 14.28 in