Hi, please help.

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# Question

# Answer

The perimeter of A(shaded) is the perimeter (p) of the square, minus one side (s) of the square, plus two of those curved segments (c).

The perimeter of A(unshaded) is two curves (c) plus one side of the square (s).

Thus, p – s + 2c = 2c + s + 140.

So p – s = s +140, p – 2s = 140. But p is the perimeter of the square = 4s. So 4s -2s = 140, s = 70.

So now we know two things: 1) each curve (c) has a radius of 35; 2) each side (s) is 70.

This means the area of a quarter circle is 1/4• πr^{2} = 1/4 x 22/7 x 35 x 35 = 962.5 cm^{2}.

This means the area of one of those curved triangular shapes is 1/4 square – 1/4 circle = 1225-962.5 = 262.5 cm^{2}.

Shaded area in B = 1/4 circle + 3 curved triangular shapes (or 1/4 square + 2 curved shapes) = 1750 cm^{2}.

It’s easier to solve purely by drawing, combining A and B. But I’m bad at drawing.