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Jeff is painting two murals on the front of an old apartment

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Jeff is painting two murals on the front of an old apartment

by Anaira Mitch » Fri Dec 30, 2016 4:38 am
Jeff is painting two murals on the front of an old apartment building that he is renovating. One mural will cover the quadrilateral face ABCD while the other will cover the circular face (shown to the right, with radius XY). Assuming that the thickness of the coats of paint is negligible, will each mural require the same amount of paint? Note: Figures are not drawn to scale.

(1) AB = BC = CD = DA and AB= XY sqrt pie

(2) AC = BD and AC= XY sqrt 2*pie

Please help with this problem.
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Source: — Data Sufficiency |
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by Anaira Mitch » Fri Dec 30, 2016 9:19 am
Anyone?
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by Matt@VeritasPrep » Fri Jan 20, 2017 1:16 am
We're essentially asked (in a more fun way) if the area of the circle = the area of the square, so let's try to answer that.

S1::

Since the quadrilateral has four equal sides, it is either a square or a rhombus. If the side is s, then the area is either s² or unknown, if we have a rhombus.

Naming the radius of the circle r, we've also got r√π = s, or r = s/√π. So we know the area of the circle is π * (s/√π)², or s².

So if the quadrilateral is a circle, the murals have equal area, but if it's only a rhombus they may not: INSUFFICIENT.

S2::

Since the quadrilateral has equal diagonals, it could be a square, a rectangle, or an isosceles trapezoid. We don't know which, so we can't find the area of the quadrilateral: INSUFFICIENT.

S1 + S2::

Together, we know from S1 that we've got either a square or a rhombus, and that each side = r√π. From S2, we know that the diagonal of the quadrilateral = r * √2π. If our quadrilateral is a square, we'd apply the Pythagorean Theorem: Side² + Side² = Diagonal². Here, that gives us

(r√π)² + (r√π)² = (r√2π)²

or

2r²π = 2r²π

Success! So it's a square, and we're set: SUFFICIENT.
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