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What is the area of square floor X?

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by VJesus12 » Mon May 28, 2018 1:55 am

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What is the area of square floor X?

(1) The perimeter of the floor is a whole-number multiple of 10. $$(2)\ \ The\ \ diagonal\ \ of\ \ the\ \ floor\ \ measures\ \ 5\sqrt{2}.$$ The OA is the option B.

I don't know how to use the length of the diagonal to find the area. Could someone clarify this to me? Thanks.
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Source: — Data Sufficiency |

by Vincen » Mon May 28, 2018 2:59 am
Hello vjesus12.

Here, we have to calculate the area of a square, which is equal to $$A=l^2$$ where l is the length of the sides of the square. Hence, all we have to do is to find the value of l.
(1) The perimeter of the floor is a whole-number multiple of 10.
The perimeter of a square is 4*l. Here, we are told that $$4\cdot l=10\cdot n,\ \ \ where\ n\in\mathbb{Z}$$ Hence we have $$4\cdot l=10\ \ \ \Rightarrow\ \ l=\frac{5}{2}\ \ \ \Rightarrow\ \ \ \ \ A=l^2=\frac{25}{4}.$$ $$4\cdot l=20\ \ \ \Rightarrow\ \ l=5\ \ \ \Rightarrow\ \ \ \ \ A=l^2=25.$$ $$4\cdot l=40\ \ \ \Rightarrow\ \ l=10\ \ \ \Rightarrow\ \ \ \ \ A=l^2=100.$$ Since we got different answers, this statement is NOT SUFFICCIENT.
$$(2)\ \ The\ \ diagonal\ \ of\ \ the\ \ floor\ \ measures\ \ 5\sqrt{2}.$$
The diagonal of a square is given by $$d=\sqrt{l^2+l^2}=\sqrt{2l^2}=l\sqrt{2}.$$ Here, we are told that $$d=5\sqrt{2}\ \ \Rightarrow\ \ \ \ l\sqrt{2}=5\sqrt{2}\ \ \ \Rightarrow\ \ \ l=5\ \ \ \Rightarrow\ \ A=25.$$ Here, we just got one answer. therefore, this statement is SUFFICCIENT.

The correct answer to this DS question is the option B.
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