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What is the perimeter of the quadrilateral shown above?
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11 Jan 2025, 00:37
11 Jan 2025, 00:37
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GRE geometry practice question.png [ 17.33 KiB | Viewed 217 times ]
(A) $\((4 \sqrt{2}) x\)$
(B) $\((4+\sqrt{2}) x\)$
(C) $\((4+2 \sqrt{2}) x\)$
(D) $\(5 x\)$
(E) $\(6 x\)$
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What is the perimeter of the quadrilateral shown above?
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19 Feb 2025, 11:22
19 Feb 2025, 11:22
Expert Reply
OFFICIAL EXPLANATION
GRE parallelogram (2).png [ 101.49 KiB | Viewed 42 times ]
In one of the isosceles right triangle in the figure, say triangle $\(A D F\)$, let the base $\(=D F=\mathrm{k}=$ height $=A F\)$, so applying Pythagoras theorem, we get $\(k^2+k^2=x^2\)$
i.e. $\(2 \mathrm{k}^2=\mathrm{x}^2 \Rightarrow \mathrm{k}=\frac{x}{\sqrt{2}}\)$ which is the value of two identical bases $\((=\mathrm{DF}=\mathrm{EC})\)$ of two isosceles right triangles \(ADF \& BEC.\) \((Pythagoras theorem - Hypotenuse ${ }^2=\)$ Perpendicular $\(^2+\)$ Base $\(^2\)$ )
Finally the perimeter of the trapezium is $\(\mathrm{AB}+\mathrm{BC}+\mathrm{FE}+\mathrm{AD}+\mathrm{DF}+\mathrm{EC}=\)$ $\(x+x+x+x+\frac{x}{\sqrt{2}}+\frac{x}{\sqrt{2}}=4 x+\sqrt{2} x=(4+\sqrt{2}) x\)$
Hence the answer is (B).
Attachment:
GRE parallelogram (2).png [ 101.49 KiB | Viewed 42 times ]
In one of the isosceles right triangle in the figure, say triangle $\(A D F\)$, let the base $\(=D F=\mathrm{k}=$ height $=A F\)$, so applying Pythagoras theorem, we get $\(k^2+k^2=x^2\)$
i.e. $\(2 \mathrm{k}^2=\mathrm{x}^2 \Rightarrow \mathrm{k}=\frac{x}{\sqrt{2}}\)$ which is the value of two identical bases $\((=\mathrm{DF}=\mathrm{EC})\)$ of two isosceles right triangles \(ADF \& BEC.\) \((Pythagoras theorem - Hypotenuse ${ }^2=\)$ Perpendicular $\(^2+\)$ Base $\(^2\)$ )
Finally the perimeter of the trapezium is $\(\mathrm{AB}+\mathrm{BC}+\mathrm{FE}+\mathrm{AD}+\mathrm{DF}+\mathrm{EC}=\)$ $\(x+x+x+x+\frac{x}{\sqrt{2}}+\frac{x}{\sqrt{2}}=4 x+\sqrt{2} x=(4+\sqrt{2}) x\)$
Hence the answer is (B).
gmatclubot
What is the perimeter of the quadrilateral shown above? [#permalink]
19 Feb 2025, 11:22
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