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In the figure above, if the area of the smaller square regio
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15 Jun 2019, 08:38
15 Jun 2019, 08:38
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In the figure above, if the area of the smaller square region is \(\frac{2}{3}\) the area of the larger square region, then the diagonal of the larger square is how many inches longer than the diagonal of the smaller square?
A. \(\sqrt{2} - \frac{2 \sqrt{3} }{3}\)
B. \(\frac{2}{3}\)
C. \(\frac{2 \sqrt{3} }{3}\)
D. \(\frac{\sqrt{2} - 2}{3}\)
E. \(\sqrt{3}\)
Re: In the figure above, if the area of the smaller square regio
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23 Jun 2019, 05:44
23 Jun 2019, 05:44
2
Area of larger square (A) = 1
Area of smaller square (a) = 1* \(\frac{2}{3}\)=\(\frac{2}{3}\)
By Pythagoras theorem length of diagonal of larger square(D) = \(\sqrt{2}\)
let the length of a side of smaller square be x then,
x*x = 2/3
hence x = \(\sqrt{\frac{2}{3}}\)
now from Pythagoras theorem again the length of diagonal of smaller square is \(\sqrt{2}\)* \(\sqrt{\frac{2}{3}}\) = \(\frac{2}{\sqrt{3}}\)
Therefore the difference between larger and smaller diagonal is \(\sqrt{2} - \frac{2}{\sqrt{3}}\)
By rationalizing the denominator we get \(\sqrt{2}\) - \(2\sqrt{3}/3\)
Area of smaller square (a) = 1* \(\frac{2}{3}\)=\(\frac{2}{3}\)
By Pythagoras theorem length of diagonal of larger square(D) = \(\sqrt{2}\)
let the length of a side of smaller square be x then,
x*x = 2/3
hence x = \(\sqrt{\frac{2}{3}}\)
now from Pythagoras theorem again the length of diagonal of smaller square is \(\sqrt{2}\)* \(\sqrt{\frac{2}{3}}\) = \(\frac{2}{\sqrt{3}}\)
Therefore the difference between larger and smaller diagonal is \(\sqrt{2} - \frac{2}{\sqrt{3}}\)
By rationalizing the denominator we get \(\sqrt{2}\) - \(2\sqrt{3}/3\)
gmatclubot
Re: In the figure above, if the area of the smaller square regio [#permalink]
23 Jun 2019, 05:44
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