Carcass wrote:
Attachment:
square.jpg
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. \({\sqrt{2} - \frac{2\sqrt3}{3}\)
B. \(\frac{2}{3}\)
C. \(\frac{2\sqrt3}{3}\)
D. \(\frac{\sqrt2 - 2}{3}\)
E. \(\sqrt{3}\)
Here,
Side of the Larger square = 1 inch
Therefore Area Large square = \(1^2 =1\)
But we know Area of a square =\(\frac{diagonal^2}{2}\)
So 1 = \(\frac{diagonal^2}{2}\)
or diagonal = \(\sqrt{2}\)
And we know
smaller square region is \(\frac{2}{3}\) the area of the larger square region
So it can be written as \(\frac{(Diagonal smaller square)^2}{2}= \frac{2}{3} * 1 = \frac{2}{3}\) (Since the area of the Larger square = 1)
or \((diagonal of smaller square)^2 = \frac{4}{3}\)
or diagonal of smaller square = \(\frac{\sqrt4}{\sqrt3} = \frac{2}{(\sqrt3)} = \frac{2}{\sqrt3} * \frac{\sqrt3}{\sqrt3} = \frac{2\sqrt3}{3}\)
Now,
Diagonal of the larger square is longer than the diagonal of the smaller square = \(\sqrt2 - \frac{2\sqrt3}{3}\)
_________________
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