Carcass wrote:
Attachment:
square.jpg
In the figure above, if the area of the smaller square region is \(\frac{1}{2}\) 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. \(\sqrt{2} - 1\)
B. \(\frac{1}{2}\)
C. \(\frac{\sqrt{2}}{2}\)
D. \(\frac{\sqrt{2} + 1}{2}\)
E. \(\sqrt{2}\)
\(\frac{{Diagonal^2}}{2} = Area of square\)
So for large square of side 1 inch
\(\frac{{Diagonal^2}}{2} = 1\) ( area of the large square with side 1 inch)
or \(Diagonal = \sqrt2\)
Now for small square
\(\frac{{Diagonal^2}}{2} = \frac{1}{2}\) ( area of the small square = \(\frac{1}{2}\) the area of larger square)
or \(Diagonal = 1\)
So the diagonal of the larger square is longer than the diagonal of the smaller square = \(\sqrt2 - 1\)
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68% (01:57) correct
