esonrev wrote:
Still don't really understand this, even with the answers. Could anyone illustrate it better/differently?
Attachment:
#GREpracticequestion In the figure above, if the area of the larger square region is twice.jpg [ 8.28 KiB | Viewed 4198 times ]
Here, see the diag. above
Now let the smaller square be ABCD, where it is mentioned the diagonal = 1 foot
Now , if a figure is square and all angles are 90 degree, hence the sides can be divided
as \(1: 1: \sqrt2\) (45-45-90 \(\triangle\))
Let us divide the square in 2 equal triangles ABC and ADC
For \(\triangle\) ABC
we have AC = 1 ( as it is a diagonal)
since the sides are in \(1: 1: \sqrt2\)
i.e diagonal AC has to be the largest side and should be equal = \(\sqrt2\)
but how to make this possible?
we can divide\(1:1:\sqrt2\) by\(\sqrt2\)
i.e\(\frac{1}{{\sqrt2}} : \frac{1}{{\sqrt2}} : 1\)
Now, the largest side (diagonal) is 1 and the other two sides in the \(\triangle\) ABC are AB = BC =\(\frac{1}{{\sqrt2 }}\)
Now we have figure out the side of the smaller square ABCD
Hence, the Area of the smaller square =\({side}^2\) = \(({\frac{1}{\sqrt2} })^2 = \frac{1}{2}\)
Now,
Larger Square = 2 * Area of the smaller square = \(2 * \frac{1}{2}= 1\)
i.e the side of the larger square = \(1\)
Ok, now we have the side of the Larger square as well as for the smaller square
Length of the side of the Larger square =\(1\)
and length of the side of the smaller square = \(\frac{1}{{\sqrt2}}\)
Then the side of the length of the larger square greater than that of the smaller square = \(1 - \frac{1}{\sqrt2} = \frac{{\sqrt2 -1}}{\sqrt2} * \frac{{\sqrt2}}{{\sqrt2}}\) = \(\frac{{(2-\sqrt2)}}{2}\)
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