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25 Mar 2023, 03:06
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36% (02:40) correct
64%
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A square UVWX is inscribed inside a square ABCD. The area of the square UVWX is 324 \(\text{unit}^2\), which of the following could be the possible value of the area of ABCD?
A. 448 \(\text{unit}^2 \)
B. 664 \(\text{unit}^2 \)
C. 796 \(\text{unit}^2 \)
D. 808 \(\text{unit}^2 \)
E. 1012 \(\text{unit}^2 \)
A. 448 \(\text{unit}^2 \)
B. 664 \(\text{unit}^2 \)
C. 796 \(\text{unit}^2 \)
D. 808 \(\text{unit}^2 \)
E. 1012 \(\text{unit}^2 \)
29 Mar 2023, 11:40
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A square UVWX is inscribed inside a square ABCD. The area of the square UVWX is 324 unit^2, which of the following could be the possible value of the area of ABCD?
Side of UVWX is sqrt(324) = 18 unit.
Diagonal is 18*sqrt(2)
Since, UVWX is inscribed inside Square ABCD, area of ABCD is Maximum when square UVWX has its diagonal equal to side of the Square ABCD.(Refer figure below - area of square UVWX is fixed but area of square ABCD is maximum for central figure.)
In that case, area of square ABCD is 2*18^2 = 648 unit^2.
Out of the five options only Option A is smaller than 648. Hence, Option A could be the value of area of square ABCD.
Answer A
Side of UVWX is sqrt(324) = 18 unit.
Diagonal is 18*sqrt(2)
Since, UVWX is inscribed inside Square ABCD, area of ABCD is Maximum when square UVWX has its diagonal equal to side of the Square ABCD.(Refer figure below - area of square UVWX is fixed but area of square ABCD is maximum for central figure.)
In that case, area of square ABCD is 2*18^2 = 648 unit^2.
Out of the five options only Option A is smaller than 648. Hence, Option A could be the value of area of square ABCD.
Answer A
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30 Mar 2023, 01:46
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gmatophobia
Note that area of a square UVWX inscribed in another square ABCD will vary from slightly less than the area of ABCD to half the area of ABCD. Why?
Look at the diagrams:
Attachment:
Screenshot 2023-03-30 at 2.12.16 PM.png [ 24.17 KiB | Viewed 1127 times ]
The ABCD square is almost fully covered in the first diagram so areas of both the squares are about the same. In the second, the side of the inside square is \(\frac{1}{\sqrt{2}}\) times the side of the outer square (assuming U is the mid point of AD).
Hence area of UVWX is half the area of ABCD.
Since area of UVWX is 324, area of ABCD can at the most be double of it. So area of ABCD cannot exceed 648. Only option (A) lies within this range.
Answer (A)
