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Figure shown above consists of two squares having the same centers.
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09 Jan 2025, 01:45
09 Jan 2025, 01:45
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100% (03:07) correct
0% (00:00) wrong based on 4 sessions
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Attachment:
GRE shaded region square.png [ 16.61 KiB | Viewed 310 times ]
Figure shown above consists of two squares having the same centers. Sides of the inner square, each of length 4 units, are parallel to the sides of the outer square each having length of 12 units.
Quantity A |
Quantity B |
Perimeter of the shaded region |
Area of the shaded region |
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Re: Figure shown above consists of two squares having the same centers.
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18 Feb 2025, 04:05
18 Feb 2025, 04:05
Expert Reply
OFFICIAL EXPLANATION
gre SQUARE.png [ 85.43 KiB | Viewed 21 times ]
The square ABCD having sides 12 unit's each, has diagonal $\(\mathrm{AC}(=\mathrm{BD})\)$ of length $\(12 \sqrt{2}\)$ (Diagonal in a square is $\(\sqrt{2}\)$ times the side).
Same way the diagonal of square PQRS having side 4 unit's each is $\(4 \sqrt{2}(=P R=Q S)\)$
Now, $\(A C-P R=12 \sqrt{2}-4 \sqrt{2}=8 \sqrt{2}\)$ is same as $\(A P+R C=2 R C\)$ (the figure has $\(A P=R C\)$ ), so we get $\(2 R C=8 \sqrt{2} \Rightarrow R C=4 \sqrt{2}\)$
Next the length of the sides of isosceles trapezium QRCB ; is $\(\mathrm{QR}=4, \mathrm{BC}=12 \& \mathrm{RC}=\mathrm{QB}=4 \sqrt{2}\)$
Finally the perimeter of the shaded region i.e. trapezium $\(Q R C B\)$ is $\(Q R+B C+R C+Q B=\)$ $\(4+12+4 \sqrt{2}+4 \sqrt{2}=16+8 \sqrt{2}(=$ Column A quantity $) \&\)$ the area of the shaded portion is $\(\frac{1}{2} \times\)($ Sum of parallel sides $) \(\times\)$ Distance between parallel sides $\(=\frac{1}{2} \times(4+12) \times 4=32(=\)$ column $\(B\)$ quantity)
Clearly $\(16+8 \sqrt{2}<32\)$ i.e. column $B$ has higher quantity when compared with column $\(A\)$.
Hence the answer is (B).
Attachment:
gre SQUARE.png [ 85.43 KiB | Viewed 21 times ]
The square ABCD having sides 12 unit's each, has diagonal $\(\mathrm{AC}(=\mathrm{BD})\)$ of length $\(12 \sqrt{2}\)$ (Diagonal in a square is $\(\sqrt{2}\)$ times the side).
Same way the diagonal of square PQRS having side 4 unit's each is $\(4 \sqrt{2}(=P R=Q S)\)$
Now, $\(A C-P R=12 \sqrt{2}-4 \sqrt{2}=8 \sqrt{2}\)$ is same as $\(A P+R C=2 R C\)$ (the figure has $\(A P=R C\)$ ), so we get $\(2 R C=8 \sqrt{2} \Rightarrow R C=4 \sqrt{2}\)$
Next the length of the sides of isosceles trapezium QRCB ; is $\(\mathrm{QR}=4, \mathrm{BC}=12 \& \mathrm{RC}=\mathrm{QB}=4 \sqrt{2}\)$
Finally the perimeter of the shaded region i.e. trapezium $\(Q R C B\)$ is $\(Q R+B C+R C+Q B=\)$ $\(4+12+4 \sqrt{2}+4 \sqrt{2}=16+8 \sqrt{2}(=$ Column A quantity $) \&\)$ the area of the shaded portion is $\(\frac{1}{2} \times\)($ Sum of parallel sides $) \(\times\)$ Distance between parallel sides $\(=\frac{1}{2} \times(4+12) \times 4=32(=\)$ column $\(B\)$ quantity)
Clearly $\(16+8 \sqrt{2}<32\)$ i.e. column $B$ has higher quantity when compared with column $\(A\)$.
Hence the answer is (B).
Re: Figure shown above consists of two squares having the same centers.
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18 Feb 2025, 04:08
18 Feb 2025, 04:08
Expert Reply
OFFICIAL EXPLANATION
As $\(\mathrm{AB}=\mathrm{AD}=10 \& \mathrm{BC}=\mathrm{CD}=5\)$, so the opposite angles must also be equal.
Let us assume the equal opposite angles as $\(x\)$ and $\(y\)$, we get the figure as
GRE PARALLELOGRAM.png [ 44.71 KiB | Viewed 17 times ]
If angle BAD is 60 degrees, the triangle ABD becomes equilateral and side BD comes 10 , which in triangle $\(B C D\)$ gives $\(B C+C D=5+5=10=B D\)$, which is not possible as the sum of the sides in a triangle is always greater than the third side.
To satisfy $\(\mathrm{BC}+\mathrm{CD}>\mathrm{BD}\)$, the length of BD must be less than 10 which implies the measure of angle opposite to side BD in triangle ABD must be less than 60 .
Hence only option (A) i.e. 45 degrees can be the measure of angle BAD, so is correct.
As $\(\mathrm{AB}=\mathrm{AD}=10 \& \mathrm{BC}=\mathrm{CD}=5\)$, so the opposite angles must also be equal.
Let us assume the equal opposite angles as $\(x\)$ and $\(y\)$, we get the figure as
Attachment:
GRE PARALLELOGRAM.png [ 44.71 KiB | Viewed 17 times ]
If angle BAD is 60 degrees, the triangle ABD becomes equilateral and side BD comes 10 , which in triangle $\(B C D\)$ gives $\(B C+C D=5+5=10=B D\)$, which is not possible as the sum of the sides in a triangle is always greater than the third side.
To satisfy $\(\mathrm{BC}+\mathrm{CD}>\mathrm{BD}\)$, the length of BD must be less than 10 which implies the measure of angle opposite to side BD in triangle ABD must be less than 60 .
Hence only option (A) i.e. 45 degrees can be the measure of angle BAD, so is correct.
