GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
In the figure above, $A B=B C, C I=I H$ and $D E=E F$.
[#permalink]
07 Aug 2025, 01:08
07 Aug 2025, 01:08
Expert Reply
00:00
Question Stats:
66% (00:54) correct
33% (02:05) wrong based on 3 sessions
Hide Show timer Statistics
Attachment:
GRe In the figure above, AB = BC, CI = IH.png [ 25.08 KiB | Viewed 209 times ]
In the figure above, $\(A B=B C, C I=I H\)$ and $\(D E=E F\)$.
Quantity A |
Quantity B |
Area of Rectangle \(GDEF\) |
Sum of areas of Rectangles \(ABCD\) and \(CIHE\) |
In the figure above, $A B=B C, C I=I H$ and $D E=E F$.
[#permalink]
15 Aug 2025, 09:41
15 Aug 2025, 09:41
Expert Reply
Based on the figure and the information provided, Quantity A and Quantity B are equal.
Here is the breakdown of the areas:
Quantity A: Area of Rectangle $\(G D E F\)$
The problem states that $\(D E=E F\)$. This means that the figure $\(G D E F\)$ is a square. The side length of this square is the hypotenuse of the right triangle with legs of length $a$ and $b$.
By the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides:
Area of Square $\(G D E F=(D E)^2=a^2+b^2\)$.
Quantity B: Sum of areas of Rectangles $A B C D$ and $C I H E$
The problem states that $A B=B C$ and the figure $A B C D$ is a rectangle, which means it is a square with a side length of $a$. Similarly, $\(C I=I H\)$ implies that the figure $\(C I H E\)$ is a square with a side length of $\(b\)$.
- Area of Square $\(A B C D=(\text { side })^2=a^2\)$
- Area of Square $\(C I H E=(\text { side })^2=b^2\)$
The sum of these areas is $a^2+b^2$.
Conclusion
Quantity $\(\mathrm{A}\left(a^2+b^2\right)\)$ is equal to Quantity $\(\mathrm{B}\left(a^2+b^2\right)\)$.
Here is the breakdown of the areas:
Quantity A: Area of Rectangle $\(G D E F\)$
The problem states that $\(D E=E F\)$. This means that the figure $\(G D E F\)$ is a square. The side length of this square is the hypotenuse of the right triangle with legs of length $a$ and $b$.
By the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides:
Area of Square $\(G D E F=(D E)^2=a^2+b^2\)$.
Quantity B: Sum of areas of Rectangles $A B C D$ and $C I H E$
The problem states that $A B=B C$ and the figure $A B C D$ is a rectangle, which means it is a square with a side length of $a$. Similarly, $\(C I=I H\)$ implies that the figure $\(C I H E\)$ is a square with a side length of $\(b\)$.
- Area of Square $\(A B C D=(\text { side })^2=a^2\)$
- Area of Square $\(C I H E=(\text { side })^2=b^2\)$
The sum of these areas is $a^2+b^2$.
Conclusion
Quantity $\(\mathrm{A}\left(a^2+b^2\right)\)$ is equal to Quantity $\(\mathrm{B}\left(a^2+b^2\right)\)$.
gmatclubot
In the figure above, $A B=B C, C I=I H$ and $D E=E F$. [#permalink]
15 Aug 2025, 09:41
Moderators:
