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In the figure above, $A B=B C, C I=I H$ and $D E=E F$.
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07 Aug 2025, 01:08
07 Aug 2025, 01:08
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Question Stats:
70% (01:23) correct
30% (01:42) wrong based on 10 sessions
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GRe In the figure above, AB = BC, CI = IH.png [ 25.08 KiB | Viewed 371 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 $D E=E F$.
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15 Aug 2025, 09:41
15 Aug 2025, 09:41
1
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 $\(\mathrm{B}\left(a^2+b^2\right)\)$.
In the figure above, $A B=B C, C I=I H$ and $D E=E F$.
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24 Sep 2025, 10:55
24 Sep 2025, 10:55
shouldn't the question give the angle as well? before we can conclude anything in this picture is square? or am I missing something?
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