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555-605 (Medium)|   Geometry|      
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Bunuel
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The area of square A is 131 cm^2 greater than the area of square B. If 05 May 2020, 11:07
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C
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72% (02:27) correct 28% (02:58) wrong based on 71 sessions

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The area of square A is 131 cm^2 greater than the area of square B. If the lengths of the sides of both squares are integers (in cm), then what is the area of square A?

A. 4096 cm^2
B. 4225 cm^2
C. 4356 cm^2
D. 4489 cm^2
E. 4624 cm^2


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C
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The area of square A is 131 cm^2 greater than the area of square B. If 05 May 2020, 11:17
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Bunuel
The area of square A is 131 cm^2 greater than the area of square B. If the lengths of the sides of both squares are integers (in cm), then what is the area of square A?

A. 4096 cm^2
B. 4225 cm^2
C. 4356 cm^2
D. 4489 cm^2
E. 4624 cm^2


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Given: The area of square A is 131 cm^2 greater than the area of square B.

Asked: If the lengths of the sides of both squares are integers (in cm), then what is the area of square A?

Let the sides of square A be a & side of square B be b cm

\(a^2 - b^2 = 131 = (a+b)(a-b)\)
Since 131 is prime number
a + b = 131
a - b = 1
a = 66

The area of square A = 66^2 = 4356 cm^2

IMO C
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The area of square A is 131 cm^2 greater than the area of square B. If 05 May 2020, 12:20
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Bunuel
The area of square A is 131 cm^2 greater than the area of square B. If the lengths of the sides of both squares are integers (in cm), then what is the area of square A?

A. 4096 cm^2
B. 4225 cm^2
C. 4356 cm^2
D. 4489 cm^2
E. 4624 cm^2


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Looking at the answer options, choices B and C have a difference of 131. No reason why the areas shouldn't be 4356 and 4225.

So square A must be the larger of these 2 which gives us the area of square A to be 4356

Answer is (C)

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The area of square A is 131 cm^2 greater than the area of square B. If 05 May 2020, 12:47
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We know, A =131 + B
So, B = A - 131, and also this needs to be a perfect square

If A = 4356, then B = 4225, which is a perfect square of \(65^2\)

Answer: C
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The area of square A is 131 cm^2 greater than the area of square B. If 09 May 2020, 05:59
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Bunuel
The area of square A is 131 cm^2 greater than the area of square B. If the lengths of the sides of both squares are integers (in cm), then what is the area of square A?

A. 4096 cm^2
B. 4225 cm^2
C. 4356 cm^2
D. 4489 cm^2
E. 4624 cm^2

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If we let a = the side length of square A and b = the side length of square B, we can create the equation:

a^2 - b^2 = 131

(a + b)(a - b) = 131

Since a and b are integers and 131 is a prime, we see that a + b = 131 and a - b = 1. Adding these two equations, we have:

2a = 132

a = 66

Therefore, the area of square A is 66^2 = 4356.

Answer: C
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The area of square A is 131 cm^2 greater than the area of square B. If 09 May 2020, 09:14
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The area of square A is 131 cm^2 greater than the area of square B. If the lengths of the sides of both squares are integers (in cm), then what is the area of square A?

A. 4096 cm^2
B. 4225 cm^2
C. 4356 cm^2 --> correct: a^2=131+b^2=>(a+b)(a-b)=131, because a & b are +ve integers, so (a+b)(a-b)=131*1 (131 is a prime number) i.e. a+b=131 & a-b=1 => a=66 => Area of square A=a^2=66^2=4356
D. 4489 cm^2
E. 4624 cm^2
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The area of square A is 131 cm^2 greater than the area of square B. If 09 May 2020, 10:20
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Bunuel
The area of square A is 131 cm^2 greater than the area of square B. If the lengths of the sides of both squares are integers (in cm), then what is the area of square A?

A. 4096 cm^2
B. 4225 cm^2
C. 4356 cm^2
D. 4489 cm^2
E. 4624 cm^2


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Let the sides of the Square A be " a " and the sides of the Square B be " b "

\(a^2 = b^2 + 131\)

Or, \(a^2 - b^2 = 131\)

Or, \((a + b)(a - b )= 131*1\)

Rest approach as shown by hiranmay

Answer must be (C)
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The area of square A is 131 cm^2 greater than the area of square B. If 20 Mar 2023, 01:25
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