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The perimeters of rectangle Y and rectangle Z are equal. The
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05 Aug 2020, 11:29
05 Aug 2020, 11:29
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The perimeters of rectangle Y and rectangle Z are equal. The lengths of the sides of rectangle Y are \(x^2\) + 21 and 7x − 5. The lengths of the sides of rectangle Z are 41 and 5. What is the area of rectangle Y ?
A. 245
B. 480
C. 540
D. 578
E. 720
A. 245
B. 480
C. 540
D. 578
E. 720
Re: The perimeters of rectangle Y and rectangle Z are equal. The
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05 Aug 2020, 22:55
05 Aug 2020, 22:55
1
It is given that perimeter of both rectangles are equal.
x^2 + 21 + 7x − 5 = 41 + 5
On solving this, we get x = 3 or x = -10.
Side cannot be negative and thus, x must be 3.
On substituting value of x in the above equations, we get lengths as 30 and 16.
Area = 30 * 16 = 480
x^2 + 21 + 7x − 5 = 41 + 5
On solving this, we get x = 3 or x = -10.
Side cannot be negative and thus, x must be 3.
On substituting value of x in the above equations, we get lengths as 30 and 16.
Area = 30 * 16 = 480
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Re: The perimeters of rectangle Y and rectangle Z are equal. The
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28 Apr 2021, 06:21
28 Apr 2021, 06:21
Carcass wrote:
The perimeters of rectangle Y and rectangle Z are equal. The lengths of the sides of rectangle Y are \(x^2\) + 21 and 7x − 5. The lengths of the sides of rectangle Z are 41 and 5. What is the area of rectangle Y ?
A. 245
B. 480
C. 540
D. 578
E. 720
A. 245
B. 480
C. 540
D. 578
E. 720
GIVEN: Perimeter rectangle Y = Perimeter rectangle Z
So: (x² + 21) + (x² + 21) + (7x − 5) + (7x − 5) = 41 + 41 + 5 + 5
Simplify: 2x² + 14x + 32 = 92
Subtract 92 from both sides: 2x² + 14x - 60 = 0
Divide both sides by 2 to get: x² + 7x - 30 = 0
Factor: (x + 10)(x - 3) = 0
So, EITHER x = -10 OR x = 3
Since side lengths must be positive, we know that x = 3
The given side lengths of rectangle Y are x² + 21 and + 7x − 5
Substitute to get: 3² + 21 = 30 and 7(3) − 5 = 16
Area = (30)(16) = 480
Answer: B
