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The perimeter of rectangle ABCD is 5/2 times as long as the
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23 Mar 2019, 10:24
23 Mar 2019, 10:24
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The perimeter of rectangle ABCD is \(\frac{5}{2}\) times as long as the side AB.
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Length of side AB |
Length of side BC |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
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Re: The perimeter of rectangle ABCD is 5/2 times as long as the
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10 Sep 2021, 12:59
10 Sep 2021, 12:59
1
Carcass wrote:
The perimeter of rectangle ABCD is \(\frac{5}{2}\) times as long as the side AB.
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Length of side AB |
Length of side BC |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Let x = the length of sides AB and DC
Let y = the length of sides AD and BC
The perimeter of rectangle ABCD is 5/2 times the length of side AB
We can write: \(2x + 2y = (\frac{5}{2})x\)
Multiply both sides of the equation by \(2\) to get: \(4x + 4y = 5x\)
Subtract \(4x\) from both sides to get: \(4y = x\)
So, we have:
QUANTITY A: Length of side AB = x
QUANTITY B: Length of side BC = y
Since we know that \(4y = x\), we can replace x with 4y to get:
QUANTITY A: 4y
QUANTITY B: y
Answer: A
General Discussion
Re: The perimeter of rectangle ABCD is 5/2 times as long as the
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28 Mar 2019, 23:10
28 Mar 2019, 23:10
1
Take a rectangle ABCD.
Assume length of side AB = x.
Assume length of side BC = y.
For a rectangle its perimeter is \(2(a+b)\) where a and b are adjacent sides in the rectangle.
Therefore the perimeter for this rectangle ABCD is \(2(x+y) = \frac{5}{2} * x\)
solve for x we get,
\(x = 4y\)
Hence AB = 4y while
BC = Y
since y is +ve we have AB > BC
Assume length of side AB = x.
Assume length of side BC = y.
For a rectangle its perimeter is \(2(a+b)\) where a and b are adjacent sides in the rectangle.
Therefore the perimeter for this rectangle ABCD is \(2(x+y) = \frac{5}{2} * x\)
solve for x we get,
\(x = 4y\)
Hence AB = 4y while
BC = Y
since y is +ve we have AB > BC
