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The area of a rectangle with perimeter 20
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17 Sep 2018, 16:44
17 Sep 2018, 16:44
Expert Reply
00:00
Question Stats:
69% (00:44) correct
30% (00:21) wrong based on 96 sessions
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Quantity A |
Quantity B |
The area of a rectangle with perimeter 20 |
\(30\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: The area of a rectangle with perimeter 20
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06 Apr 2020, 09:44
06 Apr 2020, 09:44
Expert Reply
The perimeter is \(2(l+w) =20
\)
\(l+w=10\)
Now, having 10 means that the base and the height are OR 2*5=10 OR 10*1=10
Either way, the Area is \(b \times h\)
30 will be always bigger than the number we could get with the combinations of l+w
B is the answer
\)
\(l+w=10\)
Now, having 10 means that the base and the height are OR 2*5=10 OR 10*1=10
Either way, the Area is \(b \times h\)
30 will be always bigger than the number we could get with the combinations of l+w
B is the answer
The area of a rectangle with perimeter 20
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24 Jun 2022, 22:27
24 Jun 2022, 22:27
2
Carcass wrote:
The perimeter is \(2(l+w) =20
\)
\(l+w=10\)
Now, having 10 means that the base and the height are OR 2*5=10 OR 10*1=10
Either way, the Area is \(b \times h\)
30 will be always bigger than the number we could get with the combinations of l+w
B is the answer
\)
\(l+w=10\)
Now, having 10 means that the base and the height are OR 2*5=10 OR 10*1=10
Either way, the Area is \(b \times h\)
30 will be always bigger than the number we could get with the combinations of l+w
B is the answer
nidhi007 :
To solve this, we should recogize the greatest product possible for two numbers summing up to half the perimeter, i.e., 10
9*1=9
8*2=16
7*3=21
6*4=24
5*5=25 Greatest
repeats
The greatest possible area for a rectangle with perimeter of 20 is therefore 25
Answer is (B)
