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The length of the diagonal of a certain rectangle is 8. If the sum of

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The length of the diagonal of a certain rectangle is 8. If the sum of  [#permalink]

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New post 28 Aug 2018, 05:26
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A
B
C
D
E

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  55% (hard)

Question Stats:

60% (01:49) correct 40% (02:39) wrong based on 101 sessions

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Re: The length of the diagonal of a certain rectangle is 8. If the sum of  [#permalink]

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New post 28 Aug 2018, 05:35
Bunuel wrote:
The length of the diagonal of a certain rectangle is 8. If the sum of the lengths of the sides is 10, what is the area of the rectangle?

A. 18
B. 36
C. 64
D. 80
E. It cannot be determined.


a+ b = 10
a^2 + b^2 = 100 -2ab
64 = 100-2ab
2ab = 36 , ab=18
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Re: The length of the diagonal of a certain rectangle is 8. If the sum of  [#permalink]

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New post 28 Aug 2018, 05:39
Bunuel wrote:
The length of the diagonal of a certain rectangle is 8. If the sum of the lengths of the sides is 10, what is the area of the rectangle?

A. 18
B. 36
C. 64
D. 80
E. It cannot be determined.


The length of the diagonal of a certain rectangle is 8 i,e d=8

a+b=10 --> (1)

(a+b)^2=100

a^2+b^2+2ab=100

we know that

a^2+b^2 = d^2

64+2ab=100

2ab=36

ab=18

Hence A
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The length of the diagonal of a certain rectangle is 8. If the sum of  [#permalink]

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New post 28 Aug 2018, 05:47
Bunuel wrote:
The length of the diagonal of a certain rectangle is 8. If the sum of the lengths of the sides is 10, what is the area of the rectangle?

A. 18
B. 36
C. 64
D. 80
E. It cannot be determined.


Given, Diagonal=\(\sqrt{l^2+w^2}=8\) Or, \(l^2+w^2=8^2\)
Given, sum of the lengths of the sides is 10 Or (l+w)=10

To find, area=l*w=?

\((l+w)^2=l^2+w^2+2lw\)
Or, \(10^2=8^2+2lw\)
Or, \(l*w=\frac{10^2-8^2}{2}\)=18

Ans. (A)
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Re: The length of the diagonal of a certain rectangle is 8. If the sum of  [#permalink]

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New post 30 Aug 2018, 18:08
Bunuel wrote:
The length of the diagonal of a certain rectangle is 8. If the sum of the lengths of the sides is 10, what is the area of the rectangle?

A. 18
B. 36
C. 64
D. 80
E. It cannot be determined.


We can create the following equations:

L^2 + W^2 = 8^2

L^2 + W^2 = 64

and

L + W = 10

Squaring both sides of the equation, we have:

(L + W)^2 = 10^2

L^2 + W^2 + 2LW = 100

Substituting, we have:

64 + 2LW = 100

2LW = 36

LW = 18, which is the area of the rectangle.

Answer: A
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Re: The length of the diagonal of a certain rectangle is 8. If the sum of  [#permalink]

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New post 07 Nov 2018, 01:17
Dear experts please help,

Shouldn't the sum of the sides of the rectangle be 2L + 2W, why is it L + W?

Sincerely, Thanks.
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Re: The length of the diagonal of a certain rectangle is 8. If the sum of  [#permalink]

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New post 10 Nov 2018, 19:56
Valhalla wrote:
Dear experts please help,

Shouldn't the sum of the sides of the rectangle be 2L + 2W, why is it L + W?

Sincerely, Thanks.


I agree the wording is slightly ambiguous but I believe it can be inferred quite clearly that sum of L + W is 10 because it forms a triangle with the diagonal, which is 8, and the sum of two sides needs to be larger than the third side. If you assume the perimeter of the rectangle to be 10 then you can't quite form the triangle with an 8 diagonal.
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The length of the diagonal of a certain rectangle is 8. If the sum of  [#permalink]

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New post 10 Aug 2019, 23:08
I understand the stated solution to the problem, but doesn't the problem make it unable for a diagonal with a length of 8 to exist within a rectangle having an area of 18. Either the two sides are 9 x 2 or 6 x 3. 9 x 2 would make 9 the diagonal of a 9,8,2 triangle and also does not add to 10. 6 x 3 only adds to 9 and would imply the diagonal is √45.

What am I missing here?

Thank you,
Bryan
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The length of the diagonal of a certain rectangle is 8. If the sum of   [#permalink] 10 Aug 2019, 23:08
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