If a rectangle has perimeter of 20 and a diagonal with length 9, what : Problem Solving (PS)

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If a rectangle has perimeter of 20 and a diagonal with length 9, what

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If a rectangle has perimeter of 20 and a diagonal with length 9, what [#permalink]

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23 Oct 2018, 03:30

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If a rectangle has perimeter of 20 and a diagonal with length 9, what is the area of the rectangle?

A. 7.5
B. 8
C. 8.5
D. 9
E. 9.5

Spoiler: OA

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Re: If a rectangle has perimeter of 20 and a diagonal with length 9, what [#permalink]

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23 Oct 2018, 03:47

Bunuel wrote:

If a rectangle has perimeter of 20 and a diagonal with length 9, what is the area of the rectangle?

A. 7.5
B. 8
C. 8.5
D. 9
E. 9.5

Rectangle perimeter = 2*(L+W)

Area = L*W

20 = 2 (L+W)

L+W = 10

Given we have the diagonal length = 9

by Pythagorean rule we know $D^2 = L^2 + W^2$

$81 = L^2 + W^2$

$(L+W)^2 = L^2 + 2LW + W^2$

Given we know $L+W = 10$ then $(L+W)^2 = 100$

$100 = 81 + 2LW$

$2LW = 19$

$LW = \frac{19}{2} = 9.5$

Answer choice E

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If a rectangle has perimeter of 20 and a diagonal with length 9, what [#permalink]

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23 Oct 2018, 05:33

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Any other shorter way to solve this question?

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Re: If a rectangle has perimeter of 20 and a diagonal with length 9, what [#permalink]

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23 Oct 2018, 05:53

Bunuel wrote:

If a rectangle has perimeter of 20 and a diagonal with length 9, what is the area of the rectangle?

A. 7.5
B. 8
C. 8.5
D. 9
E. 9.5

$2( L + B ) = 20$

So, $L + B = 10$

$D = \sqrt{L^2 + B^2}$

Or, $9 = \sqrt{L^2 + B^2}$

Or, $L^2 + B^2 = 81$

$( L + B)^2 = L^2 + B^2 +2LB$

Or, $10^2 = 81 + 2LB$

Or, $2LB = 19$

So, $LB = 9.50$ , Answer Area must be (E) 9.50
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Re: If a rectangle has perimeter of 20 and a diagonal with length 9, what [#permalink]

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25 Oct 2018, 08:24

Bunuel wrote:

If a rectangle has perimeter of 20 and a diagonal with length 9, what is the area of the rectangle?

A. 7.5
B. 8
C. 8.5
D. 9
E. 9.5

We can let L = the length and W = the width of the rectangle. Our goal is to determine the value of LW since that is the area of the rectangle. We can create the equations:

By the Pythagorean theorem:

L^2 + W^2 = 9^2 = 81

and

by the formula for the perimeter of a rectangle:

2L + 2W = 20

L + W = 10

Squaring both sides of L + W = 10, we have:

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

Substituting, we have:

81 + 2LW = 100

2LW = 19

LW = 9.5

Answer: E
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Re: If a rectangle has perimeter of 20 and a diagonal with length 9, what [#permalink]

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26 Oct 2018, 23:31

One question: When I solved this question by quadratic equation - (l^2 + (10-l)^2) =81, I got l=(20+/-Sq.root (248))/(4).
Which results in l~= 9 or 9.1 and w~=1.

Please help me understand where I went wrong?