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

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Math Expert
Joined: 02 Sep 2009
Posts: 58320
If a rectangle has perimeter of 20 and a diagonal with length 9, what  [#permalink]

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23 Oct 2018, 04:30
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55% (hard)

Question Stats:

66% (01:54) correct 34% (02:15) wrong based on 93 sessions

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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

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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, 04:47
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 $$(L+W)^2 = 100$$

$$100 = 81 + 2LW$$

$$2LW = 19$$

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

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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, 06:33
Posted from my mobile device

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, 06:53
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, 09:24
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

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

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27 Oct 2018, 00:31
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.

Re: If a rectangle has perimeter of 20 and a diagonal with length 9, what   [#permalink] 27 Oct 2018, 00:31
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