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555-605 Level|   Geometry|      
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Bunuel
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If a rectangle has perimeter of 20 and a diagonal with length 9, what 23 Oct 2018, 04:30
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E
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35% (medium)

Question Stats:

72% (01:55) correct 28% (02:21) wrong based on 192 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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E
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If a rectangle has perimeter of 20 and a diagonal with length 9, what 23 Oct 2018, 04:47
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Bunuel
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 23 Oct 2018, 06:33
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Posted from my mobile device

Any other shorter way to solve this question?
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If a rectangle has perimeter of 20 and a diagonal with length 9, what 23 Oct 2018, 06:53
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Bunuel
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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If a rectangle has perimeter of 20 and a diagonal with length 9, what 25 Oct 2018, 09:24
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Bunuel
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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If a rectangle has perimeter of 20 and a diagonal with length 9, what 27 Oct 2018, 00:31
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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?
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If a rectangle has perimeter of 20 and a diagonal with length 9, what 31 Dec 2022, 15:08
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Quick solution:

If the perimeter=20, then 2 adjacent sides will sum to 10. Using the diagonal, we have a triangle with hypotenuse 9 and legs that sum to 10. 1 and 9 will make a hypotenuse just barely bigger than 9. 2 and 8 will make a hypotenuse closer to 8. So our actual area is between 9*1=9 and 8*2=16. That leaves only E.

If there were a larger answer, it would help to notice that 9 and 1 are very close to the actual values, but that the area increases as the sides draw closer to each other in value.
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