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A certain rectangle has an area of 36 and a perimeter of 26. If the le

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Joined: 02 Sep 2009

Posts: 53066

A certain rectangle has an area of 36 and a perimeter of 26. If the le [#permalink]

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17 Jan 2019, 22:41

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25% (medium)

Question Stats:

79% (01:19) correct

21% (02:11) wrong

based on 23 sessions

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A certain rectangle has an area of 36 and a perimeter of 26. If the length of the rectangle is greater than the width, what is the rectangle’s length?

A 4
B 6
C 9
D 10
E 13

Spoiler: OA

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A certain rectangle has an area of 36 and a perimeter of 26. If the le [#permalink]

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Updated on: 18 Jan 2019, 05:54

Bunuel wrote:

A certain rectangle has an area of 36 and a perimeter of 26. If the length of the rectangle is greater than the width, what is the rectangle’s length?

A 4
B 6
C 9
D 10
E 13

l*b= 13
2 ( l+b) = 26
( l+b)= 13
solve
we get b= 9,4
since l>b
so b = 4, l = 9 IMO C
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Originally posted by Archit3110 on 18 Jan 2019, 04:44.
Last edited by Archit3110 on 18 Jan 2019, 05:54, edited 2 times in total.

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Re: A certain rectangle has an area of 36 and a perimeter of 26. If the le [#permalink]

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18 Jan 2019, 05:52

Archit3110 wrote:

Bunuel wrote:

A certain rectangle has an area of 36 and a perimeter of 26. If the length of the rectangle is greater than the width, what is the rectangle’s length?

A 4
B 6
C 9
D 10
E 13

l*b= 13
2 ( l+b) = 26
( l+b)= 13
solve
we get b= 9,4
since l>b
so b = 4 IMO A

Question is asking for the length.

Posted from my mobile device

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Re: A certain rectangle has an area of 36 and a perimeter of 26. If the le [#permalink]

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18 Jan 2019, 05:54

nmccull wrote:

Archit3110 wrote:

Bunuel wrote:

A certain rectangle has an area of 36 and a perimeter of 26. If the length of the rectangle is greater than the width, what is the rectangle’s length?

A 4
B 6
C 9
D 10
E 13

l*b= 13
2 ( l+b) = 26
( l+b)= 13
solve
we get b= 9,4
since l>b
so b = 4 IMO A

Question is asking for the length.

Posted from my mobile device

edited thanks nmccull
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Re: A certain rectangle has an area of 36 and a perimeter of 26. If the le [#permalink]

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18 Jan 2019, 06:19

let the sides of rectangle be X and Y where X>y

Area of rectangle = X*Y = 36...............i

Perimeter = 2X+2Y = 26
X+Y = 13..................ii
We want to know X
So substitute Y=13-X in first equation -i

X(13-X)=36
-X^2+13X-36 =0
OR
X^2-13X+36 = 0

(X-9)(X-4)=0
X=9 OR X=4....Substitute these values in equation ii
If X=9, then Y = 4, OR X=4 and Y=9

In any scenario, the greater of the two sides = 9

Answer: C
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