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

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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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New post 17 Jan 2019, 23:41
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

82% (01:21) correct 18% (02:11) wrong based on 26 sessions

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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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New post Updated on: 18 Jan 2019, 06:54
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, 05:44.
Last edited by Archit3110 on 18 Jan 2019, 06:54, 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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New post 18 Jan 2019, 06:52
1
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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New post 18 Jan 2019, 06:54
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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New post 18 Jan 2019, 07:19
1
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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Re: A certain rectangle has an area of 36 and a perimeter of 26. If the le   [#permalink] 18 Jan 2019, 07:19
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