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# The length of a certain rectangle is 4 inches longer than its width.

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Math Expert
Joined: 02 Sep 2009
Posts: 43917
The length of a certain rectangle is 4 inches longer than its width. [#permalink]

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09 Sep 2015, 23:27
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The length of a certain rectangle is 4 inches longer than its width. If the area of the rectangle is 221 square inches, then the length of the rectangle’s diagonal, in inches, is

(A) between 19 and 20
(B) between 20 and 21
(C) between 21 and 22
(D) between 22 and 23
(E) between 23 and 24

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Joined: 07 Apr 2015
Posts: 179
The length of a certain rectangle is 4 inches longer than its width. [#permalink]

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10 Sep 2015, 00:38
1
KUDOS
l = w+4

$$l*w = 221$$
$$w+4*w = 221$$
$$w^2+4w-221=0$$

$$(w+17)(w-13)$$

w is -17 or 13, only 13 is possible.
Hence L = 13+4 = 17

$$13^2+17^2 = x^2$$
$$169+289 = x^2$$
$$458 = x^2$$

20*20 is 400, therefore x has to be greater 20. 22*22 (just tried that one) is 484, therefore it has to be below that. Hence C

(P.S.: No Idea how to do that task in <2 minutes, to much calculation involved)
Manager
Joined: 10 Aug 2015
Posts: 103
The length of a certain rectangle is 4 inches longer than its width. [#permalink]

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10 Sep 2015, 03:51
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Solution: a(a+4) = 221. 221 is near to 15^2. So, look for two no. around 15 whose product will have 1 in its uits place.You dont have to look far, its 13 and 17.
Length required = sqrt(13^2 + 17^2) = sqrt(169 + 289) = sqrt(458)
21^2 = 441 and 22^2 = 484.

Option C
Manager
Joined: 29 Jul 2015
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Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]

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10 Sep 2015, 06:10
Bunuel wrote:
The length of a certain rectangle is 4 inches longer than its width. If the area of the rectangle is 221 square inches, then the length of the rectangle’s diagonal, in inches, is

(A) between 19 and 20
(B) between 20 and 21
(C) between 21 and 22
(D) between 22 and 23
(E) between 23 and 24

Kudos for a correct solution.

Let the width of the rectangle be x,then the length will be x+4.
The area is 221.
so,
x(x+4)=221
or $$x^2+4x-221=0$$
or $$(x-13)(x+17)=0$$
or x=13

width is 13 so the lenght will 13+4= 17
So, The diagonal of the rectangle will be
$$\sqrt{13^2 + 17^2}$$ = $$\sqrt{458}$$

Clearly, the sum lies between $$21^2$$ i.e 441 and $$22^2$$ i.e 484

Intern
Joined: 20 Apr 2015
Posts: 10
Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]

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10 Sep 2015, 23:34
1
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L = 17, B = 13
169+289 = 458 is between 441 and 484 hence between 21 and 22. answer C
Director
Joined: 21 May 2013
Posts: 581
Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]

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11 Sep 2015, 02:01
Bunuel wrote:
The length of a certain rectangle is 4 inches longer than its width. If the area of the rectangle is 221 square inches, then the length of the rectangle’s diagonal, in inches, is

(A) between 19 and 20
(B) between 20 and 21
(C) between 21 and 22
(D) between 22 and 23
(E) between 23 and 24

Kudos for a correct solution.

Width=x, Length=x+4
x*x+4=221
Solving for x, x=13,x+4=17
Therefore using pythagoras theorem, Diagonal^2=169+289=458
Diagonal=More than 21 but less than 22
Math Expert
Joined: 02 Sep 2009
Posts: 43917
Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]

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13 Sep 2015, 08:00
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Expert's post
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Bunuel wrote:
The length of a certain rectangle is 4 inches longer than its width. If the area of the rectangle is 221 square inches, then the length of the rectangle’s diagonal, in inches, is

(A) between 19 and 20
(B) between 20 and 21
(C) between 21 and 22
(D) between 22 and 23
(E) between 23 and 24

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

First, set up an equation for the area of the rectangle. If x is the width, then we have

x(x + 4) = 221

Note that if we put this into standard quadratic form and then try to factor, we wind up back where we started, in some sense: we are looking for two numbers that multiply to 221 and that differ by 4.

x^2 + 4x – 221 = 0

Beyond pure trial and error, we can look for nearby squares. 221 is nearly 225, which equals 15^2. So we might try numbers near 15. As it turns out, 221 = 13 × 17. We might even get there by noticing a difference of squares:

221 = 225 – 4 = 15^2 – 2^2 = (15 – 2)(15 + 2) = 13 × 17.

As a last resort, we could always use the quadratic formula, which gets us the roots of the equation as well.

Now, the diagonal of the rectangle will be given by the Pythagorean Theorem:

d^2 = 13^2 + 17^2
= 169 + 289
= 458.

The square root of 458 is definitely larger than 20, since 20^2 = 400. Going up, we can compute 21^2 = 441 < 458, whereas 22^2 = 484 > 458. So the length of the diagonal must be between 21 and 22.

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Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]

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16 May 2017, 19:19
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Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]

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16 May 2017, 20:11
SHORTER METHOD

Let the dimensions be L & B
Given: L - B= 4 & L.B= 221
Length of diagonal = (L^2 + B^2)^(1/2)

L^2 + B^2= √{(L-B)^2 + 2L.B}=√{(4)^2 + 2.(221)}= √{16 + 442}= √(458)

Now 21^2=441 & 22^2= 484.

Hence 21< Diagonal length< 22. Hence C
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Re: The length of a certain rectangle is 4 inches longer than its width. [#permalink]

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17 May 2017, 10:45
Bunuel wrote:
The length of a certain rectangle is 4 inches longer than its width. If the area of the rectangle is 221 square inches, then the length of the rectangle’s diagonal, in inches, is

(A) between 19 and 20
(B) between 20 and 21
(C) between 21 and 22
(D) between 22 and 23
(E) between 23 and 24

Kudos for a correct solution.

W* (w+4)= 221

221 has four factors 1, 13, 17,and liself

W= 13
length= 13 +4 = 17

According to , Pythagorean theoram
C^2= a^2 + b^2
C^2= 13^2+ 17^2

thus Diagonal could be \sqrt{458}
and could be Between 21 and 22.
Re: The length of a certain rectangle is 4 inches longer than its width.   [#permalink] 17 May 2017, 10:45
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