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# What is the area of the rectangular region above?

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Senior Manager
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What is the area of the rectangular region above? [#permalink]

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Updated on: 30 Jan 2018, 21:00
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Question Stats:

74% (00:42) correct 26% (00:47) wrong based on 236 sessions

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What is the area of the rectangular region above?

(1) l + w = 6.
(2) d^2 = 20

Attachment:

untitled.PNG [ 1.94 KiB | Viewed 7144 times ]

Originally posted by heygirl on 24 Feb 2011, 11:39.
Last edited by Bunuel on 30 Jan 2018, 21:00, edited 2 times in total.
Edited the question
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Re: What is the area of the rectangular region above? [#permalink]

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24 Feb 2011, 11:46
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(1)
$$l+w = 6$$
$$(l+w)^2 = l^2+w^2+2w*l=36$$
$$w*l=\frac{36-l^2-w^2}{2}$$
Not Sufficient.

(2)
$$d^2=l^2+w^2=20$$
Not Sufficient.

Combining both;
$$w*l=\frac{36-(l^2+w^2)}{2}$$
$$w*l=\frac{36-20}{2}$$
$$w*l=\frac{16}{2}$$
$$w*l=8$$(Area is w*l)

Sufficient.

Ans: "C"
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Re: What is the area of the rectangular region above? [#permalink]

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24 Feb 2011, 11:49
Thanks!
Actually, can we not use the pythagorean theorem and 45-45-90 triangle rule to get l and w using (2)?
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What is the area of the rectangular region above? [#permalink]

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24 Feb 2011, 12:10
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What is the area of the rectangular region above?

$$area=lw=?$$

(1) l + w = 6. Not sufficient to get the value of $$lw$$.
(2) d^2 = 20 --> $$l^2 +w^2 = 20$$. Not sufficient to get the value of $$lw$$.

(1)+(2) Square (1): $$l^2+2lw+w^2=36$$, as from (2) $$l^2 +w^2 = 20$$ then $$2lw+20=36$$ --> $$lw=8$$. Sufficient.

heygirl wrote:
Thanks!
Actually, can we not use the pythagorean theorem and 45-45-90 triangle rule to get l and w using (2)?

Usually the diagonal does not divide a rectangle into two 45-45-90 triangles (it'll be correct only for squares, so when l=w).

Similar questions:
http://gmatclub.com/forum/need-some-hel ... 05414.html
http://gmatclub.com/forum/og12-d48-102246.html
http://gmatclub.com/forum/one-more-geometry-96381.html
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Re: What is the area of the rectangular region above? [#permalink]

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24 Feb 2011, 12:12
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Thanks for letting me know Bunuel. I shall follow the rules henceforth!!
I actually thought b could be the right answer here. I made a wrong assumption : diag of a rect make 45 degrees!
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Re: What is the area of the rectangular region above? [#permalink]

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09 Mar 2011, 00:19
Lolaergasheva wrote:
What is the area of the rectangular region ?
(1) l + w = 6
(2) d^2 = 20

Area of rectangular region is given by $$l*w$$.

Statement 1 gives us,$$l+w=6$$, but l and w can take any values, so insufficient

Statement 2 gives us $$d^2 = 20$$ . Assuming that d refers to length of diagonal, we have $$d^2 = l^2 + w^2 = 20$$. Again l and w can take multiple values, so insufficient.

Combining 1 and 2,

we get $$l+w=6$$ ... (1)

and $$l^2 + w^2 = 20$$.... (2)

Squaring both sides of (1), we get

so, $$l^2 + w^2 + 2l*w = 36$$

Putting $$l^2 + w^2 = 20$$ in above we get $$2l*w = 16$$ or $$l*w = 8$$

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Re: What is the area of the rectangular region above? [#permalink]

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07 Sep 2015, 07:28
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Expert's post
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

What is the area of the rectangular region above?

(1) l + w = 6. Not sufficient to get the value of .
(2) d^2 = 20

In the original condition we have 2 variables for the rectangle(width, length) and thus we need 2 variables to match the number of variables and equations. Since there is 1 each in 1) and 2), C is likely the answer.
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Re: What is the area of the rectangular region above? [#permalink]

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30 Jan 2018, 19:03
Bunuel wrote:
heygirl wrote:
What is the area of a rectangle with length l and width w?
(1)l+w=6
(2)d^2=20(d is the diagonal)

Make sure you type the question in exactly as it was stated from the source. Yuo should not reword or/and shorten the questions.

Original question:

Attachment:
untitled.PNG
What is the area of the rectangular region above?

$$area=lw=?$$

(1) l + w = 6. Not sufficient to get the value of $$lw$$.
(2) d^2 = 20 --> $$l^2 +w^2 = 20$$. Not sufficient to get the value of $$lw$$.

(1)+(2) Square (1): $$l^2+2lw+w^2=36$$, as from (2) $$l^2 +w^2 = 20$$ then $$2lw+20=36$$ --> $$lw=8$$. Sufficient.

heygirl wrote:
Thanks!
Actually, can we not use the pythagorean theorem and 45-45-90 triangle rule to get l and w using (2)?

Usually the diagonal does not divide a rectangle into two 45-45-90 triangles (it'll be correct only for squares, so when l=w).

Similar questions:
http://gmatclub.com/forum/need-some-hel ... 05414.html
http://gmatclub.com/forum/og12-d48-102246.html
http://gmatclub.com/forum/one-more-geometry-96381.html

Bunuel, I screwed up and assumed I was to use the 30:60:90 triangle rule for statement #2 (since it's a rectangle).

How am I supposed to know that rule won't work for this question? Is it because d^2 = 20 -> d= 4 square root 5, but not 4 square root 3?
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Posts: 45455
Re: What is the area of the rectangular region above? [#permalink]

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30 Jan 2018, 21:07
OCDianaOC wrote:
Bunuel wrote:
heygirl wrote:
What is the area of a rectangle with length l and width w?
(1)l+w=6
(2)d^2=20(d is the diagonal)

Make sure you type the question in exactly as it was stated from the source. Yuo should not reword or/and shorten the questions.

Original question:

Attachment:
untitled.PNG
What is the area of the rectangular region above?

$$area=lw=?$$

(1) l + w = 6. Not sufficient to get the value of $$lw$$.
(2) d^2 = 20 --> $$l^2 +w^2 = 20$$. Not sufficient to get the value of $$lw$$.

(1)+(2) Square (1): $$l^2+2lw+w^2=36$$, as from (2) $$l^2 +w^2 = 20$$ then $$2lw+20=36$$ --> $$lw=8$$. Sufficient.

heygirl wrote:
Thanks!
Actually, can we not use the pythagorean theorem and 45-45-90 triangle rule to get l and w using (2)?

Usually the diagonal does not divide a rectangle into two 45-45-90 triangles (it'll be correct only for squares, so when l=w).

Similar questions:
http://gmatclub.com/forum/need-some-hel ... 05414.html
http://gmatclub.com/forum/og12-d48-102246.html
http://gmatclub.com/forum/one-more-geometry-96381.html

Bunuel, I screwed up and assumed I was to use the 30:60:90 triangle rule for statement #2 (since it's a rectangle).

How am I supposed to know that rule won't work for this question? Is it because d^2 = 20 -> d= 4 square root 5, but not 4 square root 3?

Knowing only the length of a diagonal of a rectangle does not allow to find its adjacent angles. You can squeeze or stretch a rectangle with any length of a diagonal along one of its sides to get any adjacent lengths from 0 to 90 not inclusive.
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Re: What is the area of the rectangular region above? [#permalink]

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22 Apr 2018, 04:38
Bunuel wrote:

What is the area of the rectangular region above?

$$area=lw=?$$

(1) l + w = 6. Not sufficient to get the value of $$lw$$.
(2) d^2 = 20 --> $$l^2 +w^2 = 20$$. Not sufficient to get the value of $$lw$$.

(1)+(2) Square (1): $$l^2+2lw+w^2=36$$, as from (2) $$l^2 +w^2 = 20$$ then $$2lw+20=36$$ --> $$lw=8$$. Sufficient.

heygirl wrote:
Thanks!
Actually, can we not use the pythagorean theorem and 45-45-90 triangle rule to get l and w using (2)?

Usually the diagonal does not divide a rectangle into two 45-45-90 triangles (it'll be correct only for squares, so when l=w).

Similar questions:
http://gmatclub.com/forum/need-some-hel ... 05414.html
http://gmatclub.com/forum/og12-d48-102246.html
http://gmatclub.com/forum/one-more-geometry-96381.html

Bunuel why are you squaring statement one l + w = 6 ?? the formula for area of rectangle is L*W , isnt it
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Re: What is the area of the rectangular region above? [#permalink]

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22 Apr 2018, 11:17
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dave13 wrote:
Bunuel wrote:

What is the area of the rectangular region above?

$$area=lw=?$$

(1) l + w = 6. Not sufficient to get the value of $$lw$$.
(2) d^2 = 20 --> $$l^2 +w^2 = 20$$. Not sufficient to get the value of $$lw$$.

(1)+(2) Square (1): $$l^2+2lw+w^2=36$$, as from (2) $$l^2 +w^2 = 20$$ then $$2lw+20=36$$ --> $$lw=8$$. Sufficient.

heygirl wrote:
Thanks!
Actually, can we not use the pythagorean theorem and 45-45-90 triangle rule to get l and w using (2)?

Usually the diagonal does not divide a rectangle into two 45-45-90 triangles (it'll be correct only for squares, so when l=w).

Similar questions:
http://gmatclub.com/forum/need-some-hel ... 05414.html
http://gmatclub.com/forum/og12-d48-102246.html
http://gmatclub.com/forum/one-more-geometry-96381.html

Bunuel why are you squaring statement one l + w = 6 ?? the formula for area of rectangle is L*W , isnt it

Hello Dave

I think Bunuel is doing this because he sees two equations from two statements: one equation says that l + w = 6 and the other says that l^2 + w^2 = 20. But he has to find l*w. So how to find l*w from l+w and l^2 + w^2. Immediately the formula which comes to mind is:
(l+w)^2 = l^2 + w^2 + 2*l*w.
We can substitute the values of l+w and l^2 + w^2 in the above formula and we can get the value of l*w. Hope this is clear.
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Re: What is the area of the rectangular region above? [#permalink]

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24 Apr 2018, 01:44
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Need to find area = l*w
1. l+w=6 --> sqrn both sides --> l^2 + w^2 + 2wl=36 -eqn(a)
INSUFFICIENT
2. d^2=20 --> l^2+w^2=d^2=20 -eqn(b)
INSUFFICIENT

Consider eqn (a) and (b)
20 +2wl=36 --> Solve for wl

Ans -> C
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Re: What is the area of the rectangular region above? [#permalink]

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08 May 2018, 07:59
Area=LW
We got L*L + W*W=D*D
Converted to

=(L+W)^2 -2LW=20

=36-20=2LW

2LW=16

LW =8

Both statements are required

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Re: What is the area of the rectangular region above?   [#permalink] 08 May 2018, 07:59
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