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What is the area of the rectangular region depicted above with sides

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What is the area of the rectangular region depicted above with sides [#permalink]

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New post 05 Aug 2016, 04:26
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What is the area of the rectangular region depicted above with sides A and B and diagonal C?

(1) C^2 = 9 + A^2

(2) A + B = 7

[Reveal] Spoiler:
Attachment:
2016-08-05_1625.png
2016-08-05_1625.png [ 2.57 KiB | Viewed 1319 times ]
[Reveal] Spoiler: OA

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What is the area of the rectangular region depicted above with sides [#permalink]

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New post 05 Aug 2016, 05:57
Bunuel wrote:
Image
What is the area of the rectangular region depicted above with sides A and B and diagonal C?

(1) C^2 = 9 + A^2

(2) A + B = 7

[Reveal] Spoiler:
Attachment:
2016-08-05_1625.png


stat 1: C^2 = 9 + A^2. From fig we can get C^2 = B^2 + A^2.

=> sub this C^2 value in theorem. 9 + A^2 = B^2 + A^2 => B^2 = 9 and B is +/-3. Side can't be negative.Still we don't know C and A values. Insufficient to get the area.

Stat 2: A + B = 7 ...we are not sure to get the A and B values independently.

Stat 1 + Stat 2: When B is 3 from Stat 2, A is 4...Then we'll get the area of triangle as 12..

IMO option C is correct answer...

OA please..will correct if I missed anything..
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Re: What is the area of the rectangular region depicted above with sides [#permalink]

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New post 05 Aug 2016, 10:51
Bunuel wrote:
Image
What is the area of the rectangular region depicted above with sides A and B and diagonal C?

(1) C^2 = 9 + A^2

(2) A + B = 7

[Reveal] Spoiler:
Attachment:
2016-08-05_1625.png

(1) B=3
But value of A could be anything and thus we have varying value of C as well area of rectangle.
Not suff...
(2) A=1 then B=6
A=2 then B=5
thus many a combination of A and B makes varying area of rectangle
Not suff..

combining both if B=3 then A=4
thus we have area =3*4=12
suff...

Ans C
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What is the area of the rectangular region depicted above with sides [#permalink]

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New post 05 Aug 2016, 11:30
Bunuel wrote:
What is the area of the rectangular region depicted above with sides A and B and diagonal C?
[Reveal] Spoiler:
Image
Attachment:
2016-08-05_1625.png


We require \(A\) and \(B\) to determine the area of the rectangle.

\(\textbf{(1) } C^2 = 9 + A^2\)

By Pythagoras: \(C^2 = A^2 + B^2\)

\(B^2 = 9\\
B = 3\)

We do not know \(A\).

Insufficient

\(\textbf{(2) } A + B = 7\)

\(A\) and \(B\) could be any values that add up to 7, we can determine the perimeter, but not the area.

Insufficient

\(\textbf{(1 + 2)}\)

\(B = 3 \\
A + B = 7\\
A = 4\)

Sufficient As we have \(A\) and \(B\).

[Reveal] Spoiler:
(C) both statements taken together are sufficient to answer the question, but neither statement alone is sufficient

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Re: What is the area of the rectangular region depicted above with sides [#permalink]

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New post 05 Aug 2016, 11:48
shouldn't the answer be A.
As C^2-A^2=9
So (C-A)(C+A)=9
which has only two values which satisfy C=5 and A=4

Answer A
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Re: What is the area of the rectangular region depicted above with sides [#permalink]

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kplusprinja1988 wrote:
shouldn't the answer be A.
As C^2-A^2=9


The variables are not restricted to integers. \(C = 10, A = \sqrt{91}\) would satisfy the equation.
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Re: What is the area of the rectangular region depicted above with sides [#permalink]

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New post 06 Aug 2016, 23:49
What is the area of the rectangular region depicted above with sides A and B and diagonal C?

(1) C^2 = 9 + A^2

(2) A + B = 7

A) Since C is the hyp : C^2 = 3^2 + A^2
(3,4,5) is a the pyth triplet. We can consider it : in that case: A = 4 and and area is 12
But it is not given that the lengths of the sides are int's ------> hence we could have some fractions for A and C satisfying C^2 = 9 + A^2 -----> insufficient

B) A + B = 7 ------> Could be (A, B) ----> (1 , 7) : Area = 7
(2, 5) : Area = 10 or (3, 4) Area : 12

C) B = 3 and A + B = 7 hence A = 4 hence area is 12 -----> suff
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Re: What is the area of the rectangular region depicted above with sides [#permalink]

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New post 14 Mar 2017, 16:47
What is the area of the rectangular region depicted above with sides A and B and diagonal C?

(1) C^2 = 9 + A^2

(2) A + B = 7


In order to solve this question we need the width and length of the rectangle; in other words, we need to know the value of A and B.

Statement (1) gives us two variables so we cannot solve for either C or A. There is also no way to calculate B from this statement.

Statement (2) tells us that A + B = 7 - we can assume that B corresponds to the opposite side in the rectangle and use the Pythagorean theorem to calculate the hypotenuse

A^2 + B^2 =49 therefore C equals 7. Though we still have two variables, and no angles or angle ratios in the triangle and therefore cannot calculate either side A or B.

Statement (1) and (2) satisfy each other's missing variable. Therefore, Statement (1) and Statement (2) are sufficient.
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Re: What is the area of the rectangular region depicted above with sides [#permalink]

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New post 15 Mar 2017, 02:29
area = AB
and C^2 =A^2 + B^2

St 1
C^2 = 9 +A^2
clearly we can see that B^2 = 9 or B=3. no idea about A. INSUFFICIENT

St 2: A+B=7. two variable, one equation. INSUFFICIENT

St 1 & St 2: A+B=7 and B=3. therefore A=4. hence AB = 12. ANSWER

Option C
Re: What is the area of the rectangular region depicted above with sides   [#permalink] 15 Mar 2017, 02:29
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