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ABCD is a parallelogram which has a right angle. If the

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ABCD is a parallelogram which has a right angle. If the  [#permalink]

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New post 30 Oct 2011, 03:01
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ABCD is a parallelogram which has a right angle. If the diagonals intersect in point E, what is the area of ABCD?

1. side AB equals 5
2. Area of AEB triangle is 5

any inputs?
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Re: Area of Rectangle - 700+  [#permalink]

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New post 30 Oct 2011, 04:39
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SonyGmat wrote:
ABCD is a parallelogram which has a right angle. If the diagonals intersect in point E, what is the area of ABCD?

1. side AB equals 5
2. Area of AEB triangle is 5

any inputs?


We know ABCD is a rectangle. And the intersection of the diagonal is at half the distance from either side.

2. Lets draw a perpendicular from E on AB and call the intersection P.
Area of AEB= 1/2*AB*PE
PE=1/2*BC
Area of AEB= 1/2*AB*1/2*BC=5
AB*BC=5*4=20
Sufficient.

Ans: "B"
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Re: Area of Rectangle - 700+  [#permalink]

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New post 30 Oct 2011, 10:11
fluke wrote:
SonyGmat wrote:
ABCD is a parallelogram which has a right angle. If the diagonals intersect in point E, what is the area of ABCD?

1. side AB equals 5
2. Area of AEB triangle is 5

any inputs?


We know ABCD is a rectangle. And the intersection of the diagonal is at half the distance from either side.

2. Lets draw a perpendicular from E on AB and call the intersection P.
Area of AEB= 1/2*AB*PE
PE=1/2*BC
Area of AEB= 1/2*AB*1/2*BC=5
AB*BC=5*4=20
Sufficient.

Ans: "B"


why can't it be a square?? How did you arrive to the point that its a rectangle ?
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New post 30 Oct 2011, 10:33
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gmatcracker24 wrote:
why can't it be a square?? How did you arrive to the point that its a rectangle ?


A rectangle includes square i.e. a square is a specialized rectangle and it is perfectly alright to call a square rectangle; however the reverse may not be true. That's beside the point though. Square or rectangle, the point of intersection of the two diagonals is midway between both sides in either of the quadrilaterals. In other words, the distance between the point of intersection and side of the rectangle would either be 1/2*length OR be 1/2*width. You can prove it with similar triangles;

Say,
ABCD is the rectangle. AC and BD intersect at O. Draw a perpendicular from O on AB such that it intersects at P. Now,

ABC ~ APO
AC/AO=BC/PO
AC=2AO(Diagonals bisect each other)
BC/PO=2
PO=1/2*BC
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Re: Area of Rectangle - 700+  [#permalink]

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New post 30 Oct 2011, 11:22
fluke wrote:
Say,
ABCD is the rectangle. AC and BD intersect at O. Draw a perpendicular from O on AB such that it intersects at P. Now,

ABC ~ APO
AC/AO=BC/PO
AC=2AO(Diagonals bisect each other)
BC/PO=2
PO=1/2*BC


But we must prove that the triangles, ABC, APO are similar using the SAS theorem. (2 sides in proportion and the angle between them will be equal)

1.AP and AB are in proportion (because the altitude, AP (of APO triangle) will also be median, since AOB is isosceles triangle. It is isosceles because the diagonals bisect each other and AO=BO)

2.AO and AC are in proportion (since in rectangle the diagonals bisect each other)

3.they share the included angle CAB=OAP

Therefore the sides of triangles of APO:ABC will be 1:2, therefore the ratio of PO:BC will be 1:2 also.

Or trying to prove similarity, we could say that they have 2 angles congruent, therefore they are similar.

Correct?

Thanks so much fluke for your input!
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Re: Area of Rectangle - 700+  [#permalink]

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New post 30 Oct 2011, 13:04
SonyGmat wrote:
fluke wrote:
Say,
ABCD is the rectangle. AC and BD intersect at O. Draw a perpendicular from O on AB such that it intersects at P. Now,

ABC ~ APO
AC/AO=BC/PO
AC=2AO(Diagonals bisect each other)
BC/PO=2
PO=1/2*BC


But we must prove that the triangles, ABC, APO are similar using the SAS theorem. (2 sides in proportion and the angle between them will be equal)

1.AP and AB are in proportion (because the altitude, AP (of APO triangle) will also be median, since AOB is isosceles triangle. It is isosceles because the diagonals bisect each other and AO=BO)

2.AO and AC are in proportion (since in rectangle the diagonals bisect each other)

3.they share the included angle CAB=OAP

Therefore the sides of triangles of APO:ABC will be 1:2, therefore the ratio of PO:BC will be 1:2 also.

Or trying to prove similarity, we could say that they have 2 angles congruent, therefore they are similar.

Correct?

Thanks so much fluke for your input!


I'd prove the similarity using AA criterion.
<BAC=<PAO{:Same angle}
And
<APO=<ABC{:Alternate angle of two parallel lines OP and BC intersected by line AB}

Rest is there in my earlier post.
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Re: Area of Rectangle - 700+  [#permalink]

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New post 31 Oct 2011, 06:08
From the prompt, it is given that the parallelogram is a rectangle. Find the area of the rectangle

It cannot be a square because inorder to satisfy the criterion for a square, all the four sides have to be equal

Statement 1
Gives us one side of the rectangle. Insufficient

Statement 2
1/2*AB*height = 5. Here the height = 1/2*(the width of the rectangle, say BC)

1/2*AB*1/2*BC = 5 or AB*BC = 20. Sufficient
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Re: Area of Rectangle - 700+  [#permalink]

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New post 01 Nov 2011, 00:08
SonyGmat wrote:
ABCD is a parallelogram which has a right angle. If the diagonals intersect in point E, what is the area of ABCD?

1. side AB equals 5
2. Area of AEB triangle is 5

any inputs?


Here is my take:

I think everyone here agrees that the parallelogram is a rectangle.
Attachment:
Ques3.jpg
Ques3.jpg [ 8.04 KiB | Viewed 2705 times ]

Stmnt 2 tells us that Area of AEB = 5 = (1/2)*a*(1/2)b
What is the area of triangle BED? It is (1/2)*b*(1/2)a
It is exactly the same as the area of AEB and hence will also be 5. Similarly, the area of the other 2 triangles is also 5 each.
So area of rectangle will be 20.
Stmnt 2 is sufficient alone.

Now think of the takeaway from this question.
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Re: ABCD is a parallelogram which has a right angle. If the  [#permalink]

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New post 04 Jan 2012, 01:53
B indeed.

Question: What is the area of parallelogram ABCD?
We know that the parallelogram with a right angle is a rectangle. So, to find out the area, we just need to find out the product of the sides of the rectangle.

Statement 1:
We know the length of only one of the sides of the rectangle. We have no information on the other side. Thus INSUFFICIENT

Statement 2:
We know that E will be equidistant from (AB and CD) and (BC and AD). Thus, the height of the triangle is half the breadth of the rectangle.
From the formula of area of triangle, we can substitute the values and find out that the area of the rectangle is 20. Thus SUFFICIENT.

Answer: B
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Re: ABCD is a parallelogram which has a right angle. If the  [#permalink]

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New post 17 Jun 2017, 19:25
ABCD is a parallelogram which has a right angle. If the diagonals intersect in point E, what is the area of ABCD?

1. side AB equals 5
2. Area of AEB triangle is 5

Hi Bunuel ,

Could you please comment on this question . If the question is correct from GMAT point of view .
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Re: ABCD is a parallelogram which has a right angle. If the  [#permalink]

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