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I thought A, because if the base is 10, the side has to be 10, because Area is 10. Therefore perimeter is 40. Regardless if this is a rhombus or not. B is insuffic, gives no details for lengths of sides.
(Maybe I am confusing what a base is? I am interpreting it as one of the sides.)
The OA is C. Not sure why, can someone please explain?
Re: Parallelogram - Area and perimeter [#permalink]
29 Jun 2009, 23:27
3
This post received KUDOS
If the area of a parallelogram is 100, what is the perimeter of the parallelogram?
1. The base of the parallelogram is 10. 2. One of the angles of the parallelogram is 45 degrees.
Statement 1 base = 10 given area of a parallelogram = 100 base x height = 100 so, height = 10 height does not give the side of the parallelogram... height is only equal to side if the parallelogram is a rectangle or a square Insufficient Statement 2 one of the angle of the parallelogram is 45 degs. This does not say anything about sides. insufficient
combining we can use the Pythagoras theorem to find the side... so sufficient Answer C
Re: Parallelogram - Area and perimeter [#permalink]
02 Jul 2009, 06:57
hi there.. Sorry I cant figure out how to find the side of the parallelogram , I understand that the height is 10 which will be one side and we can use pytho to find out the hypotneous but how do you find the third side of this triangle.
Re: Parallelogram - Area and perimeter [#permalink]
28 May 2010, 13:15
I do not get this problem Ok, let's imagine that the height of the parallelogram is 10 and it's base is 10 as well. One of the angles of parallelogram is 45 degrees. That would mean, that one of the angles of the parallelogram height of the parallelogram and one of the sides of parallelogram will form right triangle. One leg of this triangle is 10, the other leg, which is part of the base of the parallelogram need to be 10 as well (Why?) Because the right triangle is an isosceles triangle with angles 45, 45 and 90 degrees respectively.
BUT, that triangle would not make sense, since the total length of the base of the parallelogram is 10, how part of the length of the parallelogram could be 10 also.
Can somebody explain the problem to me. Thank you _________________
Re: Parallelogram - Area and perimeter [#permalink]
28 May 2010, 15:57
1
This post received KUDOS
Expert's post
mirzohidjon wrote:
I do not get this problem Ok, let's imagine that the height of the parallelogram is 10 and it's base is 10 as well. One of the angles of parallelogram is 45 degrees. That would mean, that one of the angles of the parallelogram height of the parallelogram and one of the sides of parallelogram will form right triangle. One leg of this triangle is 10, the other leg, which is part of the base of the parallelogram need to be 10 as well (Why?) Because the right triangle is an isosceles triangle with angles 45, 45 and 90 degrees respectively.
BUT, that triangle would not make sense, since the total length of the base of the parallelogram is 10, how part of the length of the parallelogram could be 10 also.
Can somebody explain the problem to me. Thank you
Let parallelogram be ABCD. Base AD = 10, height BX = 10 and the angle BAD = 45 degrees. This would mean that when you draw the hight from B it will meat the base at point D (D and X will coincide). So the height will be the diagonal too. AB and CD will become hypotenuses and will be equal to \(10\sqrt{2}\).
Re: Parallelogram - Area and perimeter [#permalink]
28 May 2010, 17:16
Hi, Bunuel, I really appreciate your effort to help us. Finally after your explanation, I got the answer (since I imagined what kind of parallelogram the problem was talking about. But, based on what assumptions, you deduce that parallelogram needs to be the way you described (diagonal equal to height)? _________________
Re: Parallelogram - Area and perimeter [#permalink]
29 May 2010, 03:30
3
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
mirzohidjon wrote:
Hi, Bunuel, I really appreciate your effort to help us. Finally after your explanation, I got the answer (since I imagined what kind of parallelogram the problem was talking about. But, based on what assumptions, you deduce that parallelogram needs to be the way you described (diagonal equal to height)?
Attachment:
PointLatticeParallelograms_1000.gif [ 3.32 KiB | Viewed 10490 times ]
If the area of a parallelogram is 100, what is the perimeter of the parallelogram?
Given: \(Area=base*height=100\). Q: \(P=2b+2l=?\) (b - base, l - leg )
(1) The base of the parallelogram is 10 --> \(base=height=10\). Infinite variations are possible. Look at the diagram (let the distance between two horizontal and vertical points be 10): all 4 parallelograms have \(base=height=10\) but they have different perimeter. Not sufficient. Side notes: \(l\geq{10}\), when \(l=10=h\) we would have the square (case #3 on the diagram) and \(P=40\) (smallest possible perimeter), maximum value of perimeter is not limited.
(2) One of the angles of the parallelogram is 45 degrees. Clearly insufficient. But from this statement height BX and AX will make isosceles right triangle: \(height=BX=AX\).
(1)+(2) As from 2 we have that \(height=BX=AX\) and from (1) we have that \(base=height=10\) --> \(AX=base=AD=10\) --> X and D coincide (case #4 on the diagram) --> leg (AB) becomes hypotenuse of the isosceles right triangle with sides equal to 10 --> \(AB=10\sqrt{2}\) --> \(P=20+20\sqrt{2}\). Sufficient.
Re: Parallelogram - Area and perimeter [#permalink]
14 Jun 2010, 05:31
Hi every body, I think statement-2 alone will be suffice if we want to find the Perimeter. If one angle is 45, another angle has to be 45. In addition other two angles will be 270/2 each. Now, if you consider area of 100, you can find only one parallelogram matching these qualities. Please mind that here I dont care how i will find out the perimeter. But I do know that it can be found out.
Bunuel, Can you please throw some light on this issue? I am not sure if I am making some mistake here. But i strongly believe that statement-2 alone must be suffice.
Re: Parallelogram - Area and perimeter [#permalink]
14 Jun 2010, 09:55
1
This post received KUDOS
Expert's post
amitjash wrote:
Hi every body, I think statement-2 alone will be suffice if we want to find the Perimeter. If one angle is 45, another angle has to be 45. In addition other two angles will be 270/2 each. Now, if you consider area of 100, you can find only one parallelogram matching these qualities. Please mind that here I dont care how i will find out the perimeter. But I do know that it can be found out.
Bunuel, Can you please throw some light on this issue? I am not sure if I am making some mistake here. But i strongly believe that statement-2 alone must be suffice.
You are right about the angles, but there are many parallelograms possible with such angles and area 100: \(h=10\), \(b=10\) --> \(area=100\); \(h=5\), \(b=20\) --> \(area=100\); \(h=1\), \(b=100\) --> \(area=100\); ... --> different perimeter.
If refer to my previous post then we would have that \(height=BX=AX\) (as angle BAD is 45 degrees) and \(base=AX+XD\) --> \(Area=BX*(AX+XD)=AX*(AX+XD)=100\). You can plug different values for AX and XD to get are 100, thus the perimeter would be different for each case. _________________
Re: Perimeter of parallelogram [#permalink]
18 Oct 2010, 13:34
Orange08 wrote:
If the area of a parallelogram is 100, what is the perimeter of the parallelogram?
1.The base of the parallelogram is 10. 2.One of the angles of the parallelogram is 45 degrees.
(1) Insufficient. Base=10. Area=100. So height is 10, but we don't know the angle inside the paralellogram, so we don't actually know length of the sides. (Imagine extreme case angle=90, makes it a square. Angle=10, makes its perimeter approach a very large number)
(2) Insufficent. Angle=45. Again, we dont know the base and the height, all we know their product is 100. hxb=100. Perimeter is 2(h(sqrt(2)) + b) which can take a range of different values for different choices of h.
(1+2) In this case, base=10. Height =10. Angle=45. So the other side will be 10(sqrt(2)). Hence perimeter = 20+20sqrt(2). Sufficient
Re: Perimeter of parallelogram [#permalink]
18 Oct 2010, 22:05
shrouded1 wrote:
Orange08 wrote:
If the area of a parallelogram is 100, what is the perimeter of the parallelogram?
1.The base of the parallelogram is 10. 2.One of the angles of the parallelogram is 45 degrees.
(1) Insufficient. Base=10. Area=100. So height is 10, but we don't know the angle inside the paralellogram, so we don't actually know length of the sides. (Imagine extreme case angle=90, makes it a square. Angle=10, makes its perimeter approach a very large number)
(2) Insufficent. Angle=45. Again, we dont know the base and the height, [highlight]all we know their product is 100. hxb=100[/highlight]. Perimeter is 2(h(sqrt(2)) + b) which can take a range of different values for different choices of h.
Answer is (c)
Is the area of parallelogram not equal to 1/2 * b * h ?
Re: Parallelogram - Area and perimeter [#permalink]
20 Oct 2010, 05:34
Bunuel wrote:
mirzohidjon wrote:
Hi, Bunuel, I really appreciate your effort to help us. Finally after your explanation, I got the answer (since I imagined what kind of parallelogram the problem was talking about. But, based on what assumptions, you deduce that parallelogram needs to be the way you described (diagonal equal to height)?
Attachment:
PointLatticeParallelograms_1000.gif
If the area of a parallelogram is 100, what is the perimeter of the parallelogram?
Given: \(Area=base*height=100\). Q: \(P=2b+2l=?\) (b - base, l - leg )
(1) The base of the parallelogram is 10 --> \(base=height=10\). Infinite variations are possible. Look at the diagram (let the distance between two horizontal and vertical points be 10): all 4 parallelograms have \(base=height=10\) but they have different perimeter. Not sufficient. Side notes: \(l\geq{10}\), when \(l=10=h\) we would have the square (case #3 on the diagram) and \(P=40\) (smallest possible perimeter), maximum value of perimeter is not limited.
(2) One of the angles of the parallelogram is 45 degrees. Clearly insufficient. But from this statement height BX and AX will make isosceles right triangle: \(height=BX=AX\).
(1)+(2) As from 2 we have that \(height=BX=AX\) and from (1) we have that \(base=height=10\) --> \(AX=base=AD=10\) --> X and D coincide (case #4 on the diagram) --> leg (AB) becomes hypotenuse of the isosceles right triangle with sides equal to 10 --> \(AB=10\sqrt{2}\) --> \(P=20+20\sqrt{2}\). Sufficient.
Answer: C.
Attachment:
m10-31.png
Hope it's clear.
HI Bunuel,
I got convienced bt small doubt grilling me ... if i draw a line from B to a base not diagnolly, can n't i consider as a height... ?? Is so then equation changes??
Re: Parallelogram - Area and perimeter [#permalink]
20 Oct 2010, 07:38
Expert's post
vitamingmat wrote:
Bunuel wrote:
mirzohidjon wrote:
Hi, Bunuel, I really appreciate your effort to help us. Finally after your explanation, I got the answer (since I imagined what kind of parallelogram the problem was talking about. But, based on what assumptions, you deduce that parallelogram needs to be the way you described (diagonal equal to height)?
Attachment:
PointLatticeParallelograms_1000.gif
If the area of a parallelogram is 100, what is the perimeter of the parallelogram?
Given: \(Area=base*height=100\). Q: \(P=2b+2l=?\) (b - base, l - leg )
(1) The base of the parallelogram is 10 --> \(base=height=10\). Infinite variations are possible. Look at the diagram (let the distance between two horizontal and vertical points be 10): all 4 parallelograms have \(base=height=10\) but they have different perimeter. Not sufficient. Side notes: \(l\geq{10}\), when \(l=10=h\) we would have the square (case #3 on the diagram) and \(P=40\) (smallest possible perimeter), maximum value of perimeter is not limited.
(2) One of the angles of the parallelogram is 45 degrees. Clearly insufficient. But from this statement height BX and AX will make isosceles right triangle: \(height=BX=AX\).
(1)+(2) As from 2 we have that \(height=BX=AX\) and from (1) we have that \(base=height=10\) --> \(AX=base=AD=10\) --> X and D coincide (case #4 on the diagram) --> leg (AB) becomes hypotenuse of the isosceles right triangle with sides equal to 10 --> \(AB=10\sqrt{2}\) --> \(P=20+20\sqrt{2}\). Sufficient.
Answer: C.
Attachment:
m10-31.png
Hope it's clear.
HI Bunuel,
I got convienced bt small doubt grilling me ... if i draw a line from B to a base not diagnolly, can n't i consider as a height... ?? Is so then equation changes??
A height is a perpendicular from a vertex to a side. When we consider two statements together we get that height from B coincide with diagonal BD.
Re: Parallelogram - Area and perimeter [#permalink]
02 Feb 2011, 03:03
Expert's post
tinki wrote:
regarding the problem got question:
cant parallelogram be trapezoid with equal sides ? (in that case the area of trapezoid will be totally different and the answer will be E )
What do you mean by "cant parallelogram be trapezoid with equal sides"?
Next, given parallelogram (with one of the angles 45 degrees and base=height) can not have all 4 sides equal because it'll mean that height is equal to both base and leg which is impossible as the angle is 45 degrees (try to draw it). _________________
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