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Re: Parallelogram - Area and perimeter [#permalink]
02 Feb 2011, 05:25

Expert's post

tinki wrote:

Sorry , my question was not correclty formulated.

i somewhere read that trapezoid is a special type of parallelogram with one pare of parallel sides. Could that be right?

Common definition of a trapezoid: a quadrilateral which has at least one pair of parallel sides.

Common definition of a parallelogram: a quadrilateral with two pairs of parallel sides.

According to these definitions all parallelograms are trapezoids but not all trapezoids are parallelograms.

Other definition of a trapezoid is: a quadrilateral which has only one pair of parallel sides. So according to this definition a parallelogram can not be a trapezoid at all.

But the above has nothing to do with the original question because we are told that the figure is a prallelogram: "If the area of a parallelogram is...".

Re: Parallelogram - Area and perimeter [#permalink]
02 Feb 2011, 05:47

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 )

Trapezoid is not a parallelogram.

Parallelogram has four sides where opposite sides are always equal and parallel, whereas in a trapezoid, only one pair opposite sides is parallel.

Parallelograms: Rectangle, Square, Rhombus.

Area of parallelogram = base * height = 100

1. base = 10;

base * height = 10 * height = 100 height = 10 But, we can't find the other side of the parallelogram in order to find the perimeter. Not sufficient.

2. One of the angles is 45. Say parallelogram PQRS, where PR is parallel to QS. \angle{Q} is 45{\circ} If a line is drawn perpendicular from point P to QS and it meets at point X. We know that the altitude, distance between PR and QS, is same as the distance between Q and X. We don't know the side. Not sufficient.

Using both, as earlier explained, X is nothing but R. So, the side becomes the hypotenuse of the right triangle with base 10 and altitude 10, 10\sqrt{2}

Re: If the area of a parallelogram is 100, what is the perimeter [#permalink]
15 Jul 2012, 23:54

Shouldn't the answer be (B)

Let's say base = Height = x (because given 45 degree condition, we end up with an isosceles right triangle as shown above) then question stem says x*x = 100 which means that x= 10 and the other side = x*root(2) (as per isosceles right triangle prop.)

Then we have P = 2( 10 + 10* root(2))

Why need A when we can get the same info from only statement 2?

Re: If the area of a parallelogram is 100, what is the perimeter [#permalink]
16 Jul 2012, 04:10

Expert's post

teal wrote:

Shouldn't the answer be (B)

Let's say base = Height = x (because given 45 degree condition, we end up with an isosceles right triangle as shown above) then question stem says x*x = 100 which means that x= 10 and the other side = x*root(2) (as per isosceles right triangle prop.)

Then we have P = 2( 10 + 10* root(2))

Why need A when we can get the same info from only statement 2?

Re: Parallelogram - Area and perimeter [#permalink]
28 Feb 2013, 07:28

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.

Re: Parallelogram - Area and perimeter [#permalink]
28 Feb 2013, 08:18

Expert's post

Sachin9 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.

Re: Parallelogram - Area and perimeter [#permalink]
27 Dec 2013, 13:07

robinmrtha 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.

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]
15 May 2014, 03:46

Bunuel wrote:

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}.

P=20+20\sqrt{2}.

Hope it's clear.

I am unable to understand when the angle is 45 degree How X will coincide with D. Because angle is 45 degree not 90 degree. Can you please draw the Figure also. Thanks!

Re: Parallelogram - Area and perimeter [#permalink]
15 May 2014, 05:42

Expert's post

honchos wrote:

Bunuel wrote:

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}.

P=20+20\sqrt{2}.

Hope it's clear.

I am unable to understand when the angle is 45 degree How X will coincide with D. Because angle is 45 degree not 90 degree. Can you please draw the Figure also. Thanks!