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If the area of a parallelogram is 100, what is the perimeter

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29 Jun 2009, 19:20

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If the area of a parallelogram is 100 square units, what is its perimeter?

(1) The parallelogram has a base measuring 10 units.

(2) One of the parallelogram's angles measures 45 degrees.

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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?

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Official Answer

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robinmrtha

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30 Jun 2009, 00:27

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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

Bunuel

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29 May 2010, 04:30

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mirzohidjon

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)?

Official Solution:

If the area of a parallelogram is 100 square units, what is its perimeter?

Refer to the diagram below:

Given: \(Area = base*height = 100\). We want to find the perimeter, \(P = 2b + 2l\), where \(b\) is the base and \(l\) is the leg (side length).

(1) The parallelogram has a base measuring 10 units.

So, \(base = height = 10\). Infinitely many cases are possible. Observe the diagram (let the distance between two horizontal and vertical points be 10 units): all 4 parallelograms have \(base = height = 10\) but they have different perimeters. Therefore, this information is not sufficient.

(2) One of the angles of the parallelogram is 45 degrees.

Clearly insufficient by itself. However, from this statement, we know that height \(BX\) and \(AX\) form an isosceles right triangle: \(height = BX = AX\).

(1) + (2) Combining both statements, from (2) we have that \(height = BX = AX\), and from (1) we have that \(base = height = 10\), so \(AX = base = AD = 10\). This means that points \(X\) and \(D\) coincide (case #4 in the diagram). Thus, leg \(AB\) becomes the hypotenuse of the isosceles right triangle with sides equal to 10 units, hence \(AB = 10\sqrt{2}\). Therefore, the perimeter \(P = 20 + 20\sqrt{2}\). Both statements together are sufficient.

Answer: C

Attachment:

m10-31.png [ 1.83 KiB | Viewed 38340 times ]

Hope it's clear.

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PointLatticeParallelograms_1000.gif [ 3.32 KiB | Viewed 38336 times ]

General Discussion

gmatjon

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28 May 2010, 14:15

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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

Bunuel

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28 May 2010, 16:57

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mirzohidjon

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.

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29 May 2010, 09:12

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Thank you, it was great help!

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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.

Bunuel

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14 Jun 2010, 10:55

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amitjash

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.

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18 Oct 2010, 14:34

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Orange08

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

Answer is (c)

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20 Oct 2010, 06:34

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Bunuel

mirzohidjon

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??

Bunuel

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20 Oct 2010, 08:38

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vitamingmat

Bunuel

mirzohidjon

Attachment:

PointLatticeParallelograms_1000.gif

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.

Hope it's clear.

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02 Feb 2011, 03:37

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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 )

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02 Feb 2011, 04:03

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tinki

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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02 Feb 2011, 04:24

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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?

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02 Feb 2011, 06:25

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tinki

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...".

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02 Feb 2011, 06:47

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tinki

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, \(10\sqrt{2}\)

Perimeter: \(10+10\sqrt{2}\)

Answer: "C"

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16 Jul 2012, 08:40

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Bunuel! thanks for such good explanations, really appreciate that!! +1 to you!

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Bunuel

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Attachment:

The attachment PointLatticeParallelograms_1000.gif is no longer available

Attachment:

The attachment m10-31.png is no longer available

Hope it's clear.

Bunuel/Anyone
Why not A alone is sufficient
being known 2 sides are 10 it means angles are also same .also as height BD is making 90 deg. means the other 2 angles are 45 deg each
fig. attached)
Thus making A as ans choice

Thanks

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m10-31.png [ 5.08 KiB | Viewed 11016 times ]

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Bunuel

mirzohidjon

Attachment:

The attachment PointLatticeParallelograms_1000.gif is no longer available

Attachment:

The attachment m10-31.png is no longer available

Hope it's clear.

Hi Rohit,
Refer your PM..
Area of llgm is dependent on base and height..
Base is one of the side but height will be other side only if angle is 90°...
With height 10, various combinations can be made...

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Bunuel

mirzohidjon

Attachment:

The attachment PointLatticeParallelograms_1000.gif is no longer available

Attachment:

The attachment m10-31.png is no longer available

Hope it's clear.

Responding to a pm:

Height is the length of the altitude from a vertex to the opposite side. It will be 10 in this case. But why is it necessary that it will fall on D? Look at these two figures. The perimeter in each case will be different.

Attachment: