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Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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31 May 2008, 09:42

goalsnr wrote:

To find the area of teh ||gram we need to know the lengths od teh sides of the ||gram

Let a and b be the sides of the ||gram

1. a+b = 10 insuff

2. we know r from stat2 insuff

using info from 1 and 2 , we can use pythogrean theorem to find a and b

a^2 + b^2 = r^2

C

There is one little flaw in your assumption that a and b are adjacent sides, what if a, and b are the parallel sides. Moreover you can never be sure that a, b, and r are part of right angle triangle.

Just surfed through the forum in the search of geometry problems and found this one without solution. So, maybe somebody try? I guess it is possible to solve under 2 minutes
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Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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22 Jun 2008, 00:58

walker wrote:

Just surfed through the forum in the search of geometry problems and found this one without solution. So, maybe somebody try? I guess it is possible to solve under 2 minutes

Well, I will try to explain my theory :

my solution gave me area = 25*sqroot(3) so answer C .

First of all, this parallelogram is inscribed in a circle it will always be rectangle or square ,since it will share the same center as circle .

property of parellelogram - two opposite sides are parallel - for two opposite sides to be parallel in circle , they have to be equidistant from center and so both the diagonal will be same . If it were given that a trapezoid is inscribed inscribed in a circle our answer would be E .

1. (a+b)^2 =100 = a^2+b^2+2a^b =100 , but according to condition 2: a^2 + b^2 =100 , this leads us to 2ab=0 which is not possible , hence the first condition is talking about two opposite sides with the same length , so 2a=10 a=5

Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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23 Jun 2008, 05:00

ritula wrote:

What is the area of the parallelogram inscribed in a circle? (1) The sum of the two sides of the parallelogram is 10 cm. (2) Area of the circle is 75 cm2.

1. statement (1) alone is sufficient but statement (2) alone is not 2. statement (2) alone is sufficient but statement (1) alone is not 3. both (1) and (2) together are sufficient but none of them alone is sufficient 4. both independently are sufficient 5. both statements (1) and (2) together are not sufficient Answer : ( 3 )

individually in suff combining parallel sides are equidistant from the center so let the distance of one side from center be x and other be y

Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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23 Jun 2008, 06:45

What is the area of the parallelogram inscribed in a circle? (1) The sum of the two sides of the parallelogram is 10 cm. (2) Area of the circle is 75 cm2.

1. statement (1) alone is sufficient but statement (2) alone is not 2. statement (2) alone is sufficient but statement (1) alone is not 3. both (1) and (2) together are sufficient but none of them alone is sufficient 4. both independently are sufficient 5. both statements (1) and (2) together are not sufficient

IMO --- E..... first lets sort out the info ... parallelogram inscribed in a circle means that figure has to be rectangle...

1) The sum of the two sides of the parallelogram is 10 cm.

let the rectangle be ABCD... with AB> BC....

from the above statement its not correct to assume that two sides means AB and BC... 2 sides can also mean AD and BC.... so 1 is insufficient...

2) Area of the circle is 75 cm2. so Pi*r*r = 75 so r *r = 75 / 3.14 ====> r= some value...

AC or BD, which are the diagonals will be the diameter of the circle... so BD = AC = 2r now , we know the diagonal's length .... but we still dontknow what the lengths of the sides are... even if we combine both we wont be able to get to one solution... so E........................

Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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23 Jun 2008, 07:30

1

This post received KUDOS

Either way you can answer the question with statement (1) and (2) : answer is definitely (C).

If you assume that the question talk about two different sides of the parallelogram (which would just be logical since the question states "the two sides of the parallelogram"), then (1) and (2) are sufficient (see answers above).

If you assume that the question can talk about two opposite sides of the parallelogram (which is in fact a rectangle as shown above as well), it is even more simple. It means that theses sides are 5 cm long (since they are equal and their sum is 10 cm). Try to inscribe a rectangle with one side of 5cm in a circle of area 75cm^2 and you'll see there is just one possible answer

Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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24 Jun 2008, 10:36

rpmodi wrote:

ritula wrote:

OA is E

I still believe it is C !! chances OA is wrong ?I looked at my solution and did not find anything wrong. Can someone look at my solution and point out a flaw, if answer has to be E ?

The problem here is Stat1 is ambigious. if a and b are adjacent sides of the parallelogram then we can use deduce C as the answer

But this info is not explicitly stated..so we cannot sum the sides are adjacent. Hence E.

Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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24 Jun 2008, 12:08

goalsnr wrote:

rpmodi wrote:

ritula wrote:

OA is E

I still believe it is C !! chances OA is wrong ?I looked at my solution and did not find anything wrong. Can someone look at my solution and point out a flaw, if answer has to be E ?

The problem here is Stat1 is ambigious. if a and b are adjacent sides of the parallelogram then we can use deduce C as the answer

But this info is not explicitly stated..so we cannot sum the sides are adjacent. Hence E.

Yeah But I have a reasoning in my explanation , why this summation has to be of two opposite sides and not two adjacent sides

Yeah But I have a reasoning in my explanation , why this summation has to be of two opposite sides and not two adjacent sides

I thought you had a clever approach, but there is one small mistake. When you calculated the radius, you deduced that it was approximately 5, and that the diameter was approximately 10. You then used the equation a^2 + b^2 = 100 and the equation a^2 + 2ab + b^2 = 100 to deduce that 2ab = 0. The problem, of course, is that the diameter is not exactly equal to 10, so a^2 + b^2 is not exactly equal to 100. The diameter is equal to 10*sqroot(3/Pi) < 10, and a^2 + b^2 is equal to something a bit smaller than 100, so 2ab is not equal to zero.
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Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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24 Jun 2008, 13:22

IanStewart wrote:

rpmodi wrote:

Yeah But I have a reasoning in my explanation , why this summation has to be of two opposite sides and not two adjacent sides

I thought you had a clever approach, but there is one small mistake. When you calculated the radius, you deduced that it was approximately 5, and that the diameter was approximately 10. You then used the equation a^2 + b^2 = 100 and the equation a^2 + 2ab + b^2 = 100 to deduce that 2ab = 0. The problem, of course, is that the diameter is not exactly equal to 10, so a^2 + b^2 is not exactly equal to 100. The diameter is equal to 10*sqroot(3/Pi) < 10, and a^2 + b^2 is equal to something a bit smaller than 100, so 2ab is not equal to zero.

Ahh ! great catch .....looks like ur first post on this forum well here you go, your first Kudos +1 .....

Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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25 Jun 2008, 19:26

isnt area of a parellelogrm = d1*d2 where d1,d2 are the two diagonals. Using B, we can find the diameter. which is nothing but the diagonal and since both the diagonals are same , we can find the area using B.

isnt area of a parellelogrm = d1*d2 where d1,d2 are the two diagonals.

No, that is not true in general. I'm sure you're thinking of the rhombus, which is a special type of parallelogram- one with all sides equal; the area of a rhombus is one half of the product of the diagonals (and not the product). But you can see, by looking at a rectangle with length 4 and width 3 (and thus diagonals of 5), that if a parallelogram is not a rhombus, the area will not in general be equal to half the product of the diagonals.
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Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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25 Jun 2008, 21:50

zeenie wrote:

isnt area of a parellelogrm = d1*d2 where d1,d2 are the two diagonals. Using B, we can find the diameter. which is nothing but the diagonal and since both the diagonals are same , we can find the area using B.

Please correct me if I am wrong

Quadrilateral diagonal lengths are equal ie d1=d2?

Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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24 Jun 2015, 07:26

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Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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23 Jul 2015, 00:09

The question can be solved by logic

as it is a //gm the figure can be 1)square 2)rectangle or 3)rhombus (1) says : a+b=10...(and b can have infinite possibilities).insuff (2) from area we can get the Radius....insuff

(1)+(2) the diagonal of the //gm is the diameter of the circle therefore the diagram is a rectangle or a square in short we have now, a^2+b^2= (2r)^2. Here as a and b are real this equation should have infinite possibilities but the restriction a+b=10 (from1) forces us to get only a specific combination of a and b. from the values of a and b we can easily find the area of //gm

also if (1) represents opposite sides there is only one possibility try it by drawing a figure

Re: What is the area of the parallelogram inscribed in a circle? [#permalink]

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03 Oct 2016, 06:15

Only a square and a rectangle can be inscribed in a circle and their diagonals must be equal to the diameter of the circle.

From (1) we have many side combinations that add-up to 10. For example: 5 and 5 (area 25) and 3 and 7 (area 21). Insufficient. From (2) we can calculate the radius (and thus the diagonals). But we don't have info that could help us reveal if it's a square or a rectangle. If it was a square we could get the area. If it isn't, we couldn't.

Together we can use pythagoras and get the area. diagonal= 5, side=x, side=10-x

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