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Geometry Question

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20 Jan 2009, 17:46
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

It is a problem solving question.

Saw in an old ARCO book, but don't have the solution.

Would appreciate a detailed solution.

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Geometry Question.xls [14.5 KiB]

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21 Jan 2009, 03:30
sid3699 wrote:
It is a problem solving question.

Saw in an old ARCO book, but don't have the solution.

Would appreciate a detailed solution.

Can you post as JPG/gif FILE.
many people don't have EXCEL .
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21 Jan 2009, 04:13
Looking at the data where AB = CD = 2 and distance between the two is 2. We can imagine a square of 2x2 with in circle. which gives us the radius of the circle as half of the diagonal of square. Substrating Area of SQUARE - Area of Circle will give 4 times the area of the circle outside square. As they are equal you can divide it by half and add to the area of Square to get the area encompassing ABCD.
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21 Jan 2009, 04:39
Jpeg file attached.

What is the area encompassing ABCD?
AB is parallel to CD
AB=CD=2
Distance between AB and CD = 2
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Geometry Question.jpg [ 7.99 KiB | Viewed 1495 times ]

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21 Jan 2009, 09:57
Given AB || CD and AB = CD = 2.
Let O be the center of the cirle and EF be perpendicular from AB to CD thru O. EF = 2.

With this we have, AE = 1(Since OE bisects AB), OE = 1. <AEO = 90 ==> OA = sqrt(2). By Pythogorous theorem.

Hence Radius of the circle = sqrt(2)

Now Area of the region formed by ABCD = Area of the sectors OCA + Area of the sector OBD + Area of the triangle OAB + Area of the triangle OCD.

Area of the sector OCA = 90 / 360 * Pi * 2 = pi / 2. -----> i
Area of the sector OBD = 90 / 360 * Pi * 2 = pi / 2. -----> ii
Area of the triangle OAB = 1/2 * 1 * 2 = 1 -----> iii
Area of the triangle OCD = 1/2 * 1 * 2 = 1 -----> iv

ie. Total area of the region formed by ABCD = Pi + 2 (i + ii + iii + iv)

My apologies for not having a diagram. Not sure how to soft copy diagram.
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21 Jan 2009, 10:33
mrsmarthi wrote:
Given AB || CD and AB = CD = 2.
Let O be the center of the cirle and EF be perpendicular from AB to CD thru O. EF = 2.

With this we have, AE = 1(Since OE bisects AB), OE = 1. <AEO = 90 ==> OA = sqrt(2). By Pythogorous theorem.

Hence Radius of the circle = sqrt(2)

Now Area of the region formed by ABCD = Area of the sectors OCA + Area of the sector OBD + Area of the triangle OAB + Area of the triangle OCD.

Area of the sector OCA = 90 / 360 * Pi * 2 = pi / 2. -----> i
Area of the sector OBD = 90 / 360 * Pi * 2 = pi / 2. -----> ii
Area of the triangle OAB = 1/2 * 1 * 2 = 1 -----> iii
Area of the triangle OCD = 1/2 * 1 * 2 = 1 -----> iv

ie. Total area of the region formed by ABCD = Pi + 2 (i + ii + iii + iv)

My apologies for not having a diagram. Not sure how to soft copy diagram.

same approach as yours.

but got AREA= 2+pi sqrt(2)

= Area of triangle I + Area of triangel II + Area of Sector III + ARea of sector IV
= 1/2 * 2*1 +1/2 * 2*1 + 2*(90/360) 2*pi*sqrt(2) (radius)
=2+pi*sqrt(2)
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21 Jan 2009, 12:38
Thanks.

I got to the part of calculating the radius, but then, the area of a sector did'nt strike me.
Went in wired directions -- triangle in a semi-circle is a right angle triangle and god knows what.

Pretty simple solution in the end.
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21 Jan 2009, 12:45
X2Suresh,

I think you calculated the Area of the sector incorrectly.

Formula for Area of the sector is x / 360 * Area of the circle.

Rather you calculated it as x / 360 * Circumferrence of the circle which gives the length of the arc.

Hope you got the difference.
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21 Jan 2009, 13:05
mrsmarthi wrote:
X2Suresh,

I think you calculated the Area of the sector incorrectly.

Formula for Area of the sector is x / 360 * Area of the circle.

Rather you calculated it as x / 360 * Circumferrence of the circle which gives the length of the arc.

Hope you got the difference.

OOPS.. thank!! you are right.
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27 Jan 2009, 07:17
my answer is --> pi + 2

the area of the square is 4

the area of the circle is 2*pi

the area of part of the circle not inclosed by the square is 2*pi-4.

the rest of the sector ABCD, beside the square is (2*pi-4) (symmetry)

thus, the area of ABCD is 4 + (2*pi-4)/2= pi + 2
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27 Jan 2009, 08:04
sid3699 wrote:
It is a problem solving question.

Saw in an old ARCO book, but don't have the solution.

Would appreciate a detailed solution.

I get 2+pi. ---> Area of square: 4.

Diameter = 2sqrt2 --> Area of circle = 2pi.

We have 4 mini semicircles (not quite semicircles, but u get the idea). So 2pi=4+4x (x= 1 semicircle).

2pi-4=4x --> (pi-2)/2=x ---> 2x+4 = Area of sector: --> (2pi-4)/2+4 --> pi+2
Re: Geometry Question   [#permalink] 27 Jan 2009, 08:04
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