AE || BD. AD is also the diameter. If the length of arc BCD is 4pi the : Problem Solving (PS)

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14 Sep 2009, 06:50

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yangsta8

- Draw a line from B to O (circle centre)
- BO = OD (both radii) and now ODB is iscoceles triangle
- EAD = ADB = DBO = 30 degrees
- angle of arc at centre = 120 degrees (180-30-30)

So 120/360 (2* Pi * r) = 4 * Pi
solving you get r = 6
Area = 36 Pi

thats one way of solving it ....u can also solve this by using the "angle at center theorm".....ie

the angle subtended by the arc at the center is twice the angle subtended by it at any other point on the circle.

arc BCD makes angle 60 deg at a : angle BAD = 60. therefore the angle made by the same arc at center :ang BOD = 120 deg.

so using ur formula....4 pi = 2pi *r* *[120/306]
r = 6
area = pi*r*r
= 36pi.

:-D

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14 Sep 2009, 09:42

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Thanks guys - both the methods are very quick ways of solving this problem.

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27 Mar 2015, 02:57

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yangsta8

- Draw a line from B to O (circle centre)
- BO = OD (both radii) and now ODB is iscoceles triangle
- EAD = ADB = DBO = 30 degrees
- angle of arc at centre = 120 degrees (180-30-30)

So 120/360 (2* Pi * r) = 4 * Pi
solving you get r = 6
Area = 36 Pi

How is Angle EAD = Angle ADB = 30. Which property is being used here?

IF anyone could help here.

Bunuel

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27 Mar 2015, 04:46

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earnit

yangsta8

- Draw a line from B to O (circle centre)
- BO = OD (both radii) and now ODB is iscoceles triangle
- EAD = ADB = DBO = 30 degrees
- angle of arc at centre = 120 degrees (180-30-30)

So 120/360 (2* Pi * r) = 4 * Pi
solving you get r = 6
Area = 36 Pi

How is Angle EAD = Angle ADB = 30. Which property is being used here?

IF anyone could help here.

When two parallel lines are cut by a third line, they form a system of angles.

Attachment:

parallell_lines2a.gif [ 1.46 KiB | Viewed 7687 times ]

In this figure, we can see that 1 and 4 are equal, as are 5 and 8. But because they are formed by two parallel lines, they are all equal to each other. The same could be said for angles 2, 3, 6, and 7.

There are many terms from geometry class you may know to describe these angles, such as “alternate interior” or “alternate exterior,” but these terms are not used on the test. For the GMAT, it is simply enough to know that all the little angles will always be equal, and all the big angles will always be equal. Additionally, you should realize that any little angle added to any big angle will always equal 180°.

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27 Mar 2015, 04:53

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Bunuel

earnit

yangsta8

- Draw a line from B to O (circle centre)
- BO = OD (both radii) and now ODB is iscoceles triangle
- EAD = ADB = DBO = 30 degrees
- angle of arc at centre = 120 degrees (180-30-30)

So 120/360 (2* Pi * r) = 4 * Pi
solving you get r = 6
Area = 36 Pi

How is Angle EAD = Angle ADB = 30. Which property is being used here?

IF anyone could help here.

When two parallel lines are cut by a third line, they form a system of angles.

Attachment:

parallell_lines2a.gif

In this figure, we can see that 1 and 4 are equal, as are 5 and 8. But because they are formed by two parallel lines, they are all equal to each other. The same could be said for angles 2, 3, 6, and 7.

There are many terms from geometry class you may know to describe these angles, such as “alternate interior” or “alternate exterior,” but these terms are not used on the test. For the GMAT, it is simply enough to know that all the little angles will always be equal, and all the big angles will always be equal. Additionally, you should realize that any little angle added to any big angle will always equal 180°.

Totally Got it. However, just in reference to the above context, how do you define 'little' angle and 'big' angle among 1,2,3,4,5....

Bunuel

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27 Mar 2015, 04:59

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earnit

Totally Got it. However, just in reference to the above context, how do you define 'little' angle and 'big' angle among 1,2,3,4,5....

One set of angles are smaller than other, so...

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15 May 2016, 06:31

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Please Bunuel
kindly illustrate your approach to solve such problem. I think it looks complex but it is simple if one recognize the concept behind it.

Bunuel

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15 May 2016, 06:40

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hatemnag

Please Bunuel
kindly illustrate your approach to solve such problem. I think it looks complex but it is simple if one recognize the concept behind it.

Here is a solution: ae-bd-ad-is-also-the-diameter-if-the-length-of-arc-bcd-is-4pi-the-83893.html#p628625 Please specify which part needs to be elaborated. Thank you.

hatemnag

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15 May 2016, 08:36

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Draw a line from B to O (circle centre)
- BO = OD (both radii) and now ODB is iscoceles triangle
- EAD = ADB = DBO = 30 degrees
- angle of arc at centre = 120 degrees (180-30-30)

So 120/360 (2* Pi * r) = 4 * Pi
solving you get r = 6
Area = 36 Pi
I am just confused that OBD is iscoceles triangle and not right triangle with 30 - 30 - 120 angles not 45 - 45 - 90 angles.

hatemnag

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15 May 2016, 08:39

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Hi Bunuel
thank you very much for responding.
I am just confused that OBD is iscoceles triangle and not right triangle with 30 - 30 - 120 angles not 45 - 45 - 90 angles.

Bunuel

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15 May 2016, 10:13

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hatemnag

Hi Bunuel
thank you very much for responding.
I am just confused that OBD is iscoceles triangle and not right triangle with 30 - 30 - 120 angles not 45 - 45 - 90 angles.

OD and OB are radii of the circle, thus OBD is an isosceles triangle.

The right triangle would be ABD, with right angle at B, because AD is a diameter (A right triangle's hypotenuse is a diameter of its circumcircle (circumscribed circle). The reverse is also true: if one of the sides of an inscribed triangle is a diameter of the circle, then the triangle is a right angled (right angel being the angle opposite the diameter/hypotenuse)).

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cr117

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20 Sep 2016, 07:20

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I had a different approach to this and got the right answer. Not sure if it's just a coincidence though.

Drawing a line AB makes a 30-60-90 triangle, with the 60 degree angle creating the arc length 4*pi. I used the ratio of \frac{60}{30}=\frac{4*pi}{x}, where x is the length of the arc created by the 30 degree angle. This gives x=2*pi, so the circumference is 2(4*pi)+2(2*pi) = 12*pi = pi*diameter. So the diameter is 12 and from there we get the area 36*pi.

Can someone tell me if this is also a correct approach?

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