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In the figure, ABC is an equilateral triangle, and DAB is

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In the figure, ABC is an equilateral triangle, and DAB is [#permalink] New post 03 Mar 2012, 12:06
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In the figure, ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?

(1) DA = 4
(2) Angle ABD = 30 degrees
[Reveal] Spoiler: OA

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Re: Inscribed Triangle [#permalink] New post 03 Mar 2012, 12:57
Yes , Angle ACB = ADB as they inscribe the same arc in the circle.
Check this out....
Hope that helps....

http://www.youtube.com/watch?v=QjXsZ5WmMZ8

The rest of the trick you know...
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Re: Inscribed Triangle [#permalink] New post 03 Mar 2012, 13:31
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In the figure, ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?

You should know the following properties to solve this question:
• All inscribed angles that subtend the same arc are equal. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle. Hence, all inscribed angles that subtend the same arc are equal.
• A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.
• In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio 1 : \sqrt{3}: 2. Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).
For more check Circles Triangles and chapters of Math Book: math-circles-87957.html and math-triangles-87197.html

So, from above we'll have that as DAB=90 degrees then DB must be a diameter of the circle. Next, as angles ACB and ADB subtend the same arc AB then they must be equal and since ACB=60 (remeber ACB is an equilateral triangle) then ADB=60 too. Thus DAB is 30-60-90 triangle and its sides are in ratio 1 : \sqrt{3}: 2.

(1) DA = 4 --> the side opposite 30 degrees is 4, then hypotenuse DB=diameter=4*2=8 --> radius=4 --> area=\pi{r^2}=16\pi. Sufficient.

(2) Angle ABD = 30 degrees --> we knew this from the stem, so nothing new. Not sufficient.

Answer: A.

Hope it's clear.
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Re: Geometry: Circle in [#permalink] New post 03 Apr 2012, 15:46
Bunuel wrote:
Merging similar topics. Please ask if anything remains unclear.


Thanks for merging!
I don't get why the Central Angle Theorem applies. In the figure, there is no central angle for that to occur with. So why would ADB be the same as ACB?
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Re: Geometry: Circle in [#permalink] New post 04 Apr 2012, 00:15
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KGG88 wrote:
Bunuel wrote:
Merging similar topics. Please ask if anything remains unclear.


Thanks for merging!
I don't get why the Central Angle Theorem applies. In the figure, there is no central angle for that to occur with. So why would ADB be the same as ACB?


All inscribed angles that subtend the same arc are equal. Since angles ACB and ADB subtend the same arc AB then they must be equal. Next, there is no central angle shown on the diagram, I just mentioned Central Angle Theorem to explain why is above property true: angles ACB and ADB have the same central angle AOB (O is the center of the circle) and since they both equal to half of it then they must be equal.

For more check Circles and Triangles chapters of Math Book: math-circles-87957.html and math-triangles-87197.html
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In the figure above, if ABC is an equilateral triangle, and [#permalink] New post 03 Apr 2013, 12:42
In the figure above, if ABC is an equilateral triangle, and DAB is a right triangle, what is the area of the circumscribed circle?

(1) Angle ABD = 30
(2) DA = 4

I came across a DS question whose explanation provides reasoning I do not follow. What am i missing?

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Re: Why is this angle 60degrees? [#permalink] New post 03 Apr 2013, 12:43
So I chose both statements together are sufficient but here is the explanation provided:
Since both angles ACB and ADB are opposite side AB, and angle ACB is part of an equilateral triangle, we can say both ACB and ADB are 60 degrees.

Why does that hold true? I get that :
1. DAB is 90
2. CAB = ACB = BAC = 60

and based on that all i can say is ABD < 60
so if we arent told ABD's value, cant ABD = 70 and ADB = 20?

In any case, if we can deduce ABD from just statement 2, then the rest is easy. if ABD = 30 degrees (as told by statement 1), then DB = diameter. we can figure out the length of DB using the 30-60-90 relationship and solve for area.

thanks.
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Re: Why is this angle 60degrees? [#permalink] New post 03 Apr 2013, 12:59
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wown wrote:
So I chose both statements together are sufficient but here is the explanation provided:
Since both angles ACB and ADB are opposite side AB, and angle ACB is part of an equilateral triangle, we can say both ACB and ADB are 60 degrees.

Why does that hold true? I get that :
1. DAB is 90
2. CAB = ACB = BAC = 60

and based on that all i can say is ABD < 60
so if we arent told ABD's value, cant ABD = 70 and ADB = 20?

In any case, if we can deduce ABD from just statement 2, then the rest is easy. if ABD = 30 degrees (as told by statement 1), then DB = diameter. we can figure out the length of DB using the 30-60-90 relationship and solve for area.

thanks.


There is a thing you are not considering: in any right triangle inscribed in a circle, the hypotenuse coincides with a diameter of the circle.
So we know that DB is the diameter, that the bisectors of the ACB triangle intersect at the center (ancient math theory if I remember...).
ABC=60° is divided in two 30°angles, DAB=90° and so ADB=60°
And now as you said "we can figure out the length of DB using the 30-60-90 relationship and solve for area."
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Re: Why is this angle 60degrees? [#permalink] New post 03 Apr 2013, 13:28
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Wown,

This is very good trap C DS questions and if running out of time on exam you want to guess anything but C.

That said, the problem is using the relationship between an inscribed angle and central angle. central angle is the angle formed by an arc at the center and inscribed angle is the angle formed by the same arc at the circle.

central angle = 2 * inscribed angle - Here <ACB and <ADB are inscribed angles formed by arc AB.

From the stem, it is given that <ACB = 60 so the central angle for arch AB is 120. Make O as the center of the circle, so <AOB = 120

Now <ADB is the inscribed angle for <AOB so it will be 60 degrees. Triangle ADB is a 30-60-90 type right angle triangle and stmt 2 gives DA = 8 using which you can calculate the diameter of the circle as 16 and circumference as 16pi. Hence stmt B is sufficient.

Thanks for sharing this problem...

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Re: Why is this angle 60degrees? [#permalink] New post 03 Apr 2013, 13:37
Oh, did not consider the hypotenuse-circle relationship. thanks for the explains.
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Re: Why is this angle 60degrees? [#permalink] New post 03 Apr 2013, 13:43
nt2010 wrote:
Wown,

This is very good trap C DS questions and if running out of time on exam you want to guess anything but C.

That said, the problem is using the relationship between an inscribed angle and central angle. central angle is the angle formed by an arc at the center and inscribed angle is the angle formed by the same arc at the circle.

central angle = 2 * inscribed angle - Here <ACB and <ADB are inscribed angles formed by arc AB.

From the stem, it is given that <ACB = 60 so the central angle for arch AB is 120. Make O as the center of the circle, so <AOB = 120

Now <ADB is the inscribed angle for <AOB so it will be 60 degrees. Triangle ADB is a 30-60-90 type right angle triangle and stmt 2 gives DA = 8 using which you can calculate the diameter of the circle as 16 and circumference as 16pi. Hence stmt B is sufficient.

Thanks for sharing this problem...

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Also did not consider this central angle-inscribed angle relationship... very interesting
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Re: Why is this angle 60degrees? [#permalink] New post 03 Apr 2013, 16:05
Did the original question show a 90 degree mark on the illustration?

I was attempting to answer the question, but without the mark I didn't want to assume it was 90 degrees
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Re: Why is this angle 60degrees? [#permalink] New post 03 Apr 2013, 20:40
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wown wrote:
I came across a DS question whose explanation provides reasoning I do not follow. What am i missing?

Image


We know that angle subtended by the same arc, on the circumference are always equal.

From F.S 1, the data given is of no value. It is so because Angle ACB = Angle ADB = 60 degrees. Thus, angle ABD has to be (90-60) = 30 degrees.Insufficient.

From F.S 2, we know the length of AD. Thus, in the case of a 60-30-90 triangle, we can easily calculate the hypotenuse = the diameter =BD and hence the area.Sufficient.

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Re: In the figure above, if ABC is an equilateral triangle, and [#permalink] New post 04 Apr 2013, 02:23
Expert's post
wown wrote:
In the figure above, if ABC is an equilateral triangle, and DAB is a right triangle, what is the area of the circumscribed circle?

(1) Angle ABD = 30
(2) DA = 4

I came across a DS question whose explanation provides reasoning I do not follow. What am i missing?

Image


Merging similar topics. Please refer to the solutions above.

Also, please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rules #1, 3, and 6.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: In the figure above, if ABC is an equilateral triangle, and   [#permalink] 04 Apr 2013, 02:23
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