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

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Manager
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In the figure, ABC is an equilateral triangle, and DAB is a [#permalink]  01 Dec 2010, 05:29
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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: in a triangle abc is equilateral [#permalink]  01 Dec 2010, 05:51
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inscribedtwice1_153.gif [ 2.27 KiB | Viewed 2353 times ]
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.

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Re: in a triangle abc is equilateral [#permalink]  01 Dec 2010, 23:13
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Bunuel. Kudos to you for writing down those principles.

However, i think, we can answer the qtn even with out knowing that the angle C = angle D

As the DB has to be diameter (cz angle A = 90), the cente of the circle has to lie on the line DB. As ABC is an equilateral triangle and inscribed in the circle, the center of ABC has to be the the centre of the circle. Hence the center of ABC is on the diameter aswell. Keeping, in mind, the above points and the verthex A is common for both the triangles , the line BD has to be a bisector for the angle B in the equilateral triangle bcz the angluar bisector for any angle in an equilateral triangle has to pass thru the center of the triangle. thus, angle B is devided into two halfs 30 and 30. hence angle D has to be 60 as 90+30+D=180. From now on, we can you the pointers specified by you(Bunuel) for the stmnt1 & 2 to prove that the ANSWER is "A".

Regards,
Murali.

Kudos?

Regards,
Murali.
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Re: in a triangle abc is equilateral [#permalink]  01 Dec 2010, 23:26
Applied the same logic, but forgot about that 1, radical 3, 2 rule for 30, 60, 90 triangles.

Nicely done.

Bunuel wrote:
Attachment:
inscribedtwice1_153.gif
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.

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Re: In the figure, ABC is an equilateral triangle, and DAB is a [#permalink]  02 Feb 2014, 02:58
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Re: In the figure, ABC is an equilateral triangle, and DAB is a [#permalink]  31 Mar 2015, 04:14
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the figure, ABC is an equilateral triangle, and DAB is a   [#permalink] 31 Mar 2015, 04:14
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