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Manager  Joined: 07 Feb 2010
Posts: 127
In the figure, ABC is an equilateral triangle, and DAB is a right  [#permalink]

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17 00:00

Difficulty:   95% (hard)

Question Stats: 48% (01:45) correct 52% (01:46) wrong based on 230 sessions

### HideShow timer Statistics 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

Attachment: Untitled.pdf [10.85 KiB]

Math Expert V
Joined: 02 Sep 2009
Posts: 53657
Re: In the figure, ABC is an equilateral triangle, and DAB is a right  [#permalink]

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Attachment: inscribedtwice1_153.gif [ 2.27 KiB | Viewed 4665 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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Manager  Joined: 30 Aug 2010
Posts: 86
Location: Bangalore, India
Re: In the figure, ABC is an equilateral triangle, and DAB is a right  [#permalink]

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1
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.
Manager  Joined: 17 Sep 2010
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GPA: 3.59
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Re: In the figure, ABC is an equilateral triangle, and DAB is a right  [#permalink]

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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.

Manager  B
Joined: 05 Dec 2015
Posts: 109
Re: In the figure, ABC is an equilateral triangle, and DAB is a right  [#permalink]

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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.

Why can't DAB be a 45-45-90 triangle?
Math Expert V
Joined: 02 Aug 2009
Posts: 7415
Re: In the figure, ABC is an equilateral triangle, and DAB is a right  [#permalink]

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mdacosta wrote:
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.

Why can't DAB be a 45-45-90 triangle?

Hi,
there are two triangles ABD and ABC having the same base AB, which is a CHORD...
RULE - ANY point on same side of chord of a circle will make SAME angle to this chord. Also if the points make angle to different BUT equal chords of the same circle, they will be equal...

Now we know angle ACB is 60 as ACB is equilateral triangle...
so angle ACB = angle ADB = 60.... Point C and D are on the SAME side of the chord and making an angle with the same chord, so angles will be same..
so DAB is 60-30-90 triangle
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Re: In the figure, ABC is an equilateral triangle, and DAB is a right  [#permalink]

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from given info we have-
angle DAB=90, so DB is the diameter of the circle.
angle ADB= angle ACB=60 (equilateral triangle)
from triangle ADB we can easily calulate all the sides. so Sufficient.
2)angle ABD= 30, we have all the angles of triangle but no info about any side. so not sufficient.

ans A.
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Re: In the figure, ABC is an equilateral triangle, and DAB is a right  [#permalink]

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

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