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# In the figure above (not drawn to scale), triangle ABC is inscribed in

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In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

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28 Jul 2015, 01:50
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35% (medium)

Question Stats:

78% (02:40) correct 22% (03:12) wrong based on 156 sessions

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In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle?

A. π
B. 8π
C. 16π
D. 48π
E. 64π

Kudos for a correct solution.

Attachment:

circle_inscribed_triangle.gif [ 8.86 KiB | Viewed 2703 times ]

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In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

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Updated on: 28 Jul 2015, 04:12
1
Bunuel wrote:

In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle?

A. π
B. 8π
C. 16π
D. 48π
E. 64π

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

AC = OB = Radius (r)
and AB = 2*r

ABC is right angle triangle as traingle drawn in Semicircle is always the Right angle triangle with right angle at vertex touching circumference and Hypotenuse at Diameter of circle

Also BC^2 = AB^2 - AC^2 = (2r)^2 - r^2 = 3r^2

i.e. BC = r√3

Area of ABC = (1/2)*BC*AC = (1/2)*r√3*r = 8√3
i.e. r = 4
Area of Circle = π*r^2 = π*4^2 = 16π

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Originally posted by GMATinsight on 28 Jul 2015, 04:10.
Last edited by GMATinsight on 28 Jul 2015, 04:12, edited 1 time in total.
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Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

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28 Jul 2015, 04:11
1
1
Bunuel wrote:

In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle?

A. π
B. 8π
C. 16π
D. 48π
E. 64π

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

IMO: C

Angle in a semi circle = 90. Thus Angle ACB = 90.
AC = OB = r
So angles of the triangle will be ABC = 30 , ACB = 90 , BAC = 60.
So sides of the triangle will be AB = 2r AC = r BC = r√3

Area of triangle is 8√3 = 1/2 b*h
base and heights will be the sides of the triangle since it is right angled triangle
8√3 = 1/2 * r * r√3
16 = r^2

Area of circle = πr^2
= 16π
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Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

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28 Jul 2015, 04:21
1
From the figure it can be seen that the triangle ACB is inscribed within a semicircle. So angle C is 90 degree.
Lets take OB = x , so OA =x (both radius) 2x, Also it is given that AC = OB = x.
For triangle OAC (draw a line ) all sides are equal (OA = AC= OC) so it is an equilateral triangle. So <A = 60degree

The right triangle ACB is thus a 30 60 90 triangle so sides are x, x√3and 2x.
Area of triangle ACB = 0.5 * x * x√3 = 8√3 So x = 4 i.e. radius of the circle is 4

Area of circle = π x * x = 16π

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Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

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28 Jul 2015, 04:29
Bunuel wrote:

In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle?

A. π
B. 8π
C. 16π
D. 48π
E. 64π

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

Given, AB is the diameter of the circle= 2r and AC = OB =r

Let BC = h

Thus is right triangle ABC, right angled at C,

$$r^2+h^2 = (2r)^2$$ and from area of the triangle = $$8\sqrt{3}$$ = $$0.5*h*r$$ ----> $$h = \frac{16\sqrt{3}}{r}$$

Thus, $$r^2+ (\frac{16\sqrt{3}}{r})^2 = 2r^2$$ ----> $$r^4 = 256$$ ---> $$r = 4$$

Thus the area of the circle = $$\pi*4^2$$ = $$16\pi$$. C is the correct answer.
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Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

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28 Jul 2015, 13:53
1
Here, $$AC = OB = OA = r$$, we can say that $$AB = 2r$$.

Given the inscribed triangle property, we can get that triangle ABC is a right triangle and so $$AB^2 = AC^2 + BC^2$$.

So, $$(2r)^2 = r^2 + BC^2$$, which means that $$BC = \sqrt{3}r$$.

Area of the triangle ABC = $$\frac{1}{2}(AC)(BC) = 8\sqrt{3}$$, means $$\frac{1}{2}(r)(\sqrt{3}r) = 8\sqrt{3}$$, which simplifies to $$r^2 = 16$$, so Area of the circle = 16$$\pi$$

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Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

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17 Aug 2015, 09:41
1
Bunuel wrote:

In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle?

A. π
B. 8π
C. 16π
D. 48π
E. 64π

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

800score Official Solution:

There are four main concepts that must be understood in order to solve this problem. The first concept is that any triangle inscribed within a circle such that any one side of the triangle is a diameter of the circle, is a right triangle. The second concept is that the proportions of the sides of a 30⁰-60⁰-90⁰ triangle are x, x√3, 2x. The third concept is that the area of a right triangle is equal to one half the product of its legs: (1/2)bh. Finally, the fourth concept is that the area of a circle is π × radius².

Since the triangle is inscribed in the circle and one of its sides constitutes a diameter of the circle, the triangle must be a right triangle. AB is the diameter so AB = 2 × OB. We know that OB = AC. Therefore AB = 2 × AC. Therefore it must be a 30°-60°-90° triangle where AC = x, AB = 2x and CB = √3 × x (This can also be calculated using the Pythagorean Theorem).

We are now ready to solve this problem. Using the third concept from above, the area must be equal to (1/2) × √3 × x × x = 8√3.
The solution of the equation is x = 4. So the radius of the circle is 4. Therefore the area of the circle is π × 4² = 16π. The correct answer is C.
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Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

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06 Feb 2016, 10:31
Bunuel wrote:

In the figure above (not drawn to scale), triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. Segments AC and OB are equal. If the area of triangle ABC is 8√3, then what is the area of the circle?

A. π
B. 8π
C. 16π
D. 48π
E. 64π

Kudos for a correct solution.

Attachment:
circle_inscribed_triangle.gif

OB is the radius, thus, AC has the same length as the radius.

since we have a triangle inscribed in a circle, and one side is the diagonal, the triangle MUST be a right triangle.
since the hypotenuse is 2R,and since one leg is R, it must be true that the triangle is 30-60-90. It must also be true, from the property of 30-60-90 triangle, that the other leg must be R*sqrt(3).
we then have the area of the triangle:
R*R(sqrt3)/2 = 8*sqrt(3)
r^2(sqrt3) = 16*sqrt(3)
r=4.
Area of the circle:
pi*r^2 = 16pi.
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Re: In the figure above (not drawn to scale), triangle ABC is inscribed in  [#permalink]

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21 Dec 2018, 17:08
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Re: In the figure above (not drawn to scale), triangle ABC is inscribed in   [#permalink] 21 Dec 2018, 17:08
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