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# If a triangle inscribed in a circle has area 40, what is the

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If a triangle inscribed in a circle has area 40, what is the [#permalink]

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22 Mar 2011, 09:46
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If a triangle inscribed in a circle has area 40, what is the area of the circle?

(1) The base of the triangle is equal to the diameter of the circle.
(2) The measure of one of the angles in the triangle is 30.
[Reveal] Spoiler: OA

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Re: If a triangle inscribed in a circle has area 40, what is the [#permalink]

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22 Mar 2011, 10:08
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rxs0005 wrote:
If a triangle inscribed in a circle has area 40, what is the area of the circle?

(1) The base of the triangle is equal to the diameter of the circle.
(2) The measure of one of the angles in the triangle is 30.

(1)
This statement tells us that the inscribed triangle is a right angled triangle.
Right angled triangle have three sides;
Two smaller sides and one hypotenuse.

Area = 1/2*(smaller side)*(bigger side)=40 if we assume that it is not an isosceles right triangle.

We know the radius = 1/2*(Diameter) = 1/2*(Hypotenuse) [Hypotenuse is referred as "BASE" in this statement]
Not Sufficient.

(2)
The measure of one of the angles in the triangle is 30.
We can make plenty such triangles with one angle as 30.
Not sufficient.

Combining both;

We know that the triangle is a right angled triangle with 30-90-60 as its angles.

The opposite sides will be in ratio: $$x:2x:\sqrt{3}x$$

$$Area = \frac{1}{2}*x*\sqrt{3}x=40$$
$$x^2=\frac{80}{\sqrt{3}}$$
$$x=\sqrt{\frac{80}{\sqrt{3}}}$$

$$Hypotenuse = Diameter = 2x= 2*\sqrt{\frac{80}{\sqrt{3}}}$$
$$Area = \pi*(\frac{Diameter}{2})^2=\pi*(\sqrt{\frac{80}{\sqrt{3}}})^2=\pi*\frac{80}{\sqrt{3}}$$

Sufficient.

Ans: "C"
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Re: If a triangle inscribed in a circle has area 40, what is the [#permalink]

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22 Mar 2011, 19:05
(1) is not sufficient, as the height can be altered, and the area = 1/2 * b * h = 40, where the base can also vary and hence the area of circle will also vary, as base of triangle = diameter.

(2) If measure of one angle is 30, still we can't decide on the other sides (or angles), so not sufficient. FOr example, the base may or may not be the diameter.

(1) and (2) together, If the base is the Diameter, and one angle is 30 degrees, the triangle is a 30.60-90 right triangle with sides in the ratio

1:sqrt(3):2. Hypotenuse is the Diameter and it can be found as area of triangle = 40.

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Re: If a triangle inscribed in a circle has area 40, what is the [#permalink]

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22 Mar 2011, 21:50
From one we know its a right angle triangle . To calculate area of circle we need to know the radius of the circle(i.e hypotenuse of a triangle (b*h=80) ).

From 2 : its 30:60 :90 triangle with ration 1: 3^1/2 :2
we know b*h=80 or x*3^1/2 x=80.. we get x here

hypotenuse =diameter of circle=2x ...once we know diameter we can calculate area of circle.

Thus c
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Re: If a triangle inscribed in a circle has area 40, what is the [#permalink]

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23 Mar 2011, 01:38
It’s C:
The answer is area of a circle, A=πr^2 =π( √(80/(√3)) )^2 here the radius is √(80/(√3))
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Re: If a triangle inscribed in a circle has area 40, what is the [#permalink]

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28 Sep 2015, 08:35
Hello from the GMAT Club BumpBot!

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Re: If a triangle inscribed in a circle has area 40, what is the [#permalink]

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28 Sep 2015, 08:44
i have a doubt in this question.

if we consider diameter as the base of the triangle, then opposite angle will be 90. in this case, two sides making the 90 degree angle will be base and height or vice versa. and in this case, diameter is the longest side. this diameter can not be the base as mentioned in the statement 1. hence E should be the option.

please explain where i am making mistake. i find statement 1 erroneous.
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Re: If a triangle inscribed in a circle has area 40, what is the [#permalink]

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10 Jan 2017, 06:31
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: If a triangle inscribed in a circle has area 40, what is the   [#permalink] 10 Jan 2017, 06:31
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