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If P is the center of the circle shown above, and BAC=30º, and the are

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If P is the center of the circle shown above, and BAC=30º, and the are  [#permalink]

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09 May 2015, 01:52
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If P is the center of the circle shown above, and BAC=30º, and the area of triangle ABC is 6, what is the area of the circle?

A. (√3)π
B. (2√3)π
C. 4π
D. 6π
E. (4√3)π

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Re: If P is the center of the circle shown above, and BAC=30º, and the are  [#permalink]

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09 May 2015, 02:55
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triangle in a circle with 1 side as diameter is a right angle triangle.
So, this is a 30-60-90 triangle with sides in the ratio 1:sqrt3:2
area of ABC= 1/2*BC*AB
= 1/2*r*sqrt3*r
and this is given as 6
So, r^2=12/sqrt3
area of circle = pi*r^2
= pi*12/sqrt3
= 4*sqrt3*pi
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Re: If P is the center of the circle shown above, and BAC=30º, and the are  [#permalink]

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09 May 2015, 04:43
Why is BA the height?
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Re: If P is the center of the circle shown above, and BAC=30º, and the are  [#permalink]

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09 May 2015, 04:47
bankerboy30 wrote:
Why is BA the height?

As angle B is 90, the side opposite to it is the largest side i.e. the hypotenuse.
So, other 2 sides are base and height.
Hope it helps..
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Re: If P is the center of the circle shown above, and BAC=30º, and the are  [#permalink]

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30 Apr 2016, 13:58
Naina1 wrote:
bankerboy30 wrote:
Why is BA the height?

As angle B is 90, the side opposite to it is the largest side i.e. the hypotenuse.
So, other 2 sides are base and height.
Hope it helps..

Agreed, that BA is the height but shouldnt the area be 1/2*AB (height) *BC??

Which should make it 1/2*2r*r =6

r^2=6
Does this not make the area to be 6PI ????

Please tell me where I'm going wrong
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Posts: 7100
Re: If P is the center of the circle shown above, and BAC=30º, and the are  [#permalink]

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30 Apr 2016, 20:25
gmater12 wrote:
Naina1 wrote:
bankerboy30 wrote:
Why is BA the height?

As angle B is 90, the side opposite to it is the largest side i.e. the hypotenuse.
So, other 2 sides are base and height.
Hope it helps..

Agreed, that BA is the height but shouldnt the area be 1/2*AB (height) *BC??

Which should make it 1/2*2r*r =6

r^2=6
Does this not make the area to be 6PI ????

Please tell me where I'm going wrong

Hi,
You are correct upto a point..
But what is Height AB, It is NOT 2r.. Because 2r means DIAMETER and that is AC..

Now how do you find AB and BC when what you know is ONLY AC as 2r..
Triangle ABC is 30-60-90 which gives sides opposite to these angles in ratio$$1:\sqrt{3}:2$$..
Now Opposite 90 angle is HYP or AC or 2r.. so other sides $$= 1*r:\sqrt{3}*r:2*r$$..
so $$AB = \sqrt{3}r$$ and BC = r..
$$area = 1/2 *AB*BC = 1/2 *r*\sqrt{3}=6$$..
$$r^2 = 12/\sqrt{3} = 4\sqrt{3}$$..
so Area of circle = $$4\sqrt{3}*pi$$
E
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If P is the center of the circle shown above, and BAC=30º, and the are  [#permalink]

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22 Feb 2018, 08:24
1
reto wrote:

If P is the center of the circle shown above, and BAC=30º, and the area of triangle ABC is 6, what is the area of the circle?

A. (√3)π
B. (2√3)π
C. 4π
D. 6π
E. (4√3)π

Recall that any triangle inscribed in a circle that has the diameter of the circle as its hypotenuse is a right triangle. Thus, triangle ABC is a 30-60-90 right triangle, with the ratio of its sides as x : 2x: x√3.. Noting that the figure is not drawn to scale, we let side AB = x and side BC = x√3, and thus:

x * x√3 * 1/2 = 6

x^2(√3) = 12

x^2 = 12/√3

x^2 = 12√3/3

x^2 = 4√3

Notice that AC, the diameter, is 2x. Thus the radius, AP or CP, is x, and hence the area of the circle, is πx^2. Since x^2 = 4√3, then the area of the circle is (4√3)π.

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If P is the center of the circle shown above, and BAC=30º, and the are &nbs [#permalink] 22 Feb 2018, 08:24
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