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What is the area of the circle?

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What is the area of the circle?

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

$$A.\ \frac{\sqrt{3}}{\pi}$$
$$B.\ \frac{2\sqrt{3}}{\pi}$$
$$C.\ 4\pi$$
$$D.\ 6\pi$$
$$E.\ \left(4\sqrt{3}\right)\pi$$

The OA is E.

I'm confused by this PS question. Experts, any suggestion about how to solve it? Thanks in advance.

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To solve, we need to know that a triangle inscribed in a circle where one side is the diameter will always be a right triangle. This means that ABC is a 30-60-90 triangle, which means its sides must have a ratio of x : x√3 : 2x, where x is BC, x√3 is AB, and 2x is AC. Since AC is the diameter of the circle, the radius of the circle will equal x. We can then use A=πr^2 to solve for the area of the circle.

We know that the area of the triangle is 6 and can be found with A=(1/2)bh, where b and h are BC and AB. Plugging in our area and the values for the sides of the triangle gives $$6=\frac{1}{2}\left(BC\right)\left(AB\right)$$ $$6=\frac{1}{2}\left(x\right)\left(x\sqrt{3}\right)$$ $$6=\frac{\sqrt{3}}{2}x^2$$ $$\frac{12}{\sqrt{3}}=x^2$$ $$4\sqrt{3}=x^2$$

Rather than solving for x, we can now plug 4√3 directly into our circle area equation, since x^2=r^2:
$$A=\pi r^2$$ $$A=\pi\left(4\sqrt{3}\right)$$
which is answer choice E.

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LUANDATO wrote:


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

$$A.\ \frac{\sqrt{3}}{\pi}$$
$$B.\ \frac{2\sqrt{3}}{\pi}$$
$$C.\ 4\pi$$
$$D.\ 6\pi$$
$$E.\ \left(4\sqrt{3}\right)\pi$$

The OA is E.

I'm confused by this PS question. Experts, any suggestion about how to solve it? Thanks in advance.
First, the area of the circle has ti be greater than the area of the inscribed triangle. So that kills A and B.

Next, if you see that triangle ABC is a right triangle - an inscribed angle that cuts off the diameter is always a right angle - you know we're dealing with a 30:60:90 triangle, meaning the sides have a ratio of x : x√3: 2x. Moreover, if 2x is the diameter the radius would be x.

If the sides of the triangle are x and x√3, we know the area would be (x^2 * √3)/2 = 6
x^2 * √3 = 12
And x^2 = 12/√3.

In other words, (radius)^2 has a √3 in it, which means that the area of the circle must have a √3 in it as well. Well, the only answer choice left with a √3 is E.

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LUANDATO wrote:


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

$$A.\ \frac{\sqrt{3}}{\pi}$$
$$B.\ \frac{2\sqrt{3}}{\pi}$$
$$C.\ 4\pi$$
$$D.\ 6\pi$$
$$E.\ \left(4\sqrt{3}\right)\pi$$
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.

Answer: E

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