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.
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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First, the area of the circle has ti be greater than the area of the inscribed triangle. So that kills A and B.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.
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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- Jeff@TargetTestPrep
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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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