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07 Nov 2010, 00:18
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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A circle is inside an equalateral triangle ABC. One side of the triangle is equal to 10, What is the area of the circle?
07 Nov 2010, 00:31
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Lets say the triangle is ABC. There for AB = BC = AC = 10
The circle has the center O and touches BC at point D.
Since ABC is equilateral,
(1) BD = DE = 5,
(2) CO bisects the angle BCA, and
(3) OD is perpendicular to DC.
From the right triangle DCO, we get sin (angle DCO) = sin 30 = 1/2. Also, sine (angle DCO) = OD/OC.
This means OD is 5, which is the radius of the circle. Therefore the area of the circle is 25π.
Is that the right answer?
The circle has the center O and touches BC at point D.
Since ABC is equilateral,
(1) BD = DE = 5,
(2) CO bisects the angle BCA, and
(3) OD is perpendicular to DC.
From the right triangle DCO, we get sin (angle DCO) = sin 30 = 1/2. Also, sine (angle DCO) = OD/OC.
This means OD is 5, which is the radius of the circle. Therefore the area of the circle is 25π.
Is that the right answer?
07 Nov 2010, 04:09
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Couple of formulae that might be handy:
(1) Formula 1: The radius of Inscribed circle in a traigle, which has sides length as a, b and c, is:
Radius = \(\frac{Area of traingle}{K}\) = \(\frac{\sqrt{K(K-a)(K-b)(K-c)}}{K}\)
(2) Formula 2: The radius of Circumcribed circle for a traingle, which has sides lengths as a, b and c, is:
Radius = \(\frac{abc}{4*Area of traingle}\) = \(\frac{abc}{4*\sqrt{K(K-a)(K-b)(K-c)}}\)
Where, K = \(\frac{a+b+c}{2}\)
So, the radius of a inscribed circle where a=b=c=10 is:
K = \(\frac{10+10+10}{2}\) = 15
Radius = \(\frac{\sqrt{15(15-10)(15-10)(15-10)}}{15}\) = \(\frac{5}{\sqrt{3}}\)
Now, the area of a traingle, which has radius \(\frac{5}{\sqrt{3}}\) is: \(pi\) * \(\frac{5}{\sqrt{3}}\) * \(\frac{5}{\sqrt{3}}\) = \(pi\) * \(\frac{25}{3}\)
(1) Formula 1: The radius of Inscribed circle in a traigle, which has sides length as a, b and c, is:
Radius = \(\frac{Area of traingle}{K}\) = \(\frac{\sqrt{K(K-a)(K-b)(K-c)}}{K}\)
(2) Formula 2: The radius of Circumcribed circle for a traingle, which has sides lengths as a, b and c, is:
Radius = \(\frac{abc}{4*Area of traingle}\) = \(\frac{abc}{4*\sqrt{K(K-a)(K-b)(K-c)}}\)
Where, K = \(\frac{a+b+c}{2}\)
So, the radius of a inscribed circle where a=b=c=10 is:
K = \(\frac{10+10+10}{2}\) = 15
Radius = \(\frac{\sqrt{15(15-10)(15-10)(15-10)}}{15}\) = \(\frac{5}{\sqrt{3}}\)
Now, the area of a traingle, which has radius \(\frac{5}{\sqrt{3}}\) is: \(pi\) * \(\frac{5}{\sqrt{3}}\) * \(\frac{5}{\sqrt{3}}\) = \(pi\) * \(\frac{25}{3}\)
