Question 1143114: Three sides of a triangle measure 12 in, 15 in and 18 in. Find the area of the triangle, the radius of inscribed circle and radius of circumscribed circle. Thanks.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Use Heron's formula for the area of a triangle given the length of all three sides
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s = (a + b + c)/2 = (12 + 15 + 18)/2 = 45/2 = 22.5
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Area(A) = square root(22.5 * (22.5-12) * (22.5-15) * (22.5-18)) = 89.29 square inches
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A = r * s, where r is the radius of the inscribed circle
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r = A/s = 89.29/22.5 = 3.97 inches
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let a = 12, b = 15, c = 18, use law of cosines to find angle A
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12^2 = 15^2 +18^2 - 2 * 15 * 18 * cos(A)
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cos(A) = (15^2 +18^2 -12^2)/(2 * 15 * 18) = 0.75
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Angle A = cos^-1 (0.75) = 41.41 degrees
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Use the law of sines, where R is the radius of the circumscribed circle
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R = a/(2 * sin A) = 12/(2 * sin (41.41) = 9.07 inches
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Area of triangle is 89.29 square inches
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radius of inscribed circle is 3.97 inches
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radius of circumscribed circle is 9.07 inches
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