Multiple Choice Question (MCQ)
If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
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×
2
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✓
8/\sqrt{15}
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×
\dfrac{5}{2}
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×
\sqrt{6}
If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
2
8/\sqrt{15}
\dfrac{5}{2}
\sqrt{6}
Using the Law of Cosines
\cos A = \dfrac{b^2+c^2-a^2}{2bc}=\dfrac{2^2+3^2-4^2}{2\times 2\times 3 } =-\dfrac{1}{4}
According to Pythagorean Identity
\sin A = \sqrt{1-\cos^2 A} = \dfrac{\sqrt{15} }{4}
Using the Law of SInes
\dfrac{a}{\sin A}=2R
R = \dfrac{4}{2\times \dfrac{\sqrt{15} }{4} } = \dfrac{8}{\sqrt{15} }
Area formula of a triangle
A = \dfrac{1}{2}ab\sin C
According g to The Law of Sines,
\dfrac{c}{\sin C} = 2R
Then
Let s =\dfrac{a+b+c}{2} , then
s=\dfrac{2+3+4}{2} = \dfrac{9}{2}
According to Heron’s formula.
A = \sqrt{s(s-a)(s-b)(s-c)} ,
=\sqrt{\dfrac{9}{2}(\dfrac{9}{2}-2)(\dfrac{9}{2}-3)(\dfrac{9}{2}-4)}
=\dfrac{1}{4}\sqrt{9\cdotp 5\cdotp 3\cdotp 1 }
=\dfrac{3}{4}\sqrt{15}
Substituting to (1) yields
R = \dfrac{2\times 3\times 4 }{4\times \dfrac{3}{4}\sqrt{15} }
=8/\sqrt{15}