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?

  1. ×

    2

  2. 8/\sqrt{15}

  3. ×

    \dfrac{5}{2}

  4. ×

    \sqrt{6}

Collected in the board: Trigonometry

Steven Zheng posted 3 weeks ago


Answer 1

  1. 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} }


Steven Zheng posted 3 weeks ago


Answer 2

  1. 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

    R = \dfrac{abc}{4A}
    (1)

    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}

Steven Zheng posted 3 weeks ago

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