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Circumradius of Triangle

CAT 2008 Slot 1 · Quantitative Ability · Medium · Geometry

This is a medium Quantitative Ability question from the CAT 2008 Slot 1 paper. It tests Geometry. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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In a triangle ABCABC, the lengths of the sides ABAB and ACAC equal 17.5 cm17.5\text{ cm} and 9 cm9\text{ cm} respectively. Let DD be a point on the line segment BCBC such that ADAD is perpendicular to BCBC. If AD=3 cmAD = 3\text{ cm}, then what is the radius (in cm) of the circle circumscribing the triangle ABCABC?

  1. A.

    17.05

  2. B.

    27.85

  3. C.

    22.45

  4. D.

    32.25

  5. E.

    26.25

Answer

E

Explanation

The circumradius RR of a triangle is given by: R=a×b×c4×Area(ABC)R = \frac{a \times b \times c}{4 \times \text{Area}(ABC)}

Here, Area(ABC)=12×BC×AD=12×a×3\text{Area}(ABC) = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times a \times 3.

Substituting this into the circumradius formula: R=a×b×c4×12×a×AD=b×c2×ADR = \frac{a \times b \times c}{4 \times \frac{1}{2} \times a \times AD} = \frac{b \times c}{2 \times AD}

Given b=AC=9 cmb = AC = 9\text{ cm}, c=AB=17.5 cmc = AB = 17.5\text{ cm}, AD=3 cmAD = 3\text{ cm}: R=9×17.52×3=157.56=26.25 cmR = \frac{9 \times 17.5}{2 \times 3} = \frac{157.5}{6} = 26.25\text{ cm}.

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