JEE Challenger
More from Circles
MathMedAdv 2007 • Paper 2

Radius of Incircle in a Right Trapezoid

Let ABCDABCD be a quadrilateral with area 18, with side ABAB parallel to the side CDCD and AB=2CDAB = 2 CD. Let ADAD be perpendicular to ABAB and CDCD. If a circle is drawn inside the quadrilateral ABCDABCD touching all the sides, then its radius is

Options

A

3

B

2

Correct
C

32\frac{3}{2}

D

1

Topics & Concepts

CirclesTangents to CirclesIncircleQuadrilateral

Step-by-Step Solution

The quadrilateral ABCDABCD is a right trapezoid:

  • Parallel sides: CDCD and ABAB with AB=2CDAB = 2CD. Let CD=xCD = x, then AB=2xAB = 2x.
  • Side ADAD is perpendicular to both ABAB and CDCD. Since the circle touches both parallel sides, the diameter of the incircle is AD=2rAD = 2r.

  1. Area of the trapezoid:
    Area=12(AB+CD)⋅AD=12(2x+x)(2r)=3rx\text{Area} = \frac{1}{2}(AB + CD) \cdot AD = \frac{1}{2}(2x + x)(2r) = 3rx
    Given Area=18\text{Area} = 18:
    3rx=18  ⟹  rx=63rx = 18 \implies rx = 6

  2. Tangential Quadrilateral Property (Pitot's Theorem):
    For a quadrilateral with an incircle touching all four sides, the sums of opposite sides are equal:
    AB+CD=AD+BC  ⟹  2x+x=2r+BC  ⟹  3x=2r+BCAB + CD = AD + BC \implies 2x + x = 2r + BC \implies 3x = 2r + BC

By the Pythagorean theorem on the non-parallel side BCBC:
BC=AD2+(AB−CD)2=(2r)2+(2x−x)2=4r2+x2BC = \sqrt{AD^2 + (AB - CD)^2} = \sqrt{(2r)^2 + (2x - x)^2} = \sqrt{4r^2 + x^2}

Equating:
3x−2r=4r2+x23x - 2r = \sqrt{4r^2 + x^2}
Squaring both sides:
(3x−2r)2=4r2+x2  ⟹  9x2−12rx+4r2=4r2+x2  ⟹  8x2=12rx  ⟹  2x=3r(3x - 2r)^2 = 4r^2 + x^2 \implies 9x^2 - 12rx + 4r^2 = 4r^2 + x^2 \implies 8x^2 = 12rx \implies 2x = 3r

  1. Solving for rr:
    From rx=6rx = 6 and x=32rx = \frac{3}{2}r:
    r(32r)=6  ⟹  32r2=6  ⟹  r2=4  ⟹  r=2r \left(\frac{3}{2}r\right) = 6 \implies \frac{3}{2}r^2 = 6 \implies r^2 = 4 \implies r = 2

Correct Answer: B

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