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Geometry of Right Angled Triangle Area

CMAT 2023 · Quantitative Techniques & Data Interpretation · Hard · Geometry

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In the following diagram, if the shaded area is one half the area of triangle ABC and angle ABC is right angle then the length of line segment AD is

        C
       /|
      / | z
     /  |
 W  /   D
   /   /|
  /   / | y
 /   /  |
A-------B
    x

(Note: AC=WAC = W, AB=xAB = x, BD=yBD = y, DC=zDC = z)

  1. A.

    12W\frac{1}{2} W

  2. B.

    2x2+z2\sqrt{2x^2 + z^2}

  3. C.

    W2−3y2\sqrt{W^2 - 3y^2}

  4. D.

    y2−z2\sqrt{y^2 - z^2}

Answer

C

Explanation

Area of △ABC=12x(y+z)\triangle ABC = \frac{1}{2} x (y + z). Area of △ABD=12xy\triangle ABD = \frac{1}{2} xy.

Given shaded area (△ACD\triangle ACD) is half the area of △ABC\triangle ABC, so Area(△ABC\triangle ABC) =2×= 2 \times Area(△ABD\triangle ABD): 12x(y+z)=2(12xy)⇒y+z=2y⇒z=y\frac{1}{2} x (y + z) = 2 \left(\frac{1}{2} xy\right) \Rightarrow y + z = 2y \Rightarrow z = y

In right △ABC\triangle ABC: W2=x2+(y+z)2=x2+(2y)2=x2+4y2W^2 = x^2 + (y + z)^2 = x^2 + (2y)^2 = x^2 + 4y^2

In right △ABD\triangle ABD: AD2=x2+y2=(W2−4y2)+y2=W2−3y2AD^2 = x^2 + y^2 = (W^2 - 4y^2) + y^2 = W^2 - 3y^2 AD=W2−3y2AD = \sqrt{W^2 - 3y^2}

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