Area of Right Triangle in Parallelogram

CAT 2019 Slot 1 · QA · Easy · Mensuration

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In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is:

  1. A.

    18318\sqrt{3}

  2. B.

    24324\sqrt{3}

  3. C.

    32332\sqrt{3}

  4. D.

    12312\sqrt{3}

Answer

C

Explanation

Area of parallelogram = base×height=CD×AP=72\text{base} \times \text{height} = CD \times AP = 72. 9×AP=72⇒AP=89 \times AP = 72 \Rightarrow AP = 8. In right triangle APD, hypotenuse AD=16AD = 16 and leg AP=8AP = 8. PD=AD2−AP2=162−82=256−64=192=83PD = \sqrt{AD^2 - AP^2} = \sqrt{16^2 - 8^2} = \sqrt{256 - 64} = \sqrt{192} = 8\sqrt{3}. Area of △APD=12×AP×PD=12×8×83=323\triangle APD = \frac{1}{2} \times AP \times PD = \frac{1}{2} \times 8 \times 8\sqrt{3} = 32\sqrt{3}.

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