TheCATExam
Sign in

CAT 2018 Slot 1 — QA Question 18

Basics of QuadrilateralsEasy

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

Answer & solution

Correct answer: 32√3

  • A

    12√3

  • B

    24√3

  • C

    18√3

  • 32√3

Solution

Easy

The parallelogram's area with base CDCD gives the height APAP. Then △APD\triangle APD is right-angled at PP with hypotenuse ADAD, so Pythagoras gives PDPD, and the triangle's area follows.

A B C D P AP
1

Find the height APAP. Area == base ×\times height, with base CD=9CD=9:

72=9×AP⇒ AP=8\begin{aligned} &72 = 9\times AP\\ &\Rightarrow\ AP = 8 \end{aligned}
2

Find PDPD. In right triangle APDAPD, the hypotenuse is AD=16AD=16 and one leg is AP=8AP=8:

PD=162−82=256−64=192=83\begin{aligned} &PD = \sqrt{16^2 - 8^2} = \sqrt{256-64} = \sqrt{192} = 8\sqrt3 \end{aligned}
3

Area of △APD\triangle APD. Legs AP=8AP=8 and PD=83PD=8\sqrt3:

Area=12×PD×AP=12×83×8=323\begin{aligned} &\text{Area} = \tfrac12\times PD \times AP = \tfrac12\times 8\sqrt3 \times 8 = 32\sqrt3 \end{aligned}
Area(△APD)=323 sq cm\text{Area}(\triangle APD) = \mathbf{32\sqrt3}\ \text{sq cm}

Related Basics of Quadrilaterals questions

See all Quadrilaterals & Polygons questions →
CAT 2018 Slot 1 QA Q18: In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P — Solution | TheCATExam