CAT 2018

Slot 1

QA

Question & Solution

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

Given that area of parallelogram = 72 sq cm

Area of triangle ABC = $\dfrac{1}{2}$*area of parallelogram

(1/2)*AB*BC*sinABC = $\dfrac{1}{2}$*72

sinABC = $\dfrac{1}{2}$

$\angle$ ABC = 30°

Let us draw a perpendicular CQ from C to AB. 

$\angle\ B\ =\ \angle\ D\ =\ 30^{\circ\ }$ ( opposite angles in a parallelogram)

$\angle\ QCB\ =\ \angle\ DAP\ =\ 60^{\circ\ }$  ( Sum of angles in a triangle = 180)

In triangles APD and CQB, AP= CQ , AD=CB ( opposite sides of a parallelogram) , $\angle\ QCB\ =\ \angle\ DAP\ =\ 60^{\circ\ }$

Hence APD and CQB are congruent triangles using SAS property.

Therefore, we can say that area of triangle APD = area of triangle CQB

In right angle triangle CQB,

QB = CBcos30° = $16*\dfrac{\sqrt{3}}{2}$ = $8\sqrt{3}$ cm

CQ = CBsin30° = $16*\dfrac{1}{2}$ = $8$ cm

Therefore, area of triangle CQB = 1/2*CQ*QB = $1/2*8*8\sqrt{3}$ = $32\sqrt{3}$

Hence, we can say that area of triangle APD = $32\sqrt{3}$.