ABCD is a parallelogram of area 162 sq. cm. P is a point on AB such that AP : PB = 1 : 2. Calculate :

(i) the area of △APD

(ii) the ratio PA : DC.

(i) Given: ABCD is a parallelogram and P is a point on AB such that AP : PB = 1 : 2.

DB is the diagonal and in a parallelogram, a diagonal divides it into two congruent triangles.

$\dfrac{1}{2}$ Ar.(//gm ABCD) = Ar.(△ ADB) = Ar.(△ BDC)

⇒ Ar.(△ ADB) = $\dfrac{1}{2}$ x 162

= 81 sq. cm

P is divides the line AB in the ratio 1 : 2. So, Ar.(△ APD) :Ar.(△ ADB) = 1:3.

Ar.(△ APD) = $\dfrac{1}{3}$ x Ar.(△ ADB)

= $\dfrac{1}{3}$ x 81 sq. cm

= 27 sq. cm

Hence, the area of △APD = 27 sq. cm.

(ii) Let AP = x and PB = 2x

AB = AP + PB (From the figure)

= x + 2x

= 3x

∴ $\dfrac{AP}{AB} = \dfrac{1}{3}$

As we know that AB = CD,

∴ $\dfrac{AP}{CD} = \dfrac{1}{3}$

Hence, the ratio PA : DC = 1 : 3.