ABCD is a parallelogram of area $162{\text{c}}{{\text{m}}^2}$. P is a point on AB such that AP: PB = 1: 2. Calculate the area of ∆ APD.
$
  {\text{A}}{\text{. 20c}}{{\text{m}}^2} \\
  {\text{B}}{\text{. 27c}}{{\text{m}}^2} \\
  {\text{C}}{\text{. 24c}}{{\text{m}}^2} \\
  {\text{D}}{\text{. 25c}}{{\text{m}}^2} \\
$

Hint: To find the area of the triangle, we first begin by computing the area of the parallelogram. Then we write the formula of area of the triangle and deduce its length of base and height by comparing it with the dimensions of parallelogram. The height of the triangle is the same as the parallelogram, its base is divided by the base of the parallelogram in a ratio, and we use this to find the length of its base.

Complete step-by-step answer:

Given Data – AP: PB = 1: 2
⟹PB = 2AP
Also,
AP + PB =AB -- (Since, given P is a point on AB such that AP: PB = 1: 2)
⟹AP + 2AP = AB
⟹AP = $\dfrac{{{\text{AB}}}}{3}$
Now let the altitude of the parallelogram be CL.
Now,
Given the area of parallelogram (ABCD) = 162${\text{c}}{{\text{m}}^2}$
We know area of a parallelogram = base × height
From the figure,
⟹AB × CL = 162${\text{c}}{{\text{m}}^2}$

And we know the area of a triangle = $\dfrac{1}{2}$ × base × height.
So, from the figure
Area of ΔAPD = $\dfrac{1}{2}$× (AP) × (CL) -- (height is same throughout the entire figure)
= $\dfrac{1}{2}$× ($\dfrac{{{\text{AB}}}}{3}$) × (CL) --- (P divides AB in the ratio 1: 2 hence AP =$\dfrac{{{\text{AB}}}}{3}$)
= $\dfrac{{{\text{AB }} \times {\text{ CL}}}}{6}$
= $\dfrac{{162}}{6}$= 27${\text{c}}{{\text{m}}^2}$
Hence the area of triangle APD is 27${\text{c}}{{\text{m}}^2}$. Option B is the correct answer

Note:The key in solving such problems is to get the quantities we require using the ones we already know. In this case writing the area of the triangle in terms quantities in area of parallelogram, the ratio given is the relation between them. Altitude of parallelogram is the perpendicular distance from the base to its opposite side.