Question

In a parallelogram ABCD, M is the midpoint of side CD. What is the ratio of Area(ABCD):Area(ΔABM)?

• A. 2

Answer 1

The correct option is A 2
Area of a triangle = $\frac{1}{2}\times$ base $\times$ height
Area of a parallelogram = base $\times$ height

Now $\Delta$ABM and parallelogram ABCD are between the same set of parallel lines, so their heights are equal.

They also have the same base AB.

So, Area($\Delta$ABM)= $\frac{1}{2}\times$ AB $\times$ height
=$\frac{1}{2}$ Area(ABCD)

$⟹$ Area(ABCD) = 2 $\times$ Area ($\Delta$ABM)

$\therefore$ Area(ABCD) : Area($\Delta$ABM) =
2 $\times$ Area ($\Delta$ABM) : Area ($\Delta$ABM)
= 2:1 = 2