Question

Question 78

In the given figure, area of $\Delta AFB$ is equal to the area of parallelogram ABCD. If altitude EF is 16 cm long, find the altitude of the parallelogram to the base AB of length 10 cm. What is the area of $\Delta DAO$, where O is the mid-point of DC?

Answer 1

Given,
Area of $\Delta AFB$ = Area of parallelogram ABCD
$\Rightarrow \frac{1}{2}\times AB\times EF=CD\times$ (Corresponding height)
$[\because$ area of trianlge = base $\times$ height and area of parallelogram = base $\times$ corresponding height]
$\Rightarrow \frac{1}{2}\times AB\times EF=CD\times EG$

Let the corresponding height be h.
Then, $\frac{1}{2}\times 10\times 16=10\times h$
$[\because$ Altitude, EF = 16 cm and base, AB = 10 cm, given] $[\because AB=CD]$
$\Rightarrow h=8cm$
In $\Delta DAO$, DO = 5 cm [$\therefore$ O is the mid-point of CD]
$\therefore$ Area of $\Delta DAO=\frac{1}{2}\times OD\times h$
$=\frac{1}{2}\times 5\times 8=20c{m}^2$