Question

$ABCD$ is a parallelogram. $AB=8cm,AD=4cm,\angle B={120}^o$
a. What is $\angle A$
b. What is the perpendicular distance from $D$ to $AB$?
c. What is the area of $ABCD$?

Answer 1

Given $ABCD$ is parallelogram, $AB=8cm,AD=4cm,\angle B={120}^o$
a. $\angle B=\angle D={120}^o$
(opposite angles equal)
$\angle A+\angle C={360}^o-({120}^o+{120}^o)$
$=360-240={120}^o$
$\therefore \angle A=\angle A=\angle C=\frac{120}{2}=60$
$\therefore A={60}^o$
b. Draw DE as perpendicular from $D$ to $AB$. From the figure.
$DE=2\sqrt{3}cm.[1:\sqrt{3}:2]$
c. The are of $ABCD=bh$ (formula)
$=8\times 2\sqrt{3}=\sqrt{3}c{m}^2$