Question

Construct a parallelogram $ABCD$, when diagonal $AC=6.4cm$, diagonal $BD=8.2cm$ and angle between the diagonals $={60}^{\circ }$. Then the perimeter of the parallelogram (in $cm$) is:

(Round off your answer to the nearest integer)

Answer 1

We know that the diagonals of a parallelogram bisect each other.
Steps of Construction:

(i) Draw $AC=6.4cm.$
(ii) Bisect $AC$ at $O$.
(iii) Make $\angle COX={60}^o$ and produce $XO$ to $Y$.
(iv) Set off $OB=\frac{1}{2}\left(8.2\right)=4.1cm$ and $OD=\frac{1}{2}\left(8.2\right)=4.1cm$ as shown.
(v) Join $AB,BC,CD$ and $DA$.

Then, ABCD is the required parallelogram.

Measure the sides of the parallelogram. The sides are of length $3.7cm$ and $6.3cm$. Therefore, perimeter $=2(3.7+6.3)=20cm$