Question

What is the perimeter of a parallelogram, if its area is 24 cm2 and and the distances between the point of intersection of the diagonals and the sides are 2 cm and 3 cm?

Answer 1

Answer:

p=20cm

Step-by-step explanation:

Area of parallelogram is :$24cm^{2}$

Since, it is a parallelogram, the four triangles are created by drawing diagonals of equal area=$\frac{24}{4}=6cm^{2}$.

Now, Height of the one triangle is 2cm and the other is 3 cm.

Therefore, From ΔODC,

Area=$\frac{1}{2}{\times}b_{1}{\times}h_{1}$

$6$=$\frac{1}{2}{\times}b_{1}{\times}2$

$b_{1}=6$

$DC=6cm$

Again, from ΔOAD,

Area=$\frac{1}{2}{\times}b_{2}{\times}h_{2}$

$6=\frac{1}{2}{\times}b_{2}{\times}3$

$b_{2}=4cm$

$AD=4cm$

Now, opposite sides of parallelogram are equal, therefore,AD=BC=4cm and DC=AB=6cm.

Perimeter of parallelogram= Sum of all the sides

=AB+BC+CD+AD

=6+4+6+4

=20cm