Question

The ratio of the height and base of a parallelogram having area p square units is 3:5,then the perpendicular distance between parallel sides of parallelogram can be?​

Answer 1

I feel this is the answer !!

Answer 2

Given,

Area of the parallelogram = p units²

The ratio of the height to the base of the parallelogram = 3:5

To find,

The value of the height (perpendicular distance between parallel sides of a parallelogram).

Solution,

The height of the parallelogram will be

$\sqrt{ \frac{3p}{5} }$ units.

According to the question,

Area of the parallelogram = p units²

The ratio of the height to the base of the parallelogram = 3:5

Let's take the height to be h units and the base to be b units.

h/b = 3/5

Using the cross multiplication method,

5h = 3b

b = 5h/3

We know that the area of a parallelogram is:

A = base × height

p = (5h/3) × h

5h²/3 = p

Using the cross multiplication method,

5h² = 3p

h² = 3p/5

h = $\sqrt{ \frac{3p}{5} }$ units

Hence, the height of the parallelogram is $\sqrt{ \frac{3p}{5} }$ units.