Question

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

$1) \: 3 \sqrt{ \frac{p}{15} }$
$2) \: 5 \sqrt{ \frac{p}{15} }$
$3) \: 5\sqrt{ \frac{p}{3} }$
$4) \: 3\sqrt{ \frac{p}{5} }$



​

Answer 1

Answer:

the perpendicular distance between parallel sides of the parallelogram can be 3$\sqrt{P/15}$ units, Option (1)

Step-by-step explanation:

Given : area of parallelogram is P square units

            Height : Base = 3 :5

           Distance between parallal sides = ?

Solution : Let Height = 3a units and Base = 5a units

       ∴ Area = Height x base

             P = 3a x 5a

             P = 15 a²

             a² = P/15

             a = $\sqrt{P/15}$ units

∴        Distance between parallal sides = Height

                                                                = 3 x a

                                                                = 3$\sqrt{P/15}$ units