Find the area of a trapezium (in sq. unit) with parallel sides of length 3 unit and 5 unit and the shortest distance between its parallel sides is 6 unit.

Correct option is C

Given:

a = 3 units

b = 5 units

h = 6 units 

Formula Used: 

To find the area of a trapezium, we use the following formula:

​$$\text{Area} = \frac{1}{2} \times (a + b) \times h$$ 

​where:

$a$ and $b$ are the lengths of the parallel sides

$h$ is the height (the shortest distance between the parallel sides).

Solution:

substitute the values into the formula:

​$$\text{Area} = \frac{1}{2} \times (3 + 5) \times 6$$ 

$$\text{Area} = \frac{1}{2} \times 8 \times 6$$ 

$$\text{Area} = 4 \times 6 = 24 \, \text{square units}$$ 

​Thus, the area of the trapezium is $24$ square units.​