Question

In a trapezium with parallel sides measuring 20cm and 12cm , enclosing area
288cm?. Find the distance between the parallel sides of the trapezium.​

Answer 1

Given :

A trapezium with parallel sides measuring 20cm and 12cm , enclosing area 288cm².

To Find :

The distance between the parallel sides of the trapezium.

Solution :

Analysis :

Here the formula of area of trapezium is used. By using the length of the parallel sides and area of the trapezium we can find the distance between them.

Required Formula :

Area of trapezium = 1/2 × (a + b) × d

where,

• a = First parallel side
• b = Second parallel side
• d = Distance between the parallel sides

Explanation :

Let us assume that the distance between the parallel sides be "d" cm.

We know that if we are given the area of the trapezium and the two parallel sides and is asked to find the distance between them then our required formula is,

Area of trapezium = 1/2 × (a + b) × d

where,

• a = 20 cm
• b = 12 cm
• d = d cm
• Area = 288 cm²

Using the required formula and substituting the required values,

⇒ Area of trapezium = 1/2 × (a + b) × d

⇒ 288 = 1/2 × (20 + 12) × d

⇒ 288 = 1/2 × (32) × d

⇒ 288 = 1/2 × 32 × d

⇒ 288 = 1 × 16 × d

⇒ 288 = 16 × d

⇒ 288/16 = d

⇒ 18 = d

∴ Distance = 18 cm.

The distance between the parallel sides is 18 cm.

Verification :

⇒ Area of trapezium = 1/2 × (a + b) × d

⇒ 288 = 1/2 × (20 + 12) × 18

⇒ 288 = 1/2 × (32) × 18

⇒ 288 = 1/2 × 32 × 18

⇒ 288 = 1 × 16 × 18

⇒ 288 = 288

∴ LHS = RHS.

• Hence verified.

Answer 2

Step-by-step explanation:

$\frak {Given} \begin{cases} &\sf{Parallel\ sides\ =\ 20cm\ and\ 12cm.} \\ &\sf{Area\ =\ 288cm^2.} \end{cases}$

To find:- We have to find the distance between the parallel sides ?

__________________

$\frak{\underline{\underline{\dag As\ we\ know\ that:-}}}$

$\sf : \implies {\underline{\boxed{\bf Area\ of\ trapezium\ =\ \dfrac{1}{2}\ \times\ (a\ +\ b)\ \times\ d.}}}$

Here:-

• a, is for first parallel side.
• b, is for second parallel side.
• d, is for distance between parallel sides.

__________________

$\frak{\underline{\underline{\dag By\ substituting\ the\ values,\ we\ get:-}}}$

$\sf : \implies {Area\ of\ trapezium\ =\ \dfrac{1}{2}\ \times\ (a\ +\ b)\ \times\ d} \\ \\ \\ \sf : \implies {288\ =\ \dfrac{1}{2}\ \times\ (20\ +\ 12)\ \times\ d} \\ \\ \\ \sf : \implies {288\ =\ \dfrac{1}{2}\ \times\ (32)\ \times\ d} \\ \\ \\ \sf : \implies {288\ =\ \dfrac{1}{\cancel 2}\ \times\ \cancel {32}\ \times\ d} \\ \\ \\ \sf : \implies {288\ =\ 1\ \times\ 16\ \times\ d} \\ \\ \\ \sf : \implies {288\ =\ 16\ \times\ d} \\ \\ \\ \sf : \implies {\cancel \dfrac{288}{16}\ =\ d} \\ \\ \\ \sf : \implies {18\ =\ d} \\ \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf d\ =\ 18cm.}}}}\bigstar$

Hence:-

$\sf \therefore {\underline{The\ required\ distance\ between\ the\ parallel\ sides\ is\ 18cm.}}$