Question:
Construct a trapezium ABCD in which AB = 6 cm, BC = 4 cm, CD = 3.2 cm, ∠B = 75° and DC||AB.
Solution:
Steps of construction:
Step 1: Draw $A B=6 \mathrm{~cm}$
Step 2: Make $\angle A B X=75^{\circ}$
Step 3: With $B$ as the centre, draw an arc at $4 \mathrm{~cm}$. Name that point as $C$.
Step 4: $A B \| C D$
$\therefore \angle A B X+\angle B C Y=180^{\circ}$
$\Rightarrow \angle B C Y=180^{\circ}-75^{\circ}=105^{\circ}$
Make $\angle B C Y=105^{\circ}$
At $C$, draw an arc of length $3.2 \mathrm{~cm}$.
Step 5: Join A and D.
Thus, ABCD is the required trapezium.