Construct a trapezium ABCD in which AB = 6 cm,

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.

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