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