Question
Draw a trapezium $ABCD$ in which $AB || DC, AB = 7\ cm, BC = 5\ cm, AD = 6.5\ cm$ and $\angle\text{B}=60^\circ.$
✓
Answer
Steps of construction:
Step 1: Draw $AB$ equal to $7\ cm.$
Step 2: Make an angle, $\angle\text{ABX},$ equal to $60^\circ .$
Step 3: With $B$ as the centre, draw an arc of $5\ cm.$ Name that point as $C$. Join $B$ and $C.$
Step 4: $AB || DC$
$\therefore\angle\text{ABX}+\angle\text{BCY}=180^\circ$
$\Rightarrow\angle\text{BCY}=180^\circ-60^\circ=120^\circ$
Draw an angle, $\angle\text{BCY},$ equal to $120^\circ .$
Step 5: With $A$ as the centre, draw an arc of length $6.5\ cm,$ which cuts $CY.$ Mark that point as $D.$
Step 6: Join $A$ and $D.$
Thus, $ABCD$ is the required trapezium.
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