Question

Construct a trapezium $ABCD$. where $AB||CD,AD=BC=3.2cm,AB=6.4cm$ and $CD=9.6cm$. Find the measure of $\angle B$ and $\angle A$.


[3 Marks]

Answer 1

Steps of Construction:
Step I: Draw a line segment $DC=9.6cm$.
Step II: With $D$ as center, draw an angle measure ${60}^{\circ }$. Now, cut-off it with an arc $3.2cm$ called point A.
Step III: Now, draw a parallel $AB$ to $CD$.
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Step IV: Taking $C$ as center, cut an arc $B$ measure $3.2cm$ on previous parallel line.
Step V: Draw a line segment $BE=3.2cm$ from arc B.
Step VI: Join $B$ to $E$ and $C$.
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Thus, we have required trapezium $ABCD$ in which $\angle A={120}^{\circ }$ and $\angle B=\angle EBC+\angle ABE={60}^{\circ }+{60}^{\circ }={120}^{\circ }$. (Since, $BEC$ is an equilateral triangle and $ABED$ is a parallelogam)
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