Question

Question 203
Construct a trapezium ABCD, where $AB\parallel CD$, AD = BC = 3.2 cm, AB = 6.4 cm and CD = 9.6 cm. Measure $\angle B$ and $\angle A$.

[Hint Difference of two parallel sides gives an equilateral triangle]

Answer 1

Steps of Construction
Step I Draw a line segment DC = 9.6 cm.
Step II With D as center, draw an angle measure ${60}^{\circ }$. Now, cut-off it with an arc 3.2 cm called point A.
Step III Now, draw a parallel AB to CD.
Step IV Taking C as center, cut an arc B measure 3.2 cm on previous parallel line.
Step V Draw a line segment BE = 3.2 cm from arc B.
Step VI Join B to E and C.
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)