Construct a trapezium $A B C D$ when one of parallel sides $A B = 4.8 c m$ , height $= 2.6 c m, B C = 3.1 c m$ and $A D = 3.6 c m .$

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Solution

Step construction:
(i) Draw a line segment $A B = 4.8 c m$
(ii) At $A$ , draw a ray $A Z$ making an angle of $90^{o}$ cut off $A L = 2.6 c m$
(iii) At $L$ , draw a line $X Y$ parallel to $A B$ .
(iv) With centre $A$ and radius $3.6 c m$ and with centre $B$ and radius $3.1 c m$ , draw arcs intersecting $X Y$ at $D$ and $C$ respectively.
(v) Join $A D$ and $B C$
Thus, $A B C D$ is the required trapezium.

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