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Theory:

Working rule to construct a trapezium:

Let us discuss the working rule to construct a trapezium when the measure of all the four sides of a trapezium are given.
Example:
Construct a trapezium $$ABCD$$ in which $$\overline{AB}$$ is parallel to $$\overline{CD}$$, $$AB$$ $$=$$ $$9$$ $$cm$$, $$BC$$ $$=$$ $$5$$ $$cm$$, $$CD$$ $$=$$ $$6$$ $$cm$$ and $$DA$$ $$=$$ $$6.5$$ $$cm$$. Also, find its area.

Construction:

Step 1: Draw a line segment $$AB$$ $$=$$ $$9$$ $$cm$$.

Step 2: Mark a point $$P$$ on the line segment $$AB$$ such that $$AP$$ $$=$$ $$6cm$$.

Step 3: With $$P$$ and $$B$$ as centres, draw two arcs of radii $$6.5$$ $$cm$$ and $$5$$ $$cm$$ respectively such that they intersect each other at $$C$$.

Step 4: Join $$PC$$ and $$BC$$.

Step 5: With $$C$$ and $$A$$ as centres, draw two arcs of radii $$6$$ $$cm$$ and $$6.5$$ $$cm$$ respectively such that they intersect each other at $$D$$.

Step 6: Join $$AD$$ and $$CD$$.

Step 7: $$ABCD$$ is the required trapezium. The measure of $$DQ$$ gives the height of the trapezium $$ABCD$$.

Area calculation:

Area of the trapezium $$ABCD$$ $$=$$ $$\frac{1}{2} \times h \times (a + b)$$ square units

$$=$$ $$\frac{1}{2} \times 5.8 \times (9 + 6)$$

$$=$$ $$\frac{1}{2} \times 5.8 \times 15$$

$$=$$ $$\frac{87}{2}$$

$$=$$ $$43.50$$ $$cm^2$$