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Working rule to construct a trapezium:

Let us discuss the working rule to construct a trapezium when the measure of two sides and two angles of a trapezium are given.
Example:
Construct a trapezium $$PQRS$$ in which $$\overline{PQ}$$ is parallel to $$\overline{RS}$$, $$PQ$$ $$=$$ $$8.5$$ $$cm$$, $$QR$$ $$=$$ $$6.5$$ $$cm$$, $$\angle PQR$$ $$=$$ $$75^{\circ}$$ and $$\angle QPS$$ $$=$$ $$110^{\circ}$$. Also, find its area.

Construction:

Step 1: Draw a line segment $$PQ$$ $$=$$ $$8.5$$ $$cm$$.

Step 2: With $$Q$$ as centre, mark an angle $$75^{\circ}$$ using a protractor and mark it as $$X$$. Join $$QX$$.

Step 3: With $$Q$$ as centre, draw an arc of radius $$6.5$$ $$cm$$ intersecting $$QX$$ at $$R$$.

Step 4: Draw a line $$RY$$ parallel to $$PQ$$.

Step 5: With $$P$$ as centre, mark an angle $$110^{\circ}$$ using a protractor and mark it as $$Z$$. Join $$PZ$$ intersecting $$QY$$ at $$S$$.

Step 6: $$PQRS$$ is the required trapezium. The measure of $$RA$$ gives the height of the trapezium.

Area calculation:

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

$$=$$ $$\frac{1}{2} \times 6.3 \times (8.5 + 9.1)$$

$$=$$ $$\frac{1}{2} \times 6.3 \times 17.6$$

$$=$$ $$\frac{110.88}{2}$$

$$=$$ $$55.44$$ $$cm^2$$