PUMPA - SMART LEARNING

எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

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• Depending upon the lengths of the sides and diagonal, create a triangle of $$PQR$$ based on $$SSS$$ construction.
• Make an arc (point $$S$$) at a certain distance from $$P$$.
• Make an arc at a certain distance from point $$R$$ on the earlier arc on $$S$$. Name the two points intersection as $$S$$.

• $$P$$ and $$R$$ join $$S$$. The quadrilateral $$PQRS$$ will be achieved.
Example:
Construct a quadrilateral $$ABCD$$ with the following measurements.

$$AB =$$ $$4.5 cm$$, $$BC =$$ $$5.5 cm$$, $$CD = 4cm$$, $$AD = 6cm$$, $$AC = 7cm$$.

Step 1:Draw side $$BC = 5.5 cm$$ and cut arcs above it from $$B$$ ($$4.5 cm$$) and $$C$$ ($$7 cm$$). Mark the intersection as $$A$$. Join $$AB$$ and $$AC$$.

Step 2: Draw and arc from $$A$$ equal to $$6 cm$$ which is the length of $$AD$$.

Step 3: Draw and arc from $$C$$ equal to $$4 cm$$ which is the length of $$CD$$. Mark the intersection as $$D$$ and join $$AD$$ and $$CD$$.

Thus, the $$ABCD$$ is a required quadrilateral.

Area of the quadrilateral $$ABCD$$ $$=$$ $$\frac {1}{2}$$ $$\times$$ $$d$$ $$\times$$ $$h_1 + h_2$$ sq. units

$$=$$ $$\frac{1}{2} × 10 (1.9 +2.3)$$

$$=$$ $$5\times 4.2$$

$$=$$ $$21 cm²$$.