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Question

Construct a quadrilateral PQRS, where $$\dot { P Q } = 7.7 \mathrm { cm } , Q R = 6.8 \mathrm { cm } , R S = 5.1 \mathrm { cm } , S P = 3.6 \mathrm { cm }$$ and $$\angle R = 120 ^ { \circ }$$.


Solution


$$\Rightarrow$$  Draw a line segment $$QR=6.8\,cm$$
$$\Rightarrow$$  Take $$R$$ as center and draw an angle of $$120^o.$$
$$\Rightarrow$$  Cut off $$RS=5.1\,cm$$
$$\Rightarrow$$  Take $$S$$ as center with radius $$3.6\,cm$$ and draw an arc.
$$\Rightarrow$$  Take $$Q$$ as center with radius $$7.7\,cm$$ and draw an arc, which intersects previous arc at point $$P$$.
$$\Rightarrow$$  Join $$PS$$ and $$PQ$$
$$\therefore$$  $$PQRS$$ is an required quadrilateral.

1266170_1182709_ans_b16118bd661542f3b17600922fd637ae.jpeg

Mathematics

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