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Question

Construct a quadrilateral $$ABCD$$ with $$AB=5\;cm,BC=3.6\;cm,\angle ABC=80^{\circ},\angle DAB=60^{\circ}$$ and $$\angle ADC=120^{\circ}$$.


Solution


We know that sum of four angles of a quadrilateral is $$360^o.$$
$$\therefore$$  $$\angle A+\angle B+\angle C+\angle D=360^o.$$
$$\therefore$$  $$60^o+80^o+\angle C+\angle 120^o=360^o$$
$$\therefore$$  $$260^o+\angle C=360^o.$$
$$\therefore$$  $$\angle C=100^o.$$
$$\Rightarrow$$  Draw a line segment $$AB=5\,cm$$
$$\Rightarrow$$  Take $$A$$ as a center and draw an angle of $$60^o$$
$$\Rightarrow$$  Take $$B$$ as a center and draw an angle of $$80^o$$
$$\Rightarrow$$  Cut off $$BC=3.6\,cm$$
$$\Rightarrow$$  Take $$C$$ as a center and draw an angle of $$100^o$$, which cuts of $$\angle A$$ at $$D.$$
$$\therefore$$  $$ABCD$$ is a required quadrilateral.

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Mathematics

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