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Question

The sides of a quadrilateral $$ABCD$$ $$6\ cm$$, $$8\ cm$$,$$12\ cm$$,$$14\ cm$$, respectively. The angle between the first two sides is a right angle, Find area? (taken in order)


Solution

$$\therefore AC=\sqrt{{6}^{2}+{8}^{2}}=\sqrt{100}=10cm$$
In $$\triangle ACD$$,
Semi-perimeter \left(s\right)=\dfrac{14+12+10}{12}$$
$$=18$$
$$\therefore$$ Area of quadrilateral $$ABCD=$$Area of $$\triangle ABC$$+ Area of $$\triangle ACD$$
$$=\dfrac{1}{2}\times 66\times 8+\sqrt{18\left(18-14\right) \left(18-12\right) \left(18-10\right)}$$
$$=24+\sqrt{18\times 4\times 6\times 8}$$
$$=24+\sqrt{9\times 8\times 8\times 6}$$
$$=24+24\sqrt{6}$$
$$=24\left(\sqrt{6}+1\right)sq.units$$ Ans.

1398530_1124692_ans_9a56bef4adff49e68e0b48acafa50043.png

Mathematics

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