Question Number 74293 by Mr. K last updated on 21/Nov/19
Commented by Mr. K last updated on 21/Nov/19
$${ABCD}\:{id}\:{a}\:{quadrilateral},\:{cos}\theta=\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}. \\ $$$${AE}=\mathrm{1},\:{BE}=\mathrm{4},\:{CE}=\mathrm{3}\:{and}\:{DE}=\mathrm{2}. \\ $$$${Find}\:{the}\:{area}\:{of}\:{the}\:{quadrilateral}. \\ $$
Commented by ~blr237~ last updated on 21/Nov/19
$${let}\:{named}\:{it}\:\:{S} \\ $$$${S}={Area}\left({EAD}\right)+{Area}\left({EDC}\right)+{Area}\left({ECB}\right)+{Area}\left({EBA}\right) \\ $$$${A}\:{rea}\left({EAD}\right)=\frac{\mathrm{1}}{\mathrm{2}}{EA}×{ED}×\mid{sin}\left({mesAED}\right)\mid=\mid{sin}\theta\mid \\ $$$${cause}\:\:{mesAED}=\theta\:\:\: \\ $$$${Area}\left({EDC}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ED}×{EC}×\mid{sin}\left({mesDEC}\right)\mid=\mathrm{3}\mid{sin}\theta\mid \\ $$$${cause}\:\:{mesDEC}=\pi−\theta \\ $$$${Area}\left({ECB}\right)=\frac{\mathrm{1}}{\mathrm{2}}{EC}×{EB}×\mid{sin}\theta\mid=\mathrm{6}\mid{sin}\theta\mid \\ $$$${Area}\left({EBA}\right)=\frac{\mathrm{1}}{\mathrm{2}}{EB}×{EA}×\mid{sin}\left(\pi−\theta\right)\mid=\mathrm{2}\mid{sin}\theta\mid \\ $$$${So}\:\:{S}=\mathrm{12}\mid{sin}\theta\mid \\ $$$$\:{S}^{\mathrm{2}} =\mathrm{144}\:\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right)=\mathrm{144}\left(\mathrm{1}−\frac{\mathrm{7}}{\mathrm{16}}\right)=\frac{\mathrm{144}×\mathrm{9}}{\mathrm{16}}\: \\ $$$${finally}\:\:\:{S}=\frac{\mathrm{12}×\mathrm{3}}{\mathrm{4}}=\mathrm{9} \\ $$$${check}\:{if}\:{there}\:{is}\:{not}\:{any}\:{mistake} \\ $$