# Q.5. Find the area enclosed by $2\left|x\right|+3\left|y\right|\le 6$.

Dear Student,
Here is the solution of your asked query:
$\mathrm{Given},\phantom{\rule{0ex}{0ex}}2\left|\mathrm{x}\right|+3\left|\mathrm{y}\right|\le 6\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{So}\mathrm{we}\mathrm{take}\mathrm{the}\mathrm{lines},\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2\left|\mathrm{x}\right|+3\left|\mathrm{y}\right|=6\phantom{\rule{0ex}{0ex}}\mathrm{and}\mathrm{the}\mathrm{area}\mathrm{enclosed}\mathrm{by}\mathrm{these}\mathrm{lines}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{area}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{for}1\mathrm{st}\mathrm{quadrant},\phantom{\rule{0ex}{0ex}}\mathrm{x}>0,\mathrm{y}>0\phantom{\rule{0ex}{0ex}}2\mathrm{x}+3\mathrm{y}=6\phantom{\rule{0ex}{0ex}}\mathrm{which}\mathrm{is}\mathrm{plotted}\mathrm{in}\mathrm{the}\mathrm{graph}\phantom{\rule{0ex}{0ex}}2\mathrm{nd}\mathrm{quadrant}\phantom{\rule{0ex}{0ex}}\mathrm{x}<0.\mathrm{y}>0\phantom{\rule{0ex}{0ex}}-2\mathrm{x}+3\mathrm{y}=6\phantom{\rule{0ex}{0ex}}\mathrm{which}\mathrm{is}\mathrm{plotted}\mathrm{in}\mathrm{the}\mathrm{graph}\phantom{\rule{0ex}{0ex}}\mathrm{x}<0,\mathrm{y}<0\phantom{\rule{0ex}{0ex}}-2\mathrm{x}-3\mathrm{y}=6\phantom{\rule{0ex}{0ex}}\mathrm{which}\mathrm{is}\mathrm{plotted}\mathrm{in}\mathrm{the}\mathrm{graph}\phantom{\rule{0ex}{0ex}}\mathrm{x}>0,\mathrm{y}<0\phantom{\rule{0ex}{0ex}}2\mathrm{x}-3\mathrm{y}=6\phantom{\rule{0ex}{0ex}}\mathrm{which}\mathrm{is}\mathrm{plotted}\mathrm{in}\mathrm{the}\mathrm{graph}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{hence}\mathrm{from}\mathrm{the}\mathrm{figure}\mathrm{we}\mathrm{see}\mathrm{that}\mathrm{it}\mathrm{forms}\mathrm{a}\mathbf{rhombus}\phantom{\rule{0ex}{0ex}}\mathrm{of}\mathrm{side}\mathrm{length}=\sqrt{{\left(3-0\right)}^{2}+{\left(0-2\right)}^{2}}=\sqrt{13}\mathrm{units}\phantom{\rule{0ex}{0ex}}\mathrm{length}\mathrm{of}\mathrm{diagonals}\mathrm{is}4\mathrm{and}6\mathrm{units}\phantom{\rule{0ex}{0ex}}\mathbf{area}\mathbf{=}\frac{\mathbf{pq}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{4}\mathbf{×}\mathbf{6}}{\mathbf{2}}\mathbf{=}\mathbf{12}\mathbf{}\mathbf{sq}\mathbf{.}\mathbf{}\mathbf{units}$
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