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Question Number 146701 by bramlexs22 last updated on 15/Jul/21

Commented by bramlexs22 last updated on 15/Jul/21

$$\mathrm{how}$$

Commented by som(math1967) last updated on 15/Jul/21

$$58.5squnit$$

Answered by Rasheed.Sindhi last updated on 15/Jul/21

$$△ABC:$$$$sin\mathrm{\angle }C=\frac{12}{20}=\frac{3}{5}\Rightarrow \mathrm{\angle }C=36.87°$$$$△DCE:$$$$tan\mathrm{\angle }C=\frac{DE}{EC}\Rightarrow DE=\left(EC\right)tan\mathrm{\angle }C$$$$DE=10tan36.87=7.5$$$$△BDE:$$$$BD=\sqrt{{10}^2+7.5^2}=12.5$$$$△BDE=\frac{1}{2}.BE.DE=\frac{1}{2}\left(10\right)(7.5)=37.5$$$$△ABD:$$$$AD=\sqrt{{BD}^2-{AB}^2}=\sqrt{12.5^2-{12}^2}=3.5$$$$△ABD=\frac{1}{2}.AB.AD=\frac{1}{2}\left(12\right)(3.5)=21$$$$ABED:$$$$ABED=△BDE+△ABD=37.5+21=58.5squareunits$$

Commented by bramlexs22 last updated on 15/Jul/21

$$\mathrm{yes}.{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{typo}$$

Answered by som(math1967) last updated on 15/Jul/21

$$AC=\sqrt{{20}^2-{12}^2}=16$$$$ar.△ABC=\frac{1}{2}\times 12\times 16=96squnit$$$$△DEC\sim △BAC$$$$\therefore \frac{DE}{AB}=\frac{CE}{AC}$$$$DE=12\times \frac{10}{16}=\frac{15}{2}=7.5$$$$Ar△CDE=\frac{1}{2}\times CE\times DE=\frac{1}{2}\times 10\times 7.5=37.5squnit$$$$ArABED=96-37.5=58.5squnit$$

Commented by Rasheed.Sindhi last updated on 15/Jul/21

$$\mathcal{N}ice!$$

Commented by otchereabdullai@gmail.com last updated on 16/Jul/21

$$\mathrm{nice}\mathrm{one}\mathrm{sir}!$$

Commented by som(math1967) last updated on 16/Jul/21

$$Thankyou$$

Answered by bramlexs22 last updated on 15/Jul/21

$$\mathrm{from}\mathrm{\angle }\mathrm{C}:\sin \mathrm{C}=\sin \mathrm{C}$$$$\Rightarrow \frac{\mathrm{DE}}{\sqrt{100+{\mathrm{DE}}^2}}=\frac{12}{20}=\frac{3}{5}$$$$\Rightarrow {25\mathrm{DE}}^2=900+{9\mathrm{DE}}^2$$$$\Rightarrow \mathrm{DE}=\sqrt{\frac{900}{16}}=\frac{30}{4}=\frac{15}{2}$$$$\tan \mathrm{C}=\tan \mathrm{C}$$$$\Rightarrow \frac{12}{\mathrm{AC}}=\frac{\frac{15}{2}}{10}\Rightarrow \frac{12}{\mathrm{AC}}=\frac{\cancel{15}{}^3}{\cancel{20}{}^4}$$$$\Rightarrow \mathrm{AC}=16$$$$\mathrm{Area}\mathrm{ABED}=\mathrm{area}\mathrm{ABC}-\mathrm{area}\mathrm{CED}$$$$=\frac{1}{2}\times 12\times 16-\frac{1}{2}\times \frac{15}{2}\times 10$$$$=96-\frac{75}{2}=\frac{192-75}{2}=\frac{117}{2}$$$$=58.5$$

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