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Question Number 49970 by behi83417@gmail.com last updated on 12/Dec/18

Commented by behi83417@gmail.com last updated on 12/Dec/18

$$ABC,equilateral.AD=DC,\measuredangle CDE={30}^{\bullet }$$$$\Rightarrow \frac{area\left[ABED\right]}{area\left[EDC\right]}=?$$

Answered by mr W last updated on 12/Dec/18

$$CE=\frac{CD}{2}$$$${\mathrm{\Delta }}_{CDE}=\frac{{\mathrm{\Delta }}_{CDB}}{4}=\frac{{\mathrm{\Delta }}_{ABC}}{8}=\frac{{\mathrm{\Delta }}_{CDE}+A_{ABEDA}}{8}$$$$7{\mathrm{\Delta }}_{CDE}=A_{ABEDA}$$$$\Rightarrow \frac{A_{ABEDA}}{{\mathrm{\Delta }}_{CDE}}=7$$

Commented by behi83417@gmail.com last updated on 12/Dec/18

$$yessir!itisniceandsimple.$$$$pleasesolvefor:C\stackrel{}{DE}=\alpha .$$

Answered by mr W last updated on 12/Dec/18

$$AB=a$$$$\mathrm{\angle }CDE=\alpha$$$$\frac{CE}{\sin \alpha }=\frac{a}{2\sin (\alpha +\frac{\pi }{3})}=\frac{a}{\sin \alpha +\sqrt{3}\cos \alpha }$$$$\Rightarrow CE=\frac{a}{1+\sqrt{3}\cot \alpha }$$$$\mathrm{\Delta }CBE=\frac{1}{2}\times \frac{a}{2}\times \frac{a}{1+\sqrt{3}\cot \alpha }\times \sin \frac{\pi }{3}$$$$\mathrm{\Delta }CBE=\frac{3a^2}{8(\sqrt{3}+3\cot \alpha )}$$$${\mathrm{\Delta }}_{ABC}=\frac{\sqrt{3}{a}^2}{4}$$$$\frac{{\mathrm{\Delta }}_{ABC}}{{\mathrm{\Delta }}_{CBE}}=2(1+\sqrt{3}\cot \alpha )$$$$\Rightarrow \frac{A_{ABED}}{{\mathrm{\Delta }}_{CBE}}=1+2\sqrt{3}\cot \alpha$$

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