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Question Number 208554 by efronzo1 last updated on 18/Jun/24

Answered by liuxinnan last updated on 18/Jun/24

Commented by efronzo1 last updated on 18/Jun/24

Answered by A5T last updated on 18/Jun/24

Commented by A5T last updated on 18/Jun/24

$$DF=AE=h;EF=10$$$$FC=x\Rightarrow BE=6-x$$$$h^2=8^2-x^2=6^2-{(6-x)}^2\Rightarrow x=\frac{16}{3}$$$$\Rightarrow h=\sqrt{64-{\left(\frac{16}{3}\right)}^2}=\frac{8\sqrt{5}}{3}$$$$\left[ABCD\right]=\frac{(10+16)\times \frac{8\sqrt{5}}{3}}{2}=\frac{104\sqrt{5}}{3}$$

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