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Question Number 206937 by efronzo1 last updated on 01/May/24

Answered by A5T last updated on 01/May/24

Commented by A5T last updated on 01/May/24

$$8\times 14=x(x+2r)=x^2+2rx=112$$$${14}^2={(x+r)}^2+r^2=x^2+2rx+2r^2=112+2r^2$$$$\Rightarrow r^2=42\Rightarrow r=\sqrt{42}\Rightarrow x=\sqrt{154}-\sqrt{42}$$$$sin\theta =\frac{\sqrt{42}}{14}\Rightarrow cos\theta =\frac{\sqrt{154}}{14}$$$${AB}^2=8^2+{(\sqrt{154}-\sqrt{42})}^2-16(\sqrt{154}-\sqrt{42})\frac{\sqrt{154}}{14}$$$$\Rightarrow AB=2\sqrt{21-3\sqrt{33}}$$$${BC}^2=4r^2={AB}^2+{AC}^2\Rightarrow AC=2\sqrt{21+3\sqrt{33}}$$$$\Rightarrow \frac{AB\times AC}{2}=2\sqrt{{21}^2-9\times 33}=2\times 12=24$$

Commented by A5T last updated on 01/May/24

$$Afterfindingr,onecanalsodothis:$$$$LetperpendicularfromAmeetBCatD$$$$Then,\frac{AD}{r}=\frac{8}{14}\Rightarrow AD=\frac{4r}{7}$$$$\left[ABC\right]=\frac{AD\times BC}{2}=\frac{4r^2}{7}=24$$

Answered by mr W last updated on 01/May/24

Commented by mr W last updated on 01/May/24

$$\frac{\frac{6}{2}}{r}=\frac{r}{8+6}\Rightarrow r^2=42$$$$\frac{h}{r}=\frac{8}{8+6}=\frac{4}{7}$$$$shadedarea=\frac{2rh}{2}=\frac{4r^2}{7}=\frac{4\times 42}{7}=24$$

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