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Question Number 19675 by ajfour last updated on 14/Aug/17

Commented by ajfour last updated on 14/Aug/17

$$\mathrm{Q}.19668\mathrm{Find}\mathrm{r}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{side}\mathbf{d}$$$$\mathrm{of}\mathrm{equilareral}△\mathrm{ABC}.$$

Answered by ajfour last updated on 14/Aug/17

$$\frac{\mathrm{r}}{\mathrm{x}}=\sin (90-\theta )=\cos \theta =\frac{1}{2}$$$$\Rightarrow \mathrm{x}=2\mathrm{r}$$$$\mathrm{x}+\mathrm{r}+\mathrm{R}=\frac{\mathrm{d}}{\sin \theta }\left(\mathrm{C}\mathrm{to}\mathrm{centre}\mathrm{of}\mathrm{big}\mathrm{circle}\right)$$$$\Rightarrow 3\mathrm{r}+\frac{\mathrm{d}}{\sqrt{3}}=\frac{2\mathrm{d}}{\sqrt{3}}$$$$\Rightarrow \mathrm{r}=\frac{\mathrm{d}}{3\sqrt{3}}=\frac{\mathbf{d}\sqrt{3}}{9}.\left(\mathrm{E}\right)$$

Commented by Joel577 last updated on 15/Aug/17

$$thankyouverymuch$$

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