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Question Number 178737 by cherokeesay last updated on 21/Oct/22

Answered by Rasheed.Sindhi last updated on 21/Oct/22

Commented by Rasheed.Sindhi last updated on 21/Oct/22

$$\mathrm{Let}\mathrm{AC}=\mathrm{BC}=\mathrm{r}$$$$\mathrm{blue}-\mathrm{area}=\frac{1}{4}\pi {\mathrm{r}}^2$$$$\mathrm{R}=\frac{\mathrm{AB}}{2}=\frac{\sqrt{{\mathrm{r}}^2+{\mathrm{r}}^2}}{2}=\frac{\sqrt{2}\mathrm{r}}{2}$$$$\mathrm{outer}\mathrm{circle}=\pi {\mathrm{R}}^2=\pi {\left(\frac{\sqrt{2}\mathrm{r}}{2}\right)}^2$$$$\mathrm{yello}-\mathrm{area}=\mathrm{outer}\mathrm{circle}-\mathrm{blue}-\mathrm{area}$$$$=\pi {\left(\frac{\sqrt{2}\mathrm{r}}{2}\right)}^2-\frac{1}{4}\pi {\mathrm{r}}^2=\frac{2\pi {\mathrm{r}}^2}{4}-\frac{\pi {\mathrm{r}}^2}{4}=\frac{\pi {\mathrm{r}}^2}{4}$$$$=\mathrm{blue}-\mathrm{area}$$

Commented by cherokeesay last updated on 21/Oct/22

$$Greet!thankyou$$

Commented by Tawa11 last updated on 21/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by HeferH last updated on 21/Oct/22

$$Bluearea:$$$$\frac{\pi {\left(r\sqrt{2}\right)}^2}{4}=\frac{\pi r^2}{2}$$$$Yellowarea:$$$$\pi r^2-blue=\frac{\pi r^2}{2}$$$$thenyellow=blue$$

Commented by cherokeesay last updated on 21/Oct/22

$$Nice!thankyou!$$

Commented by Tawa11 last updated on 21/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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