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Question Number 52523 by Tawa1 last updated on 09/Jan/19

Commented by Tawa1 last updated on 09/Jan/19

$$\mathrm{Please}\mathrm{show}\mathrm{me}\mathrm{steps}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

Commented by Tawa1 last updated on 09/Jan/19

$$\mathrm{please}\mathrm{sir},\mathrm{how}\mathrm{did}\mathrm{you}\mathrm{get}...\frac{\mathrm{a}}{2}+\frac{1}{2}\mathrm{r}\times \mathrm{r}=\frac{\pi {\mathrm{r}}^2}{4}$$$$\mathrm{and}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{sir}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

$$4a+4b+4c=\pi R^2.....eqn1$$$$2a+b=\pi r^2$$$$2R=4r$$$$r=\frac{R}{2}$$$$2a+b=\pi {\left(\frac{R}{2}\right)}^2$$$$2a+b=\frac{\pi R^2}{4}......eqn2$$$$\frac{a}{2}+\frac{1}{2}r\times r=\frac{\pi r^2}{4}$$$$a=\frac{\pi r^2}{2}-r^2=\frac{R^2}{4}(\frac{\pi }{2}-1)....eqn3$$$$4a=\frac{\pi R^2}{2}-R^2$$$$b=\frac{\pi R^2}{4}-2\times \frac{R^2}{4}(\frac{\pi }{2}-1)=\frac{R^2}{4}\{\pi -2(\frac{\pi }{2}-1)\}$$$$b=\frac{R^2}{4}(\pi -\pi +2)=\frac{R^2}{2}$$$$4b=2R^2$$$$4c=\pi R^2-(4a+4b)$$$$=\pi R^2-\{\frac{\pi R^2}{2}-R^2+2R^2\}$$$$=\frac{\pi R^2}{2}-R^2=R^2(\frac{\pi }{2}-1)={42}^2(\frac{\pi }{2}-1)=1006.885$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

$$plscheck...ihavesolved...$$

Commented by Tawa1 last updated on 09/Jan/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{i}\mathrm{appreciate}.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

$$\frac{1}{2}a+areaoftriangle$$$$=\frac{a}{2}+\frac{1}{2}r\times r$$$$=\frac{a}{2}+\frac{r^2}{4}isequalsto\frac{1}{4}\pi r^2$$

Commented by Tawa1 last updated on 09/Jan/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}..$$

Commented by Tawa1 last updated on 09/Jan/19

$$\mathrm{Sir},\mathrm{sorry},\mathrm{how}\mathrm{did}\mathrm{you}\mathrm{get}\frac{\mathrm{a}}{2}\mathrm{sir}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

$$aisareaofintersection[beteentwosmsll[circles$$$$ofradiusr.$$$$plsrefermy[firstpicture...$$$$nowwhen[youadd\frac{a}{2}withtrianglearea\frac{1}{2}\times r\times r$$$$youget\frac{1}{4}\pi r^2...$$

Commented by Tawa1 last updated on 09/Jan/19

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

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