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
$$\mathrm{please}\:\mathrm{sir},\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:...\:\:\:\:\frac{\mathrm{a}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{r}\:×\:\mathrm{r}\:\:=\:\:\frac{\pi\mathrm{r}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{and}\:\mathrm{equation}\:\left(\mathrm{iii}\right)\:\mathrm{sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19
$$\mathrm{4}{a}+\mathrm{4}{b}+\mathrm{4}{c}=\pi{R}^{\mathrm{2}} .....{eqn}\mathrm{1} \\ $$$$\mathrm{2}{a}+{b}=\pi{r}^{\mathrm{2}} \\ $$$$\mathrm{2}{R}=\mathrm{4}{r} \\ $$$${r}=\frac{{R}}{\mathrm{2}} \\ $$$$\mathrm{2}{a}+{b}=\pi\left(\frac{{R}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{a}+{b}=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}......{eqn}\mathrm{2} \\ $$$$\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{r}×{r}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${a}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}−{r}^{\mathrm{2}} =\frac{{R}^{\mathrm{2}} }{\mathrm{4}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right)....{eqn}\mathrm{3} \\ $$$$\mathrm{4}{a}=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}−{R}^{\mathrm{2}} \\ $$$${b}=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{2}×\frac{{R}^{\mathrm{2}} }{\mathrm{4}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right)=\frac{{R}^{\mathrm{2}} }{\mathrm{4}}\left\{\pi−\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right)\right\} \\ $$$${b}=\frac{{R}^{\mathrm{2}} }{\mathrm{4}}\left(\pi−\pi+\mathrm{2}\right)=\frac{{R}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{4}{b}=\mathrm{2}{R}^{\mathrm{2}} \\ $$$$\mathrm{4}{c}=\pi{R}^{\mathrm{2}} −\left(\mathrm{4}{a}+\mathrm{4}{b}\right) \\ $$$$\:\:=\pi{R}^{\mathrm{2}} −\left\{\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}−{R}^{\mathrm{2}} +\mathrm{2}{R}^{\mathrm{2}} \right\} \\ $$$$=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}−{R}^{\mathrm{2}} ={R}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right)=\mathrm{42}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right)=\mathrm{1006}.\mathrm{885} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19
$${pls}\:{check}...{i}\:{have}\:{solved}... \\ $$
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
$$\frac{\mathrm{1}}{\mathrm{2}}{a}+{area}\:{of}\:{triangle} \\ $$$$=\frac{{a}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{r}×{r} \\ $$$$=\frac{{a}}{\mathrm{2}}+\frac{{r}^{\mathrm{2}} }{\mathrm{4}}\:\:{is}\:{equals}\:{to}\:\frac{\mathrm{1}}{\mathrm{4}}\pi{r}^{\mathrm{2}} \\ $$
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
$$\mathrm{Sir},\:\mathrm{sorry},\:\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\:\frac{\mathrm{a}}{\mathrm{2}}\:\:\mathrm{sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19
$${a}\:{is}\:{area}\:{of}\:{intersection}\left[{beteen}\:{two}\:{smsll}\left[{circles}\right.\right. \\ $$$${of}\:{radius}\:{r}. \\ $$$${pls}\:{refer}\:{my}\left[{first}\:{picture}...\right. \\ $$$${now}\:{when}\left[{you}\:{add}\:\frac{{a}}{\mathrm{2}}\:{with}\:{triangle}\:{area}\:\frac{\mathrm{1}}{\mathrm{2}}×{r}×{r}\right. \\ $$$${you}\:{get}\:\frac{\mathrm{1}}{\mathrm{4}\:}\pi{r}^{\mathrm{2}} ... \\ $$
Commented by Tawa1 last updated on 09/Jan/19
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$