Question Number 119529 by I want to learn more last updated on 25/Oct/20
Commented by help last updated on 25/Oct/20
$${Area}\:{of}\:{shaded}\:{portion} \\ $$$$\left\{\frac{\mathrm{4}}{\mathrm{7}}×\mathrm{10}^{\mathrm{2}} \right\}=\mathrm{400}/\mathrm{7}{cm}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by benjo_mathlover last updated on 25/Oct/20
$$\frac{\mathrm{1600}}{\mathrm{7}}\:=\:\mathrm{228}.\mathrm{571429} \\ $$$$\mathrm{228}.\mathrm{571429} \\ $$$${impossible} \\ $$
Commented by help last updated on 25/Oct/20
$${this}\:{is}\:{a}\:{shortcut}\:{approach}.\:{i}\:{can}\:{prove}\:{area}\:{of}\:{one}\:{leaf}\:{to}\:{be}\:\mathrm{4}/\mathrm{7}{x}^{\mathrm{2}} \:{if}\:{needed} \\ $$
Commented by benjo_mathlover last updated on 25/Oct/20
$${your}\:{short}\:{cut}\:{is}\:{valid}\:{if}\:{radius}\:=\:\mathrm{7}{k}\: \\ $$$${where}\:{k}\:\in\mathbb{Z} \\ $$
Commented by help last updated on 25/Oct/20
$$\mathrm{400}/\mathrm{7}\:{is}\:\mathrm{100\%}\:{correct}....{there}^{\:} \:{was}\:{a}\:{typo}\:{before}\:{which}\:{was}\:{corrected}...{still}\:{learning}\:{the}\:{app}\:{usage}\:{tho} \\ $$
Commented by mr W last updated on 25/Oct/20
$${please}\:{don}'{t}\:{write}\:{all}\:{in}\:{one}\:{single}\: \\ $$$${line}! \\ $$
Commented by talminator2856791 last updated on 27/Oct/20
$$\mathrm{can}\:\mathrm{this}\:\mathrm{diagram}\:\mathrm{be}\:\mathrm{drawn}\:\mathrm{in}\:\mathrm{the}\: \\ $$$$\mathrm{drawing}\:\mathrm{tool}?\:\mathrm{i}\:\mathrm{dont}\:\mathrm{know}\:\mathrm{how}\:\mathrm{to}\:\mathrm{draw}\: \\ $$$$\mathrm{it} \\ $$
Answered by mr W last updated on 25/Oct/20
$${shaded}\:{area}=\mathrm{8}\:{sectors}= \\ $$$$\mathrm{2}\:{red}\:{circles}\:{with}\:{radius}\:\mathrm{5}\:− \\ $$$$\mathrm{2}\:{green}\:{squares}\:{with}\:{side}\:{length}\:\mathrm{5}\sqrt{\mathrm{2}} \\ $$$$=\mathrm{2}×\pi×\mathrm{5}^{\mathrm{2}} −\mathrm{2}×\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{50}\left(\pi−\mathrm{2}\right) \\ $$$$=\mathrm{57}.\mathrm{08}\:{cm}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 25/Oct/20
Commented by talminator2856791 last updated on 27/Oct/20
$$\mathrm{very}\:\mathrm{nice}\:\mathrm{drawing}.\:\mathrm{did}\:\mathrm{you}\:\mathrm{draw}\:\mathrm{it}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{drawing}\:\mathrm{tool}? \\ $$
Answered by ebi last updated on 25/Oct/20
$$ \\ $$$${lets}\:{take}\:{the}\:{first}\:{quarter}\:{of}\:{ABCD} \\ $$$${at}\:{point}\:{A} \\ $$$$ \\ $$$${area}\:{of}\:{sector}\:\left(\:{center}\:{at}\:{midpoint}\right. \\ $$$$\left.{AB},{r}=\mathrm{5}{cm}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\pi{r}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{22}}{\mathrm{7}}\right)\left(\mathrm{5}^{\mathrm{2}} \right)=\mathrm{19}.\mathrm{64}{cm}^{\mathrm{2}} \\ $$$${area}\:{of}\:{triangle}\:\left({right}−{angle}\:{triangle}\right. \\ $$$$\left.{at}\:{midpoint}\:{AB}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}\right)\left(\mathrm{5}\right)=\mathrm{12}.\mathrm{5}{cm}^{\mathrm{2}} \\ $$$${area}\:{of}\:{half}\:{of}\:{the}\:{shaded}\:{region}\:{at} \\ $$$${the}\:{first}\:{quarter} \\ $$$$=\mathrm{19}.\mathrm{64}−\mathrm{12}.\mathrm{5}=\mathrm{7}.\mathrm{14}{cm}^{\mathrm{2}} \\ $$$${area}\:{of}\:{shaded}\:{region}\:{at}\:{the}\:{first} \\ $$$${quarter} \\ $$$$=\mathrm{2}×\mathrm{7}.\mathrm{14}=\mathrm{14}.\mathrm{28}{cm}^{\mathrm{2}} \\ $$$$ \\ $$$${thus}, \\ $$$${total}\:{area}\:{of}\:{shaded}\:{region} \\ $$$$=\mathrm{4}×\mathrm{14}.\mathrm{28}=\mathrm{57}.\mathrm{12}{cm}^{\mathrm{2}} \\ $$$$ \\ $$
Answered by TANMAY PANACEA last updated on 25/Oct/20
$${area}\:{of}\:{each}\:{leaf}=\mathrm{2}{x} \\ $$$${area}\:{of}\:{each}\:{triangular}\:{curved}\:{area}\:{y} \\ $$$$\mathrm{4}\left(\mathrm{2}{x}\right)+\mathrm{4}{y}={a}^{\mathrm{2}} \\ $$$${again}\:{ares}\:{of}\:{triangle} \\ $$$$\mathrm{2}{x}+{y}=\frac{\mathrm{1}}{\mathrm{2}}×{a}×\frac{{a}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\measuredangle{DOC}=\frac{\pi}{\mathrm{2}} \\ $$$${so}\:{radius}\:{ofcurve}=\frac{{a}}{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{2}{x}\right)+{y}=\frac{\pi{a}^{\mathrm{2}} }{\mathrm{2}}\:\: \\ $$$$\mathrm{4}{x}+{y}=\frac{\pi{a}^{\mathrm{2}} }{\mathrm{2}}\:×\mathrm{4} \\ $$$$\mathrm{8}{x}+\mathrm{4}{y}={a}^{\mathrm{2}} \\ $$$$\mathrm{16}+\mathrm{4}{y}=\mathrm{2}\pi{a}^{\mathrm{2}} \\ $$$$\mathrm{8}{x}+\mathrm{4}{y}={a}^{\mathrm{2}} \\ $$$$\mathrm{8}{x}={a}^{\mathrm{2}} \left(\mathrm{2}\pi−\mathrm{1}\right) \\ $$$${area}\:{shadded}=\mathrm{8}{x} \\ $$$$ \\ $$
Commented by TANMAY PANACEA last updated on 25/Oct/20
$${Pls}\:{find}\:{error}\:{in}\:{this}\:{calculation}... \\ $$