Question Number 146840 by bramlexs22 last updated on 16/Jul/21
Answered by nimnim last updated on 16/Jul/21
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Commented by bobhans last updated on 16/Jul/21
$$\mathrm{what}\:\mathrm{your}\:\mathrm{answer}\: \\ $$
Commented by nimnim last updated on 16/Jul/21
$${I}\:{got}\:\left(\mathrm{8}\sqrt{\mathrm{3}}+\mathrm{8}\right){r}^{\mathrm{2}} .\:{but}\:{could}\:{not}\:{load}\:{the}\:{solution}!< \\ $$
Answered by EDWIN88 last updated on 16/Jul/21
$$\left(\mathrm{1}\right)\:\mathrm{TS}\:=\:\mathrm{4r}\: \\ $$$$\left(\mathrm{2}\right)\:\mathrm{UT}\:=\:\mathrm{2r}\:+\mathrm{2}\sqrt{\left(\mathrm{2r}\right)^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} }=\mathrm{2r}+\mathrm{2r}\sqrt{\mathrm{3}}\: \\ $$$$\:\mathrm{the}\:\mathrm{area}\:\mathrm{RSTU}\:=\:\mathrm{TS}\:×\:\mathrm{UT}\: \\ $$$$\:\:\:\:\:\:\:\:=\:\mathrm{4r}\:×\left(\mathrm{2r}+\mathrm{2r}\sqrt{\mathrm{3}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:=\:\mathrm{8r}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\: \\ $$
Commented by Rasheed.Sindhi last updated on 16/Jul/21
$$\mathcal{T}{he}\:{process}\:{of}\:{calculating}\:\mathrm{UT}\:{is} \\ $$$${main}\:{part}\:{of}\:{the}\:{solution},{which} \\ $$$${you}\:{have}\:{hidden}!!!\:{Solution}\:{is}\:{not} \\ $$$${much}\:{beneficiary}. \\ $$
Commented by EDWIN88 last updated on 16/Jul/21
$$\mathrm{why}\:?\:\mathrm{what}\:\mathrm{your}\:\mathrm{working}? \\ $$
Commented by Rasheed.Sindhi last updated on 17/Jul/21
$${I}\:{mean}\:{some}\:{detail}\:{of}\:{finding} \\ $$$${length}\:{of}\:{the}\:{rectangle}. \\ $$