Question Number 223066 by fantastic last updated on 13/Jul/25
Commented by fantastic last updated on 13/Jul/25
$${Area}\:{of}\:{semicircle}\:{in}\:{terms}\:{of}\:{a}\:{and}\:{b} \\ $$$$ \\ $$
Answered by mr W last updated on 13/Jul/25
Commented by mr W last updated on 14/Jul/25
Commented by mr W last updated on 14/Jul/25
$$\frac{{x}}{{z}}=\frac{{a}}{{b}} \\ $$$${xz}=\left({x}+{z}+\sqrt{{x}^{\mathrm{2}} +{z}^{\mathrm{2}} }\right){a} \\ $$$$\Rightarrow{x}={a}+\frac{{a}^{\mathrm{2}} }{{b}}+\frac{{a}}{{b}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${similarly} \\ $$$$\Rightarrow{y}={b}+\frac{{b}^{\mathrm{2}} }{{a}}+\frac{{b}}{{a}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${R}=\frac{{x}+{y}}{\mathrm{2}}=\frac{{a}+{b}}{\mathrm{2}}+\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}{ab}} \\ $$
Commented by mr W last updated on 13/Jul/25
$$\frac{{xz}}{\mathrm{2}}={a}\:\Rightarrow{x}=\frac{\mathrm{2}{a}}{{z}} \\ $$$$\frac{{yz}}{\mathrm{2}}={b}\:\Rightarrow{y}=\frac{\mathrm{2}{b}}{{z}} \\ $$$${z}^{\mathrm{2}} ={xy}=\frac{\mathrm{4}{ab}}{{z}^{\mathrm{2}} }\:\Rightarrow{z}=\sqrt{\mathrm{2}\sqrt{{ab}}} \\ $$$${R}=\frac{{x}+{y}}{\mathrm{2}}=\frac{{a}+{b}}{{z}}=\frac{{a}+{b}}{\:\sqrt{\mathrm{2}\sqrt{{ab}}}} \\ $$$${area}\:{of}\:{semi}\:{circle}\:=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{2}}=\frac{\pi\left({a}+{b}\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{4}{ab}}} \\ $$
Commented by fantastic last updated on 14/Jul/25
$${thanks}\:{sir} \\ $$
Commented by fantastic last updated on 14/Jul/25
$${thanks}\:{sir}. \\ $$
Commented by fantastic last updated on 14/Jul/25
$${now}\:{solve}\:{for}\:{the}\:{same}\:{thing}\:{if} \\ $$$${the}\:{areas}\:{of}\:{two}\:{circles}\:{inscribed} \\ $$$${in}\:{that}\:{right}\:{triangle}\:{is}\:{given} \\ $$
Commented by fantastic last updated on 14/Jul/25
$${they}\:{must}\:{not}\:{touch}\:{each}\:{other} \\ $$