Question Number 218953 by Spillover last updated on 17/Apr/25
Answered by mr W last updated on 18/Apr/25
$${eqn}.\:{of}\:{AC}: \\ $$$$\frac{{x}}{\mathrm{20}}+\frac{{y}}{\mathrm{15}}=\mathrm{1} \\ $$$${G}\left(\frac{\mathrm{20}}{\mathrm{3}},\:\frac{\mathrm{15}}{\mathrm{3}}\right) \\ $$$${r}=\frac{\mid\frac{\mathrm{1}}{\mathrm{20}}×\frac{\mathrm{20}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{15}}×\frac{\mathrm{15}}{\mathrm{3}}−\mathrm{1}\mid}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{20}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{15}^{\mathrm{2}} }}}=\mathrm{4} \\ $$$${A}_{{semicircle}} =\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{8}\pi\:\checkmark \\ $$
Answered by Spillover last updated on 18/Apr/25
Answered by Spillover last updated on 18/Apr/25
Answered by golsendro last updated on 19/Apr/25
$$\:\mathrm{AC}\:=\sqrt{\mathrm{400}+\mathrm{225}}\:=\:\mathrm{25}\: \\ $$$$\:\:\mathrm{OA}\:=\:\frac{\mathrm{20}×\mathrm{15}}{\mathrm{25}}\:=\:\mathrm{12}\: \\ $$$$\:\:\mathrm{OG}\:=\:\mathrm{r}\:=\:\frac{\mathrm{1}}{\mathrm{3}}.\mathrm{OA}=\:\mathrm{4} \\ $$$$\:\:\mathrm{A}\left(\mathrm{semicircle}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\pi.\mathrm{16}=\:\mathrm{8}\pi \\ $$
Commented by Spillover last updated on 19/Apr/25
$${thank}\:{you} \\ $$