Question-178737

Question Number 178737 by cherokeesay last updated on 21/Oct/22

Answered by Rasheed.Sindhi last updated on 21/Oct/22

Commented by Rasheed.Sindhi last updated on 21/Oct/22

Let AC=BC=r blue-area=(1/4)πr^2 R=((AB)/2)=((√(r^2 +r^2 ))/2)=(((√2) r)/2) outer circle=πR^2 =π((((√2) r)/2))^2 yello-area=outer circle−blue-area =π((((√2) r)/2))^2 −(1/4)πr^2 =((2πr^2 )/4)−((πr^2 )/4)=((πr^2 )/4) =blue-area

$$\mathrm{Let}\:\mathrm{AC}=\mathrm{BC}=\mathrm{r} \\ $$$$\mathrm{blue}-\mathrm{area}=\frac{\mathrm{1}}{\mathrm{4}}\pi\mathrm{r}^{\mathrm{2}} \\ $$$$\mathrm{R}=\frac{\mathrm{AB}}{\mathrm{2}}=\frac{\sqrt{\mathrm{r}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}\:\mathrm{r}}{\mathrm{2}} \\ $$$$\mathrm{outer}\:\mathrm{circle}=\pi\mathrm{R}^{\mathrm{2}} =\pi\left(\frac{\sqrt{\mathrm{2}}\:\mathrm{r}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{yello}-\mathrm{area}=\mathrm{outer}\:\mathrm{circle}−\mathrm{blue}-\mathrm{area} \\ $$$$=\pi\left(\frac{\sqrt{\mathrm{2}}\:\mathrm{r}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\pi\mathrm{r}^{\mathrm{2}} =\frac{\mathrm{2}\pi\mathrm{r}^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi\mathrm{r}^{\mathrm{2}} }{\mathrm{4}}=\frac{\pi\mathrm{r}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:=\mathrm{blue}-\mathrm{area} \\ $$

Commented by cherokeesay last updated on 21/Oct/22

$${Greet}\:!\:{thank}\:{you} \\ $$

Commented by Tawa11 last updated on 21/Oct/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by HeferH last updated on 21/Oct/22

Blue area: ((π(r(√2))^2 )/4) = ((πr^2 )/2) Yellow area: πr^2 −blue= ((πr^2 )/2) then yellow = blue

$${Blue}\:{area}: \\ $$$$\:\frac{\pi\left({r}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{4}}\:=\:\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}\: \\ $$$$\:{Yellow}\:{area}: \\ $$$$\:\pi{r}^{\mathrm{2}} \:−{blue}=\:\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:{then}\:{yellow}\:=\:{blue} \\ $$

Commented by cherokeesay last updated on 21/Oct/22

$${Nice}\:!\:{thank}\:{you}\:! \\ $$

Commented by Tawa11 last updated on 21/Oct/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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