Question-217802

Question Number 217802 by Ismoiljon_008 last updated on 21/Mar/25

Commented by Ismoiljon_008 last updated on 21/Mar/25

$$\:\:\:{a}+{b}\:=\:{c}\:\:\:{Then}\:\:{R}\:=\:? \\ $$$$\:\:\:{Help}\:{me},\:{please}\: \\ $$$$ \\ $$

Commented by MathematicalUser2357 last updated on 26/Mar/25

I just tried to get the intersection points where the center of the circle is (p,q) but got x=((−4(4p+3q−18)±4(√((4p+3q−18)^2 −25((q−6)^2 −r^2 +p^2 ))))/(25)) We have to solve for c like this: c=∫_((−4(4p+3q−18)−4(√((4p+3q−18)^2 −25((q−6)^2 −r^2 +p^2 ))))/(25)) ^((−4(4p+3q−18)+4(√((4p+3q−18)^2 −25((q−6)^2 −r^2 +p^2 ))))/(25)) ((√(r^2 −(x−p)^2 +))(3/4)x−6)dx But I have no idea to solve this

$$\mathrm{I}\:\mathrm{just}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{points}\:\mathrm{where}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{is}\:\left({p},{q}\right)\:\mathrm{but}\:\mathrm{got} \\ $$$${x}=\frac{−\mathrm{4}\left(\mathrm{4}{p}+\mathrm{3}{q}−\mathrm{18}\right)\pm\mathrm{4}\sqrt{\left(\mathrm{4}{p}+\mathrm{3}{q}−\mathrm{18}\right)^{\mathrm{2}} −\mathrm{25}\left(\left({q}−\mathrm{6}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)}}{\mathrm{25}} \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{for}\:{c}\:\mathrm{like}\:\mathrm{this}: \\ $$$${c}=\int_{\frac{−\mathrm{4}\left(\mathrm{4}{p}+\mathrm{3}{q}−\mathrm{18}\right)−\mathrm{4}\sqrt{\left(\mathrm{4}{p}+\mathrm{3}{q}−\mathrm{18}\right)^{\mathrm{2}} −\mathrm{25}\left(\left({q}−\mathrm{6}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)}}{\mathrm{25}}} ^{\frac{−\mathrm{4}\left(\mathrm{4}{p}+\mathrm{3}{q}−\mathrm{18}\right)+\mathrm{4}\sqrt{\left(\mathrm{4}{p}+\mathrm{3}{q}−\mathrm{18}\right)^{\mathrm{2}} −\mathrm{25}\left(\left({q}−\mathrm{6}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)}}{\mathrm{25}}} \left(\sqrt{{r}^{\mathrm{2}} −\left({x}−{p}\right)^{\mathrm{2}} +}\frac{\mathrm{3}}{\mathrm{4}}{x}−\mathrm{6}\right){dx} \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{idea}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this} \\ $$

Answered by mr W last updated on 21/Mar/25

Commented by mr W last updated on 21/Mar/25

a+b+d=triangle=((6×8)/2) c+d=(3/4) of circle + square R×R=((3πR^2 )/4)+R^2 since a+b=c, ((3πR^2 )/4)+R^2 =((6×8)/2) ⇒R=(√((96)/(3π+4))) ✓

$${a}+{b}+{d}={triangle}=\frac{\mathrm{6}×\mathrm{8}}{\mathrm{2}} \\ $$$${c}+{d}=\frac{\mathrm{3}}{\mathrm{4}}\:{of}\:{circle}\:+\:{square}\:{R}×{R}=\frac{\mathrm{3}\pi{R}^{\mathrm{2}} }{\mathrm{4}}+{R}^{\mathrm{2}} \\ $$$${since}\:{a}+{b}={c}, \\ $$$$\frac{\mathrm{3}\pi{R}^{\mathrm{2}} }{\mathrm{4}}+{R}^{\mathrm{2}} =\frac{\mathrm{6}×\mathrm{8}}{\mathrm{2}} \\ $$$$\Rightarrow{R}=\sqrt{\frac{\mathrm{96}}{\mathrm{3}\pi+\mathrm{4}}}\:\checkmark \\ $$

Commented by Ismoiljon_008 last updated on 21/Mar/25

$$\:\:\:{Thank}\:{you}\:{very}\:{much} \\ $$

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