Question-220656

Question Number 220656 by Tawa11 last updated on 17/May/25

Commented by Tawa11 last updated on 17/May/25

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{shaded}. \\ $$

Answered by Ghisom last updated on 17/May/25

the big circle has radius r y=(√(r^2 −x^2 )) with x=−2 we get y=(√(r^2 −4)) which is 2 times the radius of the small circle we now have big quarter circle minus small circle: ((πr^2 )/4)−π(((√(r^2 −4))/2))^2 =((πr^2 )/4)−((π(r^2 −4))/4)=π

$$\mathrm{the}\:\mathrm{big}\:\mathrm{circle}\:\mathrm{has}\:\mathrm{radius}\:{r} \\ $$$${y}=\sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$\mathrm{with}\:{x}=−\mathrm{2}\:\mathrm{we}\:\mathrm{get}\:{y}=\sqrt{{r}^{\mathrm{2}} −\mathrm{4}}\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{2}\:\mathrm{times}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{small}\:\mathrm{circle} \\ $$$$\mathrm{we}\:\mathrm{now}\:\mathrm{have}\:\mathrm{big}\:\mathrm{quarter}\:\mathrm{circle}\:\mathrm{minus} \\ $$$$\mathrm{small}\:\mathrm{circle}: \\ $$$$\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}}−\pi\left(\frac{\sqrt{{r}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi\left({r}^{\mathrm{2}} −\mathrm{4}\right)}{\mathrm{4}}=\pi \\ $$

Commented by Tawa11 last updated on 17/May/25

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Answered by mr W last updated on 17/May/25

R=radius of big circle r=radius of small circle (R+2r)(R−2r)=2^2 R^2 −4r^2 =2^2 shaded area=((πR^2 )/4)−πr^2 =(π/4)(R^2 −4r^2 )=π

$${R}={radius}\:{of}\:{big}\:{circle} \\ $$$${r}={radius}\:{of}\:{small}\:{circle} \\ $$$$\left({R}+\mathrm{2}{r}\right)\left({R}−\mathrm{2}{r}\right)=\mathrm{2}^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} \\ $$$${shaded}\:{area}=\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}−\pi{r}^{\mathrm{2}} =\frac{\pi}{\mathrm{4}}\left({R}^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{2}} \right)=\pi \\ $$

Commented by mr W last updated on 17/May/25

it makes no sense to assume 45°. why not 40°, 48° or any other angle? if you assume 45°, then the small circle should be exactly of the same size as the square.

$${it}\:{makes}\:{no}\:{sense}\:{to}\:{assume}\:\mathrm{45}°. \\ $$$${why}\:{not}\:\mathrm{40}°,\:\mathrm{48}°\:{or}\:{any}\:{other}\:{angle}? \\ $$$${if}\:{you}\:{assume}\:\mathrm{45}°,\:{then}\:{the}\:{small} \\ $$$${circle}\:{should}\:{be}\:{exactly}\:{of}\:{the}\:{same} \\ $$$${size}\:{as}\:{the}\:{square}. \\ $$

Commented by Tawa11 last updated on 17/May/25

Please draw lines sir. I appreciate sir.

$$\mathrm{Please}\:\mathrm{draw}\:\mathrm{lines}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 17/May/25

Commented by mr W last updated on 17/May/25

DC=R+2r CB=R−2r ((AC)/(CB))=((DC)/(AC)) ⇒DC×CB=AC^2 ⇒(R+2r)(R−2r)=2^2

$${DC}={R}+\mathrm{2}{r} \\ $$$${CB}={R}−\mathrm{2}{r} \\ $$$$\frac{{AC}}{{CB}}=\frac{{DC}}{{AC}} \\ $$$$\Rightarrow{DC}×{CB}={AC}^{\mathrm{2}} \\ $$$$\Rightarrow\left({R}+\mathrm{2}{r}\right)\left({R}−\mathrm{2}{r}\right)=\mathrm{2}^{\mathrm{2}} \\ $$

Commented by fantastic last updated on 17/May/25

$${I}\:{though}\:{like}\:{this}\:{at}\:{first} \\ $$

Commented by Tawa11 last updated on 17/May/25

$$\mathrm{Wow},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 17/May/25

Is the information sufficient to find R and r separately?

$$\mathrm{Is}\:\mathrm{the}\:\mathrm{information}\:\mathrm{sufficient}\:\mathrm{to}\:\mathrm{find} \\ $$$$\mathrm{R}\:\mathrm{and}\:\mathrm{r}\:\:\mathrm{separately}? \\ $$

Commented by Tawa11 last updated on 17/May/25

How can I set up an equation to find R and r separately sir.

$$\mathrm{How}\:\mathrm{can}\:\mathrm{I}\:\mathrm{set}\:\mathrm{up}\:\mathrm{an}\:\mathrm{equation}\:\mathrm{to}\:\mathrm{find} \\ $$$$\mathrm{R}\:\mathrm{and}\:\mathrm{r}\:\:\mathrm{separately}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 17/May/25

$$\mathrm{Sir}, \\ $$$$\mathrm{Is}\:\:\mathrm{AC}\:\:=\:\:\mathrm{OC}\:? \\ $$

Commented by mr W last updated on 17/May/25

$${no},\:{that}\:{is}\:{not}\:{given}!\:{it}\:{doesn}'{t}\:{look} \\ $$$${so}\:{either}.\:{the}\:{small}\:{circle}\:{doesn}'{t} \\ $$$${tangent}\:{OC}! \\ $$

Commented by Tawa11 last updated on 17/May/25

ohh. I assumed 45° like tan 45 = (2/(∣OC∣)) ∴ ∣OC∣ = 2 ∣OC∣ = 2r, 2 = 2r r = 1

$$\mathrm{ohh}. \\ $$$$\mathrm{I}\:\mathrm{assumed}\:\mathrm{45}° \\ $$$$\mathrm{like}\:\:\:\:\mathrm{tan}\:\mathrm{45}\:\:=\:\:\frac{\mathrm{2}}{\mid\mathrm{OC}\mid} \\ $$$$\therefore\:\:\:\:\:\mid\mathrm{OC}\mid\:\:=\:\:\mathrm{2} \\ $$$$\mid\mathrm{OC}\mid\:\:=\:\:\mathrm{2r}, \\ $$$$\:\:\:\:\mathrm{2}\:\:=\:\:\mathrm{2r} \\ $$$$\:\:\:\:\:\:\mathrm{r}\:\:=\:\:\mathrm{1} \\ $$

Commented by Tawa11 last updated on 17/May/25

Commented by Tawa11 last updated on 17/May/25

Commented by Tawa11 last updated on 17/May/25

$$\mathrm{Am}\:\mathrm{I}\:\mathrm{right}\:\mathrm{sir}? \\ $$

Commented by Tawa11 last updated on 17/May/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 17/May/25

$${following}\:{two}\:{figures}\:{fulfill}\:{also} \\ $$$${the}\:{condition}: \\ $$

Commented by Tawa11 last updated on 17/May/25

I got it, thanks sir. R = 2(√2) and r = 1

$$\mathrm{I}\:\mathrm{got}\:\mathrm{it},\:\mathrm{thanks}\:\mathrm{sir}. \\ $$$$ \\ $$$$\mathrm{R}\:\:=\:\:\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:\:\:\mathrm{and}\:\:\:\:\:\mathrm{r}\:\:=\:\:\mathrm{1} \\ $$

Commented by mr W last updated on 17/May/25

what do you think? can you determine x and y, when it is only known that y^2 −x^2 =4?

$${what}\:{do}\:{you}\:{think}?\:{can}\:{you}\:{determine} \\ $$$${x}\:{and}\:{y},\:{when}\:{it}\:{is}\:{only}\:{known}\:{that}\: \\ $$$${y}^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{4}? \\ $$

Commented by mr W last updated on 18/May/25

Commented by Tawa11 last updated on 17/May/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Commented by mr W last updated on 18/May/25

$${in}\:{both}\:{cases}\:{the}\:{circles}\:{are}\:{very} \\ $$$${different},\:{but}\:{the}\:{shaded}\:{areas}\:{are} \\ $$$${equal}. \\ $$

Answered by fantastic last updated on 17/May/25

Commented by fantastic last updated on 17/May/25

Let the radius of the bigger circle be R AO=R ,AB=2(I forgot to mention the point) ∴OB=(√((AO)^2 −(AB)^2 =))(√(R^2 −2^2 ))=(√(R^2 −4)) So P Q=(√(R^2 −4)) Area of the smaller circle=π(((√(R^2 −4))/2))^2 ...(i) Area of the quarter circle=(1/4)π(R)^2 ...(ii) Area of the shaded region=(ii)−(i) (1/4)πR^2 −π((R^2 −4)/4)=(1/4)π(R^2 −R^2 +4)=(1/4)π×4=π✓

$${Let}\:{the}\:{radius}\:{of}\:{the}\:{bigger}\:{circle}\:{be}\:{R} \\ $$$${AO}={R}\:,{AB}=\mathrm{2}\left({I}\:{forgot}\:{to}\:{mention}\:{the}\:{point}\right) \\ $$$$\therefore{OB}=\sqrt{\left({AO}\right)^{\mathrm{2}} −\left({AB}\right)^{\mathrm{2}} =}\sqrt{{R}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }=\sqrt{{R}^{\mathrm{2}} −\mathrm{4}} \\ $$$${So}\:{P}\:{Q}=\sqrt{{R}^{\mathrm{2}} −\mathrm{4}} \\ $$$${Area}\:{of}\:{the}\:{smaller}\:{circle}=\pi\left(\frac{\sqrt{{R}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\right)^{\mathrm{2}} …\left({i}\right) \\ $$$${Area}\:{of}\:{the}\:{quarter}\:{circle}=\frac{\mathrm{1}}{\mathrm{4}}\pi\left({R}\right)^{\mathrm{2}} …\left({ii}\right) \\ $$$${Area}\:{of}\:{the}\:{shaded}\:{region}=\left({ii}\right)−\left({i}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\pi{R}^{\mathrm{2}} −\pi\frac{{R}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}\pi\left({R}^{\mathrm{2}} −{R}^{\mathrm{2}} +\mathrm{4}\right)=\frac{\mathrm{1}}{\mathrm{4}}\pi×\mathrm{4}=\pi\checkmark \\ $$

Commented by Tawa11 last updated on 17/May/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate} \\ $$

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