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Question Number 5695 by FilupSmith last updated on 24/May/16

Commented by FilupSmith last updated on 24/May/16

$$\mathrm{What}\mathrm{kinds}\mathrm{of}\mathrm{methods}\mathrm{can}\mathrm{be}\mathrm{used}$$$$\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{area}\mathrm{of}ABC?$$$$\left(\mathrm{Both}\mathrm{circles}\mathrm{are}\mathrm{identical}\right)$$

Commented by FilupSmith last updated on 24/May/16

$$\mathrm{The}\mathrm{only}\mathrm{method}\mathrm{I}\mathrm{am}\mathrm{aware}\mathrm{of}\mathrm{is}\mathrm{subtracting}$$$$\mathrm{the}\mathrm{two}\mathrm{quater}\mathrm{circle}\mathrm{quadrants}\mathrm{from}\mathrm{the}$$$$\mathrm{rectancle}\mathrm{made}\mathrm{by}A,C,\mathrm{and}\mathrm{the}\mathrm{two}$$$$\mathrm{circle}\mathrm{origins}.$$$$$$$$\mathrm{What}\mathrm{other}\mathrm{methods}\mathrm{are}\mathrm{there}?$$

Commented by Rasheed Soomro last updated on 25/May/16

$$\mathrm{The}\mathrm{method}\mathrm{you}\mathrm{mentioned}\mathrm{in}\mathrm{the}$$$$\mathrm{question}\mathrm{is}\mathrm{the}\mathrm{easiest}\mathrm{one}.$$$$\mathrm{The}\mathrm{other}\mathrm{method}\mathrm{although}\mathrm{lengthy}$$$$\mathrm{and}\mathrm{not}\mathrm{preferable}\mathrm{may}\mathrm{be}:$$$$Assigningequationstocircles$$$$analyticallyandthenusingintegration.$$$$Inthiscaseparticularlyequationof$$$$onecircleisneededonly.$$

Commented by Rasheed Soomro last updated on 25/May/16

Commented by Rasheed Soomro last updated on 25/May/16

$$\mathrm{Area}\mathrm{enclosed}\mathrm{by}\mathrm{circle}{(\mathrm{x}-\mathrm{r})}^2+{(\mathrm{y}-\mathrm{r})}^2={\mathrm{r}}^2,$$$$\mathrm{x}-\mathrm{axis}\mathrm{and}\mathrm{y}-\mathrm{axis}\mathrm{can}\mathrm{be}\mathrm{determined}\mathrm{by}$$$$\mathrm{definite}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{circle}.$$$$\mathrm{Area}\mathrm{required}\mathrm{in}\mathrm{the}\mathrm{question}\mathrm{is}\mathrm{double}\mathrm{of}$$$$\mathrm{this}.$$$$$$

Answered by Rasheed Soomro last updated on 24/May/16

$$\mathrm{Drop}\mathrm{perpendicular}\mathrm{from}\mathrm{B}$$$$\mathrm{on}\mathrm{AC},\mathrm{say}\mathrm{it}\mathrm{meets}\mathrm{AC}\mathrm{at}\mathrm{D}.$$$$\mathrm{BD}\mathrm{is}\mathrm{altitude}\mathrm{of}△\mathrm{ABC}.$$$$Base\mathrm{AC}=2\mathrm{r},Altitude\mathrm{BD}=\mathrm{r}$$$$▴\mathrm{ABC}=\frac{1}{2}\times Base\times Altitude$$$$=\frac{1}{2}\times 2\mathrm{r}\times \mathrm{r}={\mathrm{r}}^2$$$$\mathrm{Area}\mathrm{enclosed}\mathrm{by}\stackrel{⌢}{\mathrm{AB}},\stackrel{⌢}{\mathrm{BC}}\mathrm{\&}\mathrm{Line}\mathrm{AC}$$$$=▴\mathrm{ABC}-\mathrm{SegmentAB}-\mathrm{SegmentAC}$$$$\mathrm{AreaSegmentAB}=\mathrm{AreaSectorAB}-▴\mathrm{AOB}\left[\mathrm{O}\mathrm{is}\mathrm{centre}\mathrm{of}\mathrm{circle}\right]$$$$=\frac{1}{2}{\mathrm{r}}^2\left(\frac{\pi }{2}\right)-\frac{1}{2}{\mathrm{r}}^2=\frac{1}{2}{\mathrm{r}}^2(\frac{\pi }{2}-1)$$$$\mathrm{AreaSegmentBC}=\mathrm{AreaSegmentAB}=\frac{1}{2}{\mathrm{r}}^2(\frac{\pi }{2}-1)$$$$\mathrm{Area}\mathrm{enclosed}\mathrm{by}\stackrel{⌢}{\mathrm{AB}},\stackrel{⌢}{\mathrm{BC}}\mathrm{\&}\mathrm{Line}\mathrm{AC}$$$$={\mathrm{r}}^2-2\times \frac{1}{2}{\mathrm{r}}^2(\frac{\pi }{2}-1)$$$$={\mathrm{r}}^2(1-\frac{\pi }{2}+1)={\mathrm{r}}^2(2-\frac{\pi }{2})$$

Commented by FilupSmith last updated on 24/May/16

$$\mathrm{But}\mathrm{would}\mathrm{ylu}\mathrm{not}\mathrm{have}\mathrm{the}\mathrm{hypotanuse}$$$$\mathrm{cutting}\mathrm{the}\mathrm{circle}?$$

Commented by Rasheed Soomro last updated on 24/May/16

$$\mathrm{Do}\mathrm{you}\mathrm{want}\mathrm{to}\mathrm{inquire}\mathrm{the}\mathrm{area}\mathrm{enclosed}$$$$\mathrm{between}\mathrm{arcs}\mathrm{AB},\mathrm{BC}\mathrm{and}\mathrm{line}\mathrm{AC}?$$$$\mathrm{Sorry}\mathrm{that}\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{calculate}\mathrm{that}.$$$$\mathrm{I}\mathrm{determined}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{ABC}.$$$$\mathrm{Here}\mathrm{is}\mathrm{way}\mathrm{to}\mathrm{calculate}\mathrm{what}\mathrm{you}$$$$\mathrm{want}.$$$$\mathrm{Area}\mathrm{enclosed}\mathrm{by}\stackrel{⌢}{\mathrm{AB}},\stackrel{⌢}{\mathrm{BC}}\mathrm{\&}\mathrm{Line}\stackrel{}{\mathrm{AC}}$$$$=▴\mathrm{ABC}-\mathrm{SegmentAB}-\mathrm{SegmentAC}$$

Commented by FilupSmith last updated on 24/May/16

$$\mathrm{Sorry},\mathrm{I}\mathrm{meant}\mathrm{arcs}:\mathrm{p}$$

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