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Question-5695

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Question Number 5695 by FilupSmith last updated on 24/May/16
Commented by FilupSmith last updated on 24/May/16
What kinds of methods can be used to find the area of ABC? (Both circles are identical)
$$\mathrm{What}\:\mathrm{kinds}\:\mathrm{of}\:\mathrm{methods}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:{ABC}? \\ $$$$\left(\mathrm{Both}\:\mathrm{circles}\:\mathrm{are}\:\mathrm{identical}\right) \\ $$
Commented by FilupSmith last updated on 24/May/16
The only method I am aware of is subtracting the two quater circle quadrants from the rectancle made by A, C, and the two circle origins. What other methods are there?
$$\mathrm{The}\:\mathrm{only}\:\mathrm{method}\:\mathrm{I}\:\mathrm{am}\:\mathrm{aware}\:\mathrm{of}\:\mathrm{is}\:\mathrm{subtracting} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{quater}\:\mathrm{circle}\:\mathrm{quadrants}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{rectancle}\:\mathrm{made}\:\mathrm{by}\:{A},\:{C},\:\mathrm{and}\:\mathrm{the}\:\mathrm{two} \\ $$$$\mathrm{circle}\:\mathrm{origins}. \\ $$$$ \\ $$$$\mathrm{What}\:\mathrm{other}\:\mathrm{methods}\:\mathrm{are}\:\mathrm{there}? \\ $$
Commented by Rasheed Soomro last updated on 25/May/16
The method you mentioned in the question is the easiest one. The other method although lengthy and not preferable may be: Assigning equations to circles analytically and then using integration. In this case particularly equation of one circle is needed only.
$$\mathrm{The}\:\mathrm{method}\:\mathrm{you}\:\mathrm{mentioned}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{question}\:\mathrm{is}\:\mathrm{the}\:\mathrm{easiest}\:\mathrm{one}. \\ $$$$\mathrm{The}\:\mathrm{other}\:\mathrm{method}\:\mathrm{although}\:\mathrm{lengthy} \\ $$$$\mathrm{and}\:\mathrm{not}\:\mathrm{preferable}\:\mathrm{may}\:\mathrm{be}: \\ $$$${Assigning}\:{equations}\:{to}\:{circles} \\ $$$${analytically}\:{and}\:{then}\:{using}\:{integration}. \\ $$$${In}\:{this}\:{case}\:{particularly}\:{equation}\:{of} \\ $$$${one}\:{circle}\:{is}\:{needed}\:{only}. \\ $$
Commented by Rasheed Soomro last updated on 25/May/16
Commented by Rasheed Soomro last updated on 25/May/16
Area enclosed by circle (x−r)^2 +(y−r)^2 =r^2 , x-axis and y-axis can be determined by definite integral of the equation of circle. Area required in the question is double of this.
$$\mathrm{Area}\:\mathrm{enclosed}\:\mathrm{by}\:\mathrm{circle}\:\left(\mathrm{x}−\mathrm{r}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{r}\right)^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} \:, \\ $$$$\mathrm{x}-\mathrm{axis}\:\mathrm{and}\:\mathrm{y}-\mathrm{axis}\:\mathrm{can}\:\mathrm{be}\:\mathrm{determined}\:\mathrm{by} \\ $$$$\mathrm{definite}\:\mathrm{integral}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{circle}. \\ $$$$\mathrm{Area}\:\mathrm{required}\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{double}\:\mathrm{of} \\ $$$$\mathrm{this}. \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 24/May/16
Drop perpendicular from B on AC, say it meets AC at D. BD is altitude of △ABC. Base AC=2r , Altitude BD=r ▲ABC=(1/2)×Base×Altitude =(1/2)×2r×r=r^2 Area enclosed by AB^(⌢) , BC^(⌢) & Line AC =▲ABC−SegmentAB−SegmentAC AreaSegmentAB=AreaSectorAB−▲AOB [O is centre of circle] =(1/2)r^2 ((π/2))−(1/2)r^2 =(1/2)r^2 ((π/2)−1) AreaSegmentBC=AreaSegmentAB=(1/2)r^2 ((π/2)−1) Area enclosed by AB^(⌢) , BC^(⌢) & Line AC =r^2 −2×(1/2)r^2 ((π/2)−1) =r^2 (1−(π/2)+1)=r^2 (2−(π/2))
$$\mathrm{Drop}\:\mathrm{perpendicular}\:\mathrm{from}\:\mathrm{B} \\ $$$$\mathrm{on}\:\mathrm{AC},\:\mathrm{say}\:\mathrm{it}\:\mathrm{meets}\:\mathrm{AC}\:\mathrm{at}\:\mathrm{D}. \\ $$$$\mathrm{BD}\:\mathrm{is}\:\mathrm{altitude}\:\mathrm{of}\:\bigtriangleup\mathrm{ABC}. \\ $$$${Base}\:\mathrm{AC}=\mathrm{2r}\:,\:{Altitude}\:\mathrm{BD}=\mathrm{r} \\ $$$$\blacktriangle\mathrm{ABC}=\frac{\mathrm{1}}{\mathrm{2}}×{Base}×{Altitude} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2r}×\mathrm{r}=\mathrm{r}^{\mathrm{2}} \\ $$$$\mathrm{Area}\:\mathrm{enclosed}\:\mathrm{by}\:\overset{\frown} {\mathrm{AB}}\:,\:\overset{\frown} {\mathrm{BC}}\:\&\:\mathrm{Line}\:\mathrm{AC} \\ $$$$\:\:\:\:\:=\blacktriangle\mathrm{ABC}−\mathrm{SegmentAB}−\mathrm{SegmentAC} \\ $$$$\mathrm{AreaSegmentAB}=\mathrm{AreaSectorAB}−\blacktriangle\mathrm{AOB}\:\left[\mathrm{O}\:\mathrm{is}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{circle}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{r}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{r}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{r}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\mathrm{AreaSegmentBC}=\mathrm{AreaSegmentAB}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{r}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\mathrm{Area}\:\mathrm{enclosed}\:\mathrm{by}\:\overset{\frown} {\mathrm{AB}}\:,\:\overset{\frown} {\mathrm{BC}}\:\&\:\mathrm{Line}\:\mathrm{AC} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{r}^{\mathrm{2}} −\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\mathrm{r}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right)\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{r}^{\mathrm{2}} \left(\mathrm{1}−\frac{\pi}{\mathrm{2}}+\mathrm{1}\right)=\mathrm{r}^{\mathrm{2}} \left(\mathrm{2}−\frac{\pi}{\mathrm{2}}\right) \\ $$
Commented by FilupSmith last updated on 24/May/16
But would ylu not have the hypotanuse cutting the circle?
$$\mathrm{But}\:\mathrm{would}\:\mathrm{ylu}\:\mathrm{not}\:\mathrm{have}\:\mathrm{the}\:\mathrm{hypotanuse} \\ $$$$\mathrm{cutting}\:\mathrm{the}\:\mathrm{circle}? \\ $$
Commented by Rasheed Soomro last updated on 24/May/16
Do you want to inquire the area enclosed between arcs AB ,BC and line AC? Sorry that I didn′t calculate that. I determined area of triangle ABC. Here is way to calculate what you want. Area enclosed by AB^(⌢) , BC^(⌢) & Line AC^() =▲ABC−SegmentAB−SegmentAC
$$\mathrm{Do}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{inquire}\:\mathrm{the}\:\mathrm{area}\:\:\mathrm{enclosed} \\ $$$$\mathrm{between}\:\mathrm{arcs}\:\:\mathrm{AB}\:,\mathrm{BC}\:\mathrm{and}\:\mathrm{line}\:\mathrm{AC}? \\ $$$$\mathrm{Sorry}\:\mathrm{that}\:\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{calculate}\:\mathrm{that}. \\ $$$$\mathrm{I}\:\mathrm{determined}\:\:\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{ABC}. \\ $$$$\:\:\:\:\:\:\mathrm{Here}\:\mathrm{is}\:\mathrm{way}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{what}\:\mathrm{you} \\ $$$$\mathrm{want}. \\ $$$$\mathrm{Area}\:\mathrm{enclosed}\:\mathrm{by}\:\overset{\frown} {\mathrm{AB}}\:,\:\overset{\frown} {\mathrm{BC}}\:\&\:\mathrm{Line}\:\overset{} {\mathrm{AC}} \\ $$$$\:\:\:=\blacktriangle\mathrm{ABC}−\mathrm{SegmentAB}−\mathrm{SegmentAC} \\ $$
Commented by FilupSmith last updated on 24/May/16
Sorry, I meant arcs :p
$$\mathrm{Sorry},\:\mathrm{I}\:\mathrm{meant}\:\mathrm{arcs}\::\mathrm{p} \\ $$

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