Question-226651

Question Number 226651 by Hanuda354 last updated on 08/Dec/25

Commented by Hanuda354 last updated on 08/Dec/25

Find the shaded area (without using integration).

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}\:\left(\mathrm{without}\:\mathrm{using}\:\mathrm{integration}\right). \\ $$

Answered by fantastic2 last updated on 10/Dec/25

Commented by fantastic2 last updated on 10/Dec/25

square side S S=4+R S=7+2b b=R−7 ⇒4+R=7+2R−14 ⇒R=11 ⇒r=2b=2(R−7)=8 ⇒S=4+R=15 tan φ=((15)/7) equation of the line y=mx+c=−((15)/7)x+15 ⇒y=15(1−(x/7)) x^2 +y^2 =11^2 x^2 +(15(1−(x/7)))^2 =11^2 x^2 +225−((450x)/7)+((225x^2 )/(49))=121 ⇒((274x^2 )/(49))−((450x)/7)+104=0 x=((1575)/(274))±((7(√(22129)))/(274)) ≈1.947[if−is taken] ≈9.458[if + is taken] y=((15(√(22129)))/(274))±((735)/(274))[+ve if x=1.947,−ve if x=9.458] in our case y>0 so −ve rejected so x≈1.947 y≈10.826 cos θ((√(11^2 −x^2 ))/(11)) α=90^0 −θ area of the teiangle containing α △=(1/2)×7×11sin (90^0 −θ) =((77)/2) ×((√(11^2 −x^2 ))/(11))=((7(√(11^2 −x^2 )))/2) area of the sector A=((90^0 −θ)/(360))π11^2 Blue area(B)+area of the intersection of two circles(C+C)=A−△ B+(C+C)=((90^0 −θ)/(360))π11^2 −((7(√(11^2 −x^2 )))/2) x^2 +y^z =11^2 & (x−15)^2 +y^2 =8^2 ⇒(2x−15)15=57 ⇒x=(((57)/(15))+15)/2 y=(√(11^2 −x^2 ))≈5.713 area of the tirangle(in the two circles) =(1/2)×b×h=(1/2)×4×y green area=(β/(360))πR^2 −(1/2)×121×sin β β=cos^(−1) (((11^2 +15^2 −8^2 )/(2×11×15)))=cos^(−1) (((47)/(55))) green area≈1.618 yellow area=(γ/(360))πr^2 −(1/2)×64×sin γ γ=cos^(−1) (((8^2 +15^2 −11^2 )/(2×8×15)))=cos^(−1) ((7/(10))) yellow area≈2.600 ∴B=(A−△)−(triangle area+green area+yellow area) B=((90^0 −cos^(−1) (((√(11^2 −(((1575)/(274))−((7(√(22129)))/(274)))^2 ))/(11))))/(360^0 ))π11^2 −((7 (√(11^2 −(((1575)/(274))−((7(√(22129)))/(274)))^2 )))/2)−(2(√(11^2 −((((((57)/(15))+15))/2))^2 ))+((cos^(−1) (((47)/(55))))/(360))π11^2 −(1/2)×121×sin(cos^(−1) ((47)/(55)) )+((cos^(−1) 0.7)/(360))π8^2 −(1/2)×64×sin (cos^(−1) 0.7)) B≈30.7274274166

$${square}\:{side}\:{S} \\ $$$${S}=\mathrm{4}+{R} \\ $$$${S}=\mathrm{7}+\mathrm{2}{b} \\ $$$${b}={R}−\mathrm{7} \\ $$$$\Rightarrow\mathrm{4}+{R}=\mathrm{7}+\mathrm{2}{R}−\mathrm{14} \\ $$$$\Rightarrow{R}=\mathrm{11} \\ $$$$\Rightarrow{r}=\mathrm{2}{b}=\mathrm{2}\left({R}−\mathrm{7}\right)=\mathrm{8} \\ $$$$\Rightarrow{S}=\mathrm{4}+{R}=\mathrm{15} \\ $$$$\mathrm{tan}\:\phi=\frac{\mathrm{15}}{\mathrm{7}} \\ $$$${equation}\:{of}\:{the}\:{line} \\ $$$${y}={mx}+{c}=−\frac{\mathrm{15}}{\mathrm{7}}{x}+\mathrm{15} \\ $$$$\Rightarrow{y}=\mathrm{15}\left(\mathrm{1}−\frac{{x}}{\mathrm{7}}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{11}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\left(\mathrm{15}\left(\mathrm{1}−\frac{{x}}{\mathrm{7}}\right)\right)^{\mathrm{2}} =\mathrm{11}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\mathrm{225}−\frac{\mathrm{450}{x}}{\mathrm{7}}+\frac{\mathrm{225}{x}^{\mathrm{2}} }{\mathrm{49}}=\mathrm{121} \\ $$$$\Rightarrow\frac{\mathrm{274}{x}^{\mathrm{2}} }{\mathrm{49}}−\frac{\mathrm{450}{x}}{\mathrm{7}}+\mathrm{104}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1575}}{\mathrm{274}}\pm\frac{\mathrm{7}\sqrt{\mathrm{22129}}}{\mathrm{274}} \\ $$$$\approx\mathrm{1}.\mathrm{947}\left[{if}−{is}\:{taken}\right] \\ $$$$\approx\mathrm{9}.\mathrm{458}\left[{if}\:+\:{is}\:{taken}\right] \\ $$$${y}=\frac{\mathrm{15}\sqrt{\mathrm{22129}}}{\mathrm{274}}\pm\frac{\mathrm{735}}{\mathrm{274}}\left[+{ve}\:{if}\:{x}=\mathrm{1}.\mathrm{947},−{ve}\:{if}\:{x}=\mathrm{9}.\mathrm{458}\right] \\ $$$${in}\:{our}\:{case}\:{y}>\mathrm{0}\:{so}\:−{ve}\:{rejected} \\ $$$${so}\: \\ $$$${x}\approx\mathrm{1}.\mathrm{947} \\ $$$${y}\approx\mathrm{10}.\mathrm{826} \\ $$$$\mathrm{cos}\:\theta\frac{\sqrt{\mathrm{11}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{\mathrm{11}} \\ $$$$\alpha=\mathrm{90}^{\mathrm{0}} −\theta \\ $$$${area}\:{of}\:{the}\:{teiangle}\:{containing} \\ $$$$\alpha \\ $$$$\bigtriangleup=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{7}×\mathrm{11sin}\:\left(\mathrm{90}^{\mathrm{0}} −\theta\right) \\ $$$$=\frac{\mathrm{77}}{\mathrm{2}}\:×\frac{\sqrt{\mathrm{11}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{\mathrm{11}}=\frac{\mathrm{7}\sqrt{\mathrm{11}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${area}\:{of}\:{the}\:{sector} \\ $$$$\:{A}=\frac{\mathrm{90}^{\mathrm{0}} −\theta}{\mathrm{360}}\pi\mathrm{11}^{\mathrm{2}} \\ $$$${Blue}\:{area}\left({B}\right)+{area}\:{of}\:{the}\:{intersection}\:{of}\:{two}\:{circles}\left({C}+{C}\right)={A}−\bigtriangleup \\ $$$${B}+\left({C}+{C}\right)=\frac{\mathrm{90}^{\mathrm{0}} −\theta}{\mathrm{360}}\pi\mathrm{11}^{\mathrm{2}} −\frac{\mathrm{7}\sqrt{\mathrm{11}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{{z}} =\mathrm{11}^{\mathrm{2}} \:\&\:\left({x}−\mathrm{15}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{2}{x}−\mathrm{15}\right)\mathrm{15}=\mathrm{57} \\ $$$$\Rightarrow{x}=\left(\frac{\mathrm{57}}{\mathrm{15}}+\mathrm{15}\right)/\mathrm{2} \\ $$$${y}=\sqrt{\mathrm{11}^{\mathrm{2}} −{x}^{\mathrm{2}} }\approx\mathrm{5}.\mathrm{713} \\ $$$${area}\:{of}\:{the}\:{tirangle}\left({in}\:{the}\:{two}\:{circles}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{b}×{h}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}×{y} \\ $$$${green}\:{area}=\frac{\beta}{\mathrm{360}}\pi{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{121}×\mathrm{sin}\:\beta \\ $$$$\beta=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{11}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }{\mathrm{2}×\mathrm{11}×\mathrm{15}}\right)=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{47}}{\mathrm{55}}\right) \\ $$$${green}\:{area}\approx\mathrm{1}.\mathrm{618} \\ $$$${yellow}\:{area}=\frac{\gamma}{\mathrm{360}}\pi{r}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{64}×\mathrm{sin}\:\gamma \\ $$$$\gamma=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{8}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} −\mathrm{11}^{\mathrm{2}} }{\mathrm{2}×\mathrm{8}×\mathrm{15}}\right)=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{7}}{\mathrm{10}}\right) \\ $$$${yellow}\:{area}\approx\mathrm{2}.\mathrm{600} \\ $$$$\therefore{B}=\left({A}−\bigtriangleup\right)−\left({triangle}\:{area}+{green}\:{area}+{yellow}\:{area}\right) \\ $$$${B}=\frac{\mathrm{90}^{\mathrm{0}} −\mathrm{cos}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{11}^{\mathrm{2}} −\left(\frac{\mathrm{1575}}{\mathrm{274}}−\frac{\mathrm{7}\sqrt{\mathrm{22129}}}{\mathrm{274}}\right)^{\mathrm{2}} }}{\mathrm{11}}\right)}{\mathrm{360}^{\mathrm{0}} }\pi\mathrm{11}^{\mathrm{2}} −\frac{\mathrm{7}\:\sqrt{\mathrm{11}^{\mathrm{2}} −\left(\frac{\mathrm{1575}}{\mathrm{274}}−\frac{\mathrm{7}\sqrt{\mathrm{22129}}}{\mathrm{274}}\right)^{\mathrm{2}} }}{\mathrm{2}}−\left(\mathrm{2}\sqrt{\mathrm{11}^{\mathrm{2}} −\left(\frac{\left(\frac{\mathrm{57}}{\mathrm{15}}+\mathrm{15}\right)}{\mathrm{2}}\right)^{\mathrm{2}} }+\frac{\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{47}}{\mathrm{55}}\right)}{\mathrm{360}}\pi\mathrm{11}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{121}×\mathrm{sin}\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{47}}{\mathrm{55}}\:\right)+\frac{\mathrm{cos}^{−\mathrm{1}} \mathrm{0}.\mathrm{7}}{\mathrm{360}}\pi\mathrm{8}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{64}×\mathrm{sin}\:\left(\mathrm{cos}^{−\mathrm{1}} \mathrm{0}.\mathrm{7}\right)\right) \\ $$$${B}\approx\mathrm{30}.\mathrm{7274274166} \\ $$

Answered by mr W last updated on 10/Dec/25

Commented by mr W last updated on 10/Dec/25

tan δ=(7/(15)) ⇒sin δ=(7/( (√(274)))), cos δ=((15)/( (√(274)))) ((sin φ)/(15))=((sin δ)/(11)) sin φ=((15)/(11))×(7/( (√(274))))=((105)/(11(√(274)))) ⇒cos φ=((√(22129))/(11(√(274)))) β=(π/2)−(φ−δ)=(π/2)+δ−φ cos ϕ=((11^2 +8^2 −15^2 )/(2×11×8))=−(5/(22)) ⇒sin ϕ=((3(√(51)))/(22)) ((sin α)/8)=((sin γ)/(11))=((sin ϕ)/(15)) ⇒sin α=(8/(15))×((3(√(51)))/(22))=((4(√(51)))/(55)) ⇒cos α=((47)/(55)) ⇒sin γ=((11)/(15))×((3(√(51)))/(22))=((√(51))/(10)) ⇒cos γ=(7/(10)) A_1 =((11^2 α)/2)−((11^2 sin α cos α)/2) =((11^2 )/2)(cos^(−1) ((47)/(55))−((4(√(51)))/(55))×((47)/(55))) =((121)/2) cos^(−1) ((47)/(55))−((94(√(51)))/(25)) A_2 =((8^2 γ)/2)−((8^2 sin γ cos γ)/2) =(8^2 /2)(cos^(−1) (7/(10))−((√(51))/(10))×(7/(10))) =32 cos^(−1) (7/(10))−((56(√(51)))/(25)) A_3 =((11×7 sin β)/2) =((11×7 cos (φ−δ))/2) =((11×7)/2)×(cos φ cos δ+sin φ sin δ) =((11×7)/2)×(((√(22129))/(11(√(274))))×((15)/( (√(274))))+((105)/(11(√(274))))×(7/( (√(274))))) =((105(49+(√(22129))))/(548)) A_(shaded) =((11^2 β)/2)−A_1 −A_2 −A_3 =((121π)/4)+((121)/2) tan^(−1) (7/(15))−((121)/2) sin^(−1) ((105)/(11(√(274))))−((121)/2) cos^(−1) ((47)/(55))+((94(√(51)))/(25))−32 cos^(−1) (7/(10))+((56(√(51)))/(25))−((105(49+(√(22129))))/(548)) =((121π)/4)+((121)/2) tan^(−1) (7/(15))−((121)/2) sin^(−1) ((105)/(11(√(274))))−((121)/2) cos^(−1) ((47)/(55))−32 cos^(−1) (7/(10))+6(√(51))−((105(49+(√(22129))))/(548)) ≈30.727427

$$\mathrm{tan}\:\delta=\frac{\mathrm{7}}{\mathrm{15}}\:\Rightarrow\mathrm{sin}\:\delta=\frac{\mathrm{7}}{\:\sqrt{\mathrm{274}}},\:\mathrm{cos}\:\delta=\frac{\mathrm{15}}{\:\sqrt{\mathrm{274}}} \\ $$$$\frac{\mathrm{sin}\:\phi}{\mathrm{15}}=\frac{\mathrm{sin}\:\delta}{\mathrm{11}} \\ $$$$\mathrm{sin}\:\phi=\frac{\mathrm{15}}{\mathrm{11}}×\frac{\mathrm{7}}{\:\sqrt{\mathrm{274}}}=\frac{\mathrm{105}}{\mathrm{11}\sqrt{\mathrm{274}}}\:\Rightarrow\mathrm{cos}\:\phi=\frac{\sqrt{\mathrm{22129}}}{\mathrm{11}\sqrt{\mathrm{274}}} \\ $$$$\beta=\frac{\pi}{\mathrm{2}}−\left(\phi−\delta\right)=\frac{\pi}{\mathrm{2}}+\delta−\phi \\ $$$$\mathrm{cos}\:\varphi=\frac{\mathrm{11}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} −\mathrm{15}^{\mathrm{2}} }{\mathrm{2}×\mathrm{11}×\mathrm{8}}=−\frac{\mathrm{5}}{\mathrm{22}}\:\Rightarrow\mathrm{sin}\:\varphi=\frac{\mathrm{3}\sqrt{\mathrm{51}}}{\mathrm{22}} \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{8}}=\frac{\mathrm{sin}\:\gamma}{\mathrm{11}}=\frac{\mathrm{sin}\:\varphi}{\mathrm{15}} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{8}}{\mathrm{15}}×\frac{\mathrm{3}\sqrt{\mathrm{51}}}{\mathrm{22}}=\frac{\mathrm{4}\sqrt{\mathrm{51}}}{\mathrm{55}}\:\Rightarrow\mathrm{cos}\:\alpha=\frac{\mathrm{47}}{\mathrm{55}} \\ $$$$\Rightarrow\mathrm{sin}\:\gamma=\frac{\mathrm{11}}{\mathrm{15}}×\frac{\mathrm{3}\sqrt{\mathrm{51}}}{\mathrm{22}}=\frac{\sqrt{\mathrm{51}}}{\mathrm{10}}\:\Rightarrow\mathrm{cos}\:\gamma=\frac{\mathrm{7}}{\mathrm{10}} \\ $$$${A}_{\mathrm{1}} =\frac{\mathrm{11}^{\mathrm{2}} \alpha}{\mathrm{2}}−\frac{\mathrm{11}^{\mathrm{2}} \mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\mathrm{2}} \\ $$$$\:\:\:=\frac{\mathrm{11}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{47}}{\mathrm{55}}−\frac{\mathrm{4}\sqrt{\mathrm{51}}}{\mathrm{55}}×\frac{\mathrm{47}}{\mathrm{55}}\right) \\ $$$$\:\:\:=\frac{\mathrm{121}}{\mathrm{2}}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{47}}{\mathrm{55}}−\frac{\mathrm{94}\sqrt{\mathrm{51}}}{\mathrm{25}} \\ $$$${A}_{\mathrm{2}} =\frac{\mathrm{8}^{\mathrm{2}} \gamma}{\mathrm{2}}−\frac{\mathrm{8}^{\mathrm{2}} \mathrm{sin}\:\gamma\:\mathrm{cos}\:\gamma}{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{\mathrm{8}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{10}}−\frac{\sqrt{\mathrm{51}}}{\mathrm{10}}×\frac{\mathrm{7}}{\mathrm{10}}\right) \\ $$$$\:\:\:\:=\mathrm{32}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{10}}−\frac{\mathrm{56}\sqrt{\mathrm{51}}}{\mathrm{25}} \\ $$$${A}_{\mathrm{3}} =\frac{\mathrm{11}×\mathrm{7}\:\mathrm{sin}\:\beta}{\mathrm{2}} \\ $$$$\:\:\:=\frac{\mathrm{11}×\mathrm{7}\:\mathrm{cos}\:\left(\phi−\delta\right)}{\mathrm{2}} \\ $$$$\:\:\:=\frac{\mathrm{11}×\mathrm{7}}{\mathrm{2}}×\left(\mathrm{cos}\:\phi\:\mathrm{cos}\:\delta+\mathrm{sin}\:\phi\:\mathrm{sin}\:\delta\right) \\ $$$$\:\:\:=\frac{\mathrm{11}×\mathrm{7}}{\mathrm{2}}×\left(\frac{\sqrt{\mathrm{22129}}}{\mathrm{11}\sqrt{\mathrm{274}}}×\frac{\mathrm{15}}{\:\sqrt{\mathrm{274}}}+\frac{\mathrm{105}}{\mathrm{11}\sqrt{\mathrm{274}}}×\frac{\mathrm{7}}{\:\sqrt{\mathrm{274}}}\right) \\ $$$$\:\:\:=\frac{\mathrm{105}\left(\mathrm{49}+\sqrt{\mathrm{22129}}\right)}{\mathrm{548}} \\ $$$${A}_{{shaded}} =\frac{\mathrm{11}^{\mathrm{2}} \beta}{\mathrm{2}}−{A}_{\mathrm{1}} −{A}_{\mathrm{2}} −{A}_{\mathrm{3}} \\ $$$$\:\:\:=\frac{\mathrm{121}\pi}{\mathrm{4}}+\frac{\mathrm{121}}{\mathrm{2}}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{15}}−\frac{\mathrm{121}}{\mathrm{2}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{105}}{\mathrm{11}\sqrt{\mathrm{274}}}−\frac{\mathrm{121}}{\mathrm{2}}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{47}}{\mathrm{55}}+\frac{\mathrm{94}\sqrt{\mathrm{51}}}{\mathrm{25}}−\mathrm{32}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{10}}+\frac{\mathrm{56}\sqrt{\mathrm{51}}}{\mathrm{25}}−\frac{\mathrm{105}\left(\mathrm{49}+\sqrt{\mathrm{22129}}\right)}{\mathrm{548}} \\ $$$$\:\:\:=\frac{\mathrm{121}\pi}{\mathrm{4}}+\frac{\mathrm{121}}{\mathrm{2}}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{15}}−\frac{\mathrm{121}}{\mathrm{2}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{105}}{\mathrm{11}\sqrt{\mathrm{274}}}−\frac{\mathrm{121}}{\mathrm{2}}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{47}}{\mathrm{55}}−\mathrm{32}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{10}}+\mathrm{6}\sqrt{\mathrm{51}}−\frac{\mathrm{105}\left(\mathrm{49}+\sqrt{\mathrm{22129}}\right)}{\mathrm{548}} \\ $$$$\:\:\:\approx\mathrm{30}.\mathrm{727427} \\ $$

Commented by fantastic2 last updated on 10/Dec/25

$${great}! \\ $$

Answered by fantastic2 last updated on 10/Dec/25

Commented by fantastic2 last updated on 10/Dec/25

C_1 ⇒x^2 +y^2 =11^2 C_2 ⇒(x−15)^2 +y^2 =8^2 equation of line y=−((15)/7)x+15 point where C_1 and line meets (x_1 ,y_1 )=( ((1575)/(274))−((7(√(22129)))/(274)),((15(√(22129)))/(274))+((735)/(274)) ) point(x co−ordinate) where C_1 &C_2 meets x_0 =9.4 Shaded area=Green −Red−yellow ⇒∫_x_1 ^x_0 (√(11^2 −x^2 )) dx−(1/2)xy−∫_7 ^x_0 (√(8^2 −(x−15)^2 )) dx ⇒⇒∫_( ((1575)/(274))−((7(√(22129)))/(274))) ^(9.4) (√(11^2 −x^2 )) dx−(1/2)(7− (((1575)/(274))−((7(√(22129)))/(274)))(((15(√(22129)))/(274))+((735)/(274)) )−∫_7 ^(9.4) (√(8^2 −(x−15)^2 )) dx ≈30.7274274165646

$${C}_{\mathrm{1}} \Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{11}^{\mathrm{2}} \\ $$$${C}_{\mathrm{2}} \Rightarrow\left({x}−\mathrm{15}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}} \\ $$$${equation}\:{of}\:{line} \\ $$$${y}=−\frac{\mathrm{15}}{\mathrm{7}}{x}+\mathrm{15} \\ $$$${point}\:{where}\:{C}_{\mathrm{1}} \:{and}\:{line}\:{meets} \\ $$$$\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)=\left(\:\frac{\mathrm{1575}}{\mathrm{274}}−\frac{\mathrm{7}\sqrt{\mathrm{22129}}}{\mathrm{274}},\frac{\mathrm{15}\sqrt{\mathrm{22129}}}{\mathrm{274}}+\frac{\mathrm{735}}{\mathrm{274}}\:\right) \\ $$$${point}\left({x}\:{co}−{ordinate}\right)\:{where}\:{C}_{\mathrm{1}} \:\&{C}_{\mathrm{2}} {meets} \\ $$$${x}_{\mathrm{0}} =\mathrm{9}.\mathrm{4} \\ $$$${Shaded}\:{area}={Green}\:−{Red}−{yellow} \\ $$$$\Rightarrow\int_{{x}_{\mathrm{1}} } ^{{x}_{\mathrm{0}} } \sqrt{\mathrm{11}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx}−\frac{\mathrm{1}}{\mathrm{2}}{xy}−\int_{\mathrm{7}} ^{{x}_{\mathrm{0}} } \sqrt{\mathrm{8}^{\mathrm{2}} −\left({x}−\mathrm{15}\right)^{\mathrm{2}} }\:{dx} \\ $$$$\Rightarrow\Rightarrow\int_{\:\frac{\mathrm{1575}}{\mathrm{274}}−\frac{\mathrm{7}\sqrt{\mathrm{22129}}}{\mathrm{274}}} ^{\mathrm{9}.\mathrm{4}} \sqrt{\mathrm{11}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{7}−\:\left(\frac{\mathrm{1575}}{\mathrm{274}}−\frac{\mathrm{7}\sqrt{\mathrm{22129}}}{\mathrm{274}}\right)\left(\frac{\mathrm{15}\sqrt{\mathrm{22129}}}{\mathrm{274}}+\frac{\mathrm{735}}{\mathrm{274}}\:\right)−\int_{\mathrm{7}} ^{\mathrm{9}.\mathrm{4}} \sqrt{\mathrm{8}^{\mathrm{2}} −\left({x}−\mathrm{15}\right)^{\mathrm{2}} }\:{dx}\right. \\ $$$$\approx\mathrm{30}.\mathrm{7274274165646} \\ $$

Commented by fantastic2 last updated on 10/Dec/25

$${by}\:{integration} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply