Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 118011 by mmmmmm1 last updated on 14/Oct/20

Answered by mr W last updated on 14/Oct/20

(x−h)^2 +(y−k)^2 =R^2 (0−h)^2 +(0−k)^2 =R^2 ...(i) (−4−h)^2 +(0−k)^2 =R^2 ...(ii) (2−h)^2 +(−3−k)^2 =R^2 ...(iii) (ii)−(i): 4(4+2h)=0 ⇒h=−2 (iii)−(i): 2(2−2h)+3(3+2k)=0 ⇒k=−(7/2) R^2 =(2)^2 +((7/2))^2 =((65)/4) ⇒R=((√(65))/2)

$$\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$ \\ $$$$\left(\mathrm{0}−{h}\right)^{\mathrm{2}} +\left(\mathrm{0}−{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:...\left({i}\right) \\ $$$$\left(−\mathrm{4}−{h}\right)^{\mathrm{2}} +\left(\mathrm{0}−{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:...\left({ii}\right) \\ $$$$\left(\mathrm{2}−{h}\right)^{\mathrm{2}} +\left(−\mathrm{3}−{k}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:...\left({iii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$$\mathrm{4}\left(\mathrm{4}+\mathrm{2}{h}\right)=\mathrm{0}\:\Rightarrow{h}=−\mathrm{2} \\ $$$$\left({iii}\right)−\left({i}\right): \\ $$$$\mathrm{2}\left(\mathrm{2}−\mathrm{2}{h}\right)+\mathrm{3}\left(\mathrm{3}+\mathrm{2}{k}\right)=\mathrm{0}\:\Rightarrow{k}=−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} =\left(\mathrm{2}\right)^{\mathrm{2}} +\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{65}}{\mathrm{4}} \\ $$$$\Rightarrow{R}=\frac{\sqrt{\mathrm{65}}}{\mathrm{2}} \\ $$

Answered by MJS_new last updated on 14/Oct/20

it′s the circumcircle of a triangle with a=4 b=(√(2^2 +3^2 ))=(√(13)) c=(√((2+4)^2 +3^2 ))=(√(45)) R=((abc)/( (√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))))=((√(65))/2) ⇒ area of circle=((65π)/4)

$$\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with} \\ $$$${a}=\mathrm{4} \\ $$$${b}=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\sqrt{\mathrm{13}} \\ $$$${c}=\sqrt{\left(\mathrm{2}+\mathrm{4}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\sqrt{\mathrm{45}} \\ $$$${R}=\frac{{abc}}{\:\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}=\frac{\sqrt{\mathrm{65}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{area}\:\mathrm{of}\:\mathrm{circle}=\frac{\mathrm{65}\pi}{\mathrm{4}} \\ $$

Answered by mr W last updated on 14/Oct/20

an other way: a=4 b=(√(2^2 +3^2 ))=(√(13)) c=(√(3^2 +(2+4)^2 ))=(√(45)) Δ=((4×3)/2)=6 R=((abc)/(4Δ))=((4×(√(13))×(√(45)))/(4×6))=((√(65))/2)

$${an}\:{other}\:{way}: \\ $$$${a}=\mathrm{4} \\ $$$${b}=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\sqrt{\mathrm{13}} \\ $$$${c}=\sqrt{\mathrm{3}^{\mathrm{2}} +\left(\mathrm{2}+\mathrm{4}\right)^{\mathrm{2}} }=\sqrt{\mathrm{45}} \\ $$$$\Delta=\frac{\mathrm{4}×\mathrm{3}}{\mathrm{2}}=\mathrm{6} \\ $$$${R}=\frac{{abc}}{\mathrm{4}\Delta}=\frac{\mathrm{4}×\sqrt{\mathrm{13}}×\sqrt{\mathrm{45}}}{\mathrm{4}×\mathrm{6}}=\frac{\sqrt{\mathrm{65}}}{\mathrm{2}} \\ $$

Answered by 1549442205PVT last updated on 15/Oct/20

Commented by 1549442205PVT last updated on 15/Oct/20

Given AC=4,CD=3,DB=2 ACD^(�) =CDB^(�) =90° first way: AC//BD⇒((DH)/(HC))=((BD)/(AC))=(2/4)=(1/2) CD=3(hypothesis)⇒HC=2 AH=(√(AC^2 +CH^2 ))=(√(4^2 +2^2 ))=2(√5) sinCAH^(�) =((CH)/(AH))=(2/(2(√5)))=(1/( (√5))) sinCLB^(�) =sinCAB^(�) =sinCAH^(�) =1/(√5) ΔBCL is right at B⇒2R=CL=((BC)/(sinCLB^(�) )) =((√(3^2 +2^2 ))/((1/(√5))))=(√(65))⇒R=(√(65))/2 ⇒S_(circle) =πR^2 =((65π)/4) second way:Suppose (CD)∩(O)=E Denote by I,F midpoint AC,CE respectively Put DF=x ⇒BF=x+3,J=(BD)∩OI ⇒BJ⊥OI(since OI⊥AC,AC//BJ) ⇒OJ=DF=x,CI=DJ=2,BJ=4 so in the righttriangles OFC and OJB we have:(x+3)^2 +2^2 =R^2 =x^2 +4^2 ⇒x^2 +6x+13=x^2 +16⇒6x=3⇒x=(1/2) ⇒R=(√(x^2 +4^2 ))=(√((1/4)+16))=(√((65)/4))=((√(65))/2) ⇒S_(circle) =((65π)/4)

$$\mathrm{Given}\:\mathrm{AC}=\mathrm{4},\mathrm{CD}=\mathrm{3},\mathrm{DB}=\mathrm{2} \\ $$$$\widehat {\mathrm{ACD}}=\widehat {\mathrm{CDB}}=\mathrm{90}° \\ $$$$\mathrm{first}\:\mathrm{way}: \\ $$$$\mathrm{AC}//\mathrm{BD}\Rightarrow\frac{\mathrm{DH}}{\mathrm{HC}}=\frac{\mathrm{BD}}{\mathrm{AC}}=\frac{\mathrm{2}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{CD}=\mathrm{3}\left(\mathrm{hypothesis}\right)\Rightarrow\mathrm{HC}=\mathrm{2} \\ $$$$\mathrm{AH}=\sqrt{\mathrm{AC}^{\mathrm{2}} +\mathrm{CH}^{\mathrm{2}} }=\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\mathrm{sin}\widehat {\mathrm{CAH}}=\frac{\mathrm{CH}}{\mathrm{AH}}=\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{5}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{sin}\widehat {\mathrm{CLB}}=\mathrm{sin}\widehat {\mathrm{CAB}}=\mathrm{sin}\widehat {\mathrm{CAH}}=\mathrm{1}/\sqrt{\mathrm{5}} \\ $$$$\Delta\mathrm{BCL}\:\mathrm{is}\:\mathrm{right}\:\mathrm{at}\:\mathrm{B}\Rightarrow\mathrm{2R}=\mathrm{CL}=\frac{\mathrm{BC}}{\mathrm{sin}\widehat {\mathrm{CLB}}} \\ $$$$=\frac{\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}{\left(\mathrm{1}/\sqrt{\mathrm{5}}\right)}=\sqrt{\mathrm{65}}\Rightarrow\mathrm{R}=\sqrt{\mathrm{65}}/\mathrm{2} \\ $$$$\Rightarrow\mathrm{S}_{\mathrm{circle}} =\pi\mathrm{R}^{\mathrm{2}} =\frac{\mathrm{65}\pi}{\mathrm{4}} \\ $$$$\mathrm{second}\:\mathrm{way}:\mathrm{Suppose}\:\left(\mathrm{CD}\right)\cap\left(\mathrm{O}\right)=\mathrm{E} \\ $$$$\mathrm{Denote}\:\mathrm{by}\:\mathrm{I},\mathrm{F}\:\mathrm{midpoint}\:\mathrm{AC},\mathrm{CE}\:\mathrm{respectively} \\ $$$$\mathrm{Put}\:\mathrm{DF}=\mathrm{x}\:\Rightarrow\mathrm{BF}=\mathrm{x}+\mathrm{3},\mathrm{J}=\left(\mathrm{BD}\right)\cap\mathrm{OI} \\ $$$$\Rightarrow\mathrm{BJ}\bot\mathrm{OI}\left(\mathrm{since}\:\mathrm{OI}\bot\mathrm{AC},\mathrm{AC}//\mathrm{BJ}\right) \\ $$$$\Rightarrow\mathrm{OJ}=\mathrm{DF}=\mathrm{x},\mathrm{CI}=\mathrm{DJ}=\mathrm{2},\mathrm{BJ}=\mathrm{4} \\ $$$$\mathrm{so}\:\mathrm{in}\:\mathrm{the}\:\mathrm{righttriangles}\:\mathrm{OFC}\:\mathrm{and}\:\mathrm{OJB} \\ $$$$\mathrm{we}\:\mathrm{have}:\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{13}=\mathrm{x}^{\mathrm{2}} +\mathrm{16}\Rightarrow\mathrm{6x}=\mathrm{3}\Rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{R}=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{16}}=\sqrt{\frac{\mathrm{65}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{65}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{S}_{\mathrm{circle}} =\frac{\mathrm{65}\pi}{\mathrm{4}} \\ $$

Terms of Service

Privacy Policy

Contact: [email protected]