Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 67254 by TawaTawa last updated on 24/Aug/19

Commented by Rasheed.Sindhi last updated on 24/Aug/19

Analytical Method x-axis:Horizanal Chord y-axis:Vertical Chord Three points on the circle: A(6,0),B(−2,0) & C(0,−3) Find out circumcentre O(x,y) of △ABC Radius=∣OA∣ or ∣OB∣ or ∣OC∣

$$\:\:\:\:\:{Analytical}\:{Method} \\ $$$${x}-{axis}:{Horizanal}\:{Chord} \\ $$$${y}-{axis}:{Vertical}\:{Chord} \\ $$$${Three}\:{points}\:{on}\:{the}\:{circle}: \\ $$$$\:{A}\left(\mathrm{6},\mathrm{0}\right),{B}\left(−\mathrm{2},\mathrm{0}\right)\:\&\:{C}\left(\mathrm{0},−\mathrm{3}\right) \\ $$$${Find}\:{out}\:{circumcentre}\:{O}\left({x},{y}\right)\:{of}\:\bigtriangleup{ABC}\: \\ $$$${Radius}=\mid{OA}\mid\:{or}\:\mid{OB}\mid\:{or}\:\mid{OC}\mid \\ $$$$ \\ $$

Commented by MJS last updated on 24/Aug/19

Sir Rasheed′s idea completed a=8 b=(√(2^2 +3^2 ))=(√(13)) c=(√(3^2 +6^2 ))=(√(45)) δ=(√((a+b+c)(−a+b+c)(a−b+c)(a+b−c))) R=((abc)/δ)=((√(65))/2) [Heron′s formula; search the web for it]

$$\mathrm{Sir}\:\mathrm{Rasheed}'\mathrm{s}\:\mathrm{idea}\:\mathrm{completed} \\ $$$${a}=\mathrm{8}\:{b}=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\sqrt{\mathrm{13}}\:\:{c}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} }=\sqrt{\mathrm{45}} \\ $$$$\delta=\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)} \\ $$$${R}=\frac{{abc}}{\delta}=\frac{\sqrt{\mathrm{65}}}{\mathrm{2}}\:\left[\mathrm{Heron}'\mathrm{s}\:\mathrm{formula};\:\mathrm{search}\:\mathrm{the}\:\mathrm{web}\:\mathrm{for}\:\mathrm{it}\right] \\ $$

Commented by TawaTawa last updated on 25/Aug/19

God bless you sirs.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sirs}.\: \\ $$

Commented by TawaTawa last updated on 25/Aug/19

Sir, please label the diagram to back up the solution. Thanks sir.

$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{label}\:\mathrm{the}\:\mathrm{diagram}\:\mathrm{to}\:\mathrm{back}\:\mathrm{up}\:\mathrm{the}\:\mathrm{solution}.\:\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Commented by Rasheed.Sindhi last updated on 25/Aug/19

T^(H^A N) Ks Sır MJS!

$$\mathcal{T}^{\mathcal{H}^{\mathcal{A}} \mathcal{N}} \mathcal{K}{s}\:\mathcal{S}\imath{r}\:\mathrm{MJS}! \\ $$

Commented by TawaTawa last updated on 25/Aug/19

Sir, how did you get a, b, c please

$$\mathrm{Sir},\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\:\:\mathrm{a},\:\mathrm{b},\:\mathrm{c}\:\:\mathrm{please} \\ $$

Commented by Tony Lin last updated on 25/Aug/19

Commented by Tony Lin last updated on 25/Aug/19

R=((abc)/(4×△Area)) =((8×(√(13))×3(√5))/(4×((1/2)×8×3))) =((√(65))/2)

$${R}=\frac{{abc}}{\mathrm{4}×\bigtriangleup{Area}} \\ $$$$=\frac{\mathrm{8}×\sqrt{\mathrm{13}}×\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{4}×\left(\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}×\mathrm{3}\right)} \\ $$$$=\frac{\sqrt{\mathrm{65}}}{\mathrm{2}} \\ $$

Commented by Tony Lin last updated on 25/Aug/19

you can remember ((abc)/(4×△Area)) this formula (a/(sinA))=(b/(sinB))=(c/(sinC))=2R ⇒((abc)/((1/2)bcsinA))=((abc)/((1/2)acsinB))=((abc)/((1/2)absinC))=4R ⇒((abc)/(4×△Area))=R

$${you}\:{can}\:{remember}\:\frac{{abc}}{\mathrm{4}×\bigtriangleup{Area}}\:{this}\:{formula} \\ $$$$\frac{{a}}{{sinA}}=\frac{{b}}{{sinB}}=\frac{{c}}{{sinC}}=\mathrm{2}{R} \\ $$$$\Rightarrow\frac{{abc}}{\frac{\mathrm{1}}{\mathrm{2}}{bcsinA}}=\frac{{abc}}{\frac{\mathrm{1}}{\mathrm{2}}{acsinB}}=\frac{{abc}}{\frac{\mathrm{1}}{\mathrm{2}}{absinC}}=\mathrm{4}{R} \\ $$$$\Rightarrow\frac{{abc}}{\mathrm{4}×\bigtriangleup{Area}}={R} \\ $$

Commented by TawaTawa last updated on 25/Aug/19

I appreciate sir. God bless you sir. Now, i understand

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{Now},\:\mathrm{i}\:\mathrm{understand} \\ $$

Answered by mr W last updated on 24/Aug/19

Commented by mr W last updated on 24/Aug/19

(h/2)=(6/3) ⇒h=4 R^2 −(((2+6)/2))^2 =(((3+4)/2)−3)^2 R^2 =16+(1/4)=((65)/4) ⇒R=((√(65))/2)

$$\frac{{h}}{\mathrm{2}}=\frac{\mathrm{6}}{\mathrm{3}} \\ $$$$\Rightarrow{h}=\mathrm{4} \\ $$$${R}^{\mathrm{2}} −\left(\frac{\mathrm{2}+\mathrm{6}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{3}+\mathrm{4}}{\mathrm{2}}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} =\mathrm{16}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{65}}{\mathrm{4}} \\ $$$$\Rightarrow{R}=\frac{\sqrt{\mathrm{65}}}{\mathrm{2}} \\ $$

Commented by TawaTawa last updated on 25/Aug/19

God bless you sir, i get the rest but i don′t understand how you used R^2 − (((2 + 6)/2))^2 = (((3 + 4)/2) − 3)^2 . R is not labelled. please help sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\:\:\:\:\mathrm{i}\:\mathrm{get}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{but}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{how}\:\mathrm{you}\:\mathrm{used} \\ $$$$\mathrm{R}^{\mathrm{2}} \:−\:\left(\frac{\mathrm{2}\:+\:\mathrm{6}}{\mathrm{2}}\right)^{\mathrm{2}} \:\:=\:\:\left(\frac{\mathrm{3}\:+\:\mathrm{4}}{\mathrm{2}}\:−\:\mathrm{3}\right)^{\mathrm{2}} . \\ $$$$\mathrm{R}\:\mathrm{is}\:\mathrm{not}\:\mathrm{labelled}.\:\:\mathrm{please}\:\mathrm{help}\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 25/Aug/19

Commented by TawaTawa last updated on 25/Aug/19

Sir, please give me any question on finding area of shaded portion i should try. Thanks sir.

$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{give}\:\mathrm{me}\:\mathrm{any}\:\mathrm{question}\:\mathrm{on}\:\mathrm{finding}\:\mathrm{area}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{portion}\:\mathrm{i}\:\mathrm{should} \\ $$$$\mathrm{try}.\:\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Commented by TawaTawa last updated on 25/Aug/19

Wow, i understand it now sir. God bless you sir.

$$\mathrm{Wow},\:\:\mathrm{i}\:\mathrm{understand}\:\mathrm{it}\:\mathrm{now}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com