Question-64829

Question Number 64829 by Tawa1 last updated on 22/Jul/19

Commented by Tawa1 last updated on 22/Jul/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by Tony Lin last updated on 22/Jul/19

let∠BDC=θ BC=(√(17^2 +11^2 −2×17×11cosθ)) =(√(410−374cosθ)) AB=(√(17^2 +31^2 −2×17×31cos(π−θ))) =(√(17^2 +31^2 +2×17×31cosθ)) =(√(1250+1054cosθ)) area of △ABC =BC×AB×(1/2) =(1/2)×(BD×DC×sinθ+BD×AD×sin(π−θ) =(1/2)×17×(11+31)sinθ =357sinθ ⇒(√((410−374cosθ)(1250+1054cosθ))) =714sinθ ⇒(√(512500−35360cosθ−394196cos^2 θ)) = 714sinθ let cosθ=t ⇒(√(512500−35360t−394196t^2 )) =714(√(1−t^2 )) ⇒512500−35360t−394196t^2 =509796−509796t^2 ⇒115600t^2 −35360t+2704=0 ⇒t=((13)/(85)) area of △BDC =(1/2)×BD×DC×sinθ =(1/2)×17×11×(√(1−(((13)/(85)))^2 )) =92.4

$${let}\angle{BDC}=\theta \\ $$$${BC}=\sqrt{\mathrm{17}^{\mathrm{2}} +\mathrm{11}^{\mathrm{2}} −\mathrm{2}×\mathrm{17}×\mathrm{11}{cos}\theta} \\ $$$$\:\:\:\:\:\:\:=\sqrt{\mathrm{410}−\mathrm{374}{cos}\theta} \\ $$$${AB}=\sqrt{\mathrm{17}^{\mathrm{2}} +\mathrm{31}^{\mathrm{2}} −\mathrm{2}×\mathrm{17}×\mathrm{31}{cos}\left(\pi−\theta\right)} \\ $$$$\:\:\:\:\:\:\:=\sqrt{\mathrm{17}^{\mathrm{2}} +\mathrm{31}^{\mathrm{2}} +\mathrm{2}×\mathrm{17}×\mathrm{31}{cos}\theta} \\ $$$$\:\:\:\:\:\:\:=\sqrt{\mathrm{1250}+\mathrm{1054}{cos}\theta} \\ $$$${area}\:{of}\:\bigtriangleup{ABC} \\ $$$$={BC}×{AB}×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\left({BD}×{DC}×{sin}\theta+{BD}×{AD}×{sin}\left(\pi−\theta\right)\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{17}×\left(\mathrm{11}+\mathrm{31}\right){sin}\theta \\ $$$$=\mathrm{357}{sin}\theta \\ $$$$\Rightarrow\sqrt{\left(\mathrm{410}−\mathrm{374}{cos}\theta\right)\left(\mathrm{1250}+\mathrm{1054}{cos}\theta\right)} \\ $$$$\:\:\:\:\:=\mathrm{714}{sin}\theta \\ $$$$\Rightarrow\sqrt{\mathrm{512500}−\mathrm{35360}{cos}\theta−\mathrm{394196}{cos}^{\mathrm{2}} \theta} \\ $$$$\:\:\:\:=\:\mathrm{714}{sin}\theta \\ $$$${let}\:{cos}\theta={t} \\ $$$$\Rightarrow\sqrt{\mathrm{512500}−\mathrm{35360}{t}−\mathrm{394196}{t}^{\mathrm{2}} } \\ $$$$\:\:\:\:=\mathrm{714}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{512500}−\mathrm{35360}{t}−\mathrm{394196}{t}^{\mathrm{2}} \\ $$$$\:\:\:\:=\mathrm{509796}−\mathrm{509796}{t}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{115600}{t}^{\mathrm{2}} −\mathrm{35360}{t}+\mathrm{2704}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{13}}{\mathrm{85}} \\ $$$${area}\:{of}\:\bigtriangleup{BDC} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{BD}×{DC}×{sin}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{17}×\mathrm{11}×\sqrt{\mathrm{1}−\left(\frac{\mathrm{13}}{\mathrm{85}}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{92}.\mathrm{4} \\ $$

Commented by mathmax by abdo last updated on 23/Jul/19

$${where}\:{are}\:{you}\:{from}\:{sir}\:{tony}… \\ $$

Commented by Tony Lin last updated on 23/Jul/19

$${Taiwan} \\ $$

Answered by MJS last updated on 22/Jul/19

put ((A+C)/2) in the center ⇒ A= (((−21)),(0) ) C= (((21)),(0) ) D= (((10)),(0) ) B lies on the circumcircle x^2 +y^2 =21^2 B= ((x),((√(441−x^2 ))) ) ∣BD∣=17 (√((x−10)^2 +((√(441−x^2 ))−0)^2 ))=17 (√(541−20x))=17 x=((63)/5) B= ((((63)/5)),(((84)/5)) ) ∣BC∣=(√((((63)/5)−21)^2 +(((84)/5)−0)^2 ))=((42(√5))/5) ⇒ the shaded triangle has sides 11, 17, ((42(√5))/5) the area of a triangle with sides a, b, c is ((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4) ⇒ shaded area = ((462)/5)=92.4

$$\mathrm{put}\:\frac{{A}+{C}}{\mathrm{2}}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center} \\ $$$$\Rightarrow\:{A}=\begin{pmatrix}{−\mathrm{21}}\\{\mathrm{0}}\end{pmatrix}\:\:\mathrm{C}=\begin{pmatrix}{\mathrm{21}}\\{\mathrm{0}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{\mathrm{10}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${B}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circumcircle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{21}^{\mathrm{2}} \\ $$$${B}=\begin{pmatrix}{{x}}\\{\sqrt{\mathrm{441}−{x}^{\mathrm{2}} }}\end{pmatrix} \\ $$$$\mid{BD}\mid=\mathrm{17} \\ $$$$\sqrt{\left({x}−\mathrm{10}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{441}−{x}^{\mathrm{2}} }−\mathrm{0}\right)^{\mathrm{2}} }=\mathrm{17} \\ $$$$\sqrt{\mathrm{541}−\mathrm{20}{x}}=\mathrm{17} \\ $$$${x}=\frac{\mathrm{63}}{\mathrm{5}} \\ $$$${B}=\begin{pmatrix}{\frac{\mathrm{63}}{\mathrm{5}}}\\{\frac{\mathrm{84}}{\mathrm{5}}}\end{pmatrix} \\ $$$$\mid{BC}\mid=\sqrt{\left(\frac{\mathrm{63}}{\mathrm{5}}−\mathrm{21}\right)^{\mathrm{2}} +\left(\frac{\mathrm{84}}{\mathrm{5}}−\mathrm{0}\right)^{\mathrm{2}} }=\frac{\mathrm{42}\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{triangle}\:\mathrm{has}\:\mathrm{sides} \\ $$$$\mathrm{11},\:\mathrm{17},\:\frac{\mathrm{42}\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{sides}\:{a},\:{b},\:{c}\:\mathrm{is} \\ $$$$\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{shaded}\:\mathrm{area}\:=\:\frac{\mathrm{462}}{\mathrm{5}}=\mathrm{92}.\mathrm{4} \\ $$$$ \\ $$

Commented by Tawa1 last updated on 22/Jul/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by ajfour last updated on 22/Jul/19

Commented by ajfour last updated on 22/Jul/19

cos θ=((21^2 −10^2 −17^2 )/(2×10×17)) = ((13)/(85)) sin θ=((√(7225−169))/(85)) = ((84)/(85)) A_(rea) =(1/2)×11×17sin θ A_(rea) =((11×17×84)/(2×85))=((11×42)/5)=((462)/5) A_(rea) = 92.4 sq units ■

$$\mathrm{cos}\:\theta=\frac{\mathrm{21}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} −\mathrm{17}^{\mathrm{2}} }{\mathrm{2}×\mathrm{10}×\mathrm{17}}\:=\:\frac{\mathrm{13}}{\mathrm{85}} \\ $$$$\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{7225}−\mathrm{169}}}{\mathrm{85}}\:=\:\frac{\mathrm{84}}{\mathrm{85}} \\ $$$${A}_{{rea}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{11}×\mathrm{17sin}\:\theta\: \\ $$$${A}_{{rea}} =\frac{\mathrm{11}×\mathrm{17}×\mathrm{84}}{\mathrm{2}×\mathrm{85}}=\frac{\mathrm{11}×\mathrm{42}}{\mathrm{5}}=\frac{\mathrm{462}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:{A}_{{rea}} \:=\:\mathrm{92}.\mathrm{4}\:{sq}\:{units}\:\blacksquare \\ $$

Commented by ajfour last updated on 22/Jul/19