Question-65128

Question Number 65128 by Tawa1 last updated on 25/Jul/19

Answered by ajfour last updated on 25/Jul/19

Commented by ajfour last updated on 25/Jul/19

OP =8 with O as origin P (4, (√(64−16))) ≡(4, 4(√3)) OQ=8(√2) eq. of OQ is y−x=0 ⊥ distance of P from line OQ p= ((4(√3)−4)/( (√2))) shaded area= Area(△OPQ) =(1/2)×8(√2)×((4((√3)−1))/( (√2))) =16((√3)−1) .

$${OP}\:=\mathrm{8} \\ $$$${with}\:{O}\:{as}\:{origin} \\ $$$${P}\:\left(\mathrm{4},\:\sqrt{\mathrm{64}−\mathrm{16}}\right)\:\equiv\left(\mathrm{4},\:\mathrm{4}\sqrt{\mathrm{3}}\right) \\ $$$${OQ}=\mathrm{8}\sqrt{\mathrm{2}} \\ $$$${eq}.\:{of}\:{OQ}\:{is}\:\:{y}−{x}=\mathrm{0} \\ $$$$\bot\:{distance}\:{of}\:{P}\:{from}\:{line}\:{OQ} \\ $$$$\:\:\:{p}=\:\frac{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{4}}{\:\sqrt{\mathrm{2}}} \\ $$$${shaded}\:{area}=\:{Area}\left(\bigtriangleup{OPQ}\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}\sqrt{\mathrm{2}}×\frac{\mathrm{4}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{16}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\:. \\ $$

Commented by Tawa1 last updated on 25/Jul/19

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

Commented by Tawa1 last updated on 25/Jul/19

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir} \\ $$

Commented by Tawa1 last updated on 25/Jul/19

Commented by Tawa1 last updated on 25/Jul/19

$$\mathrm{This}\:\mathrm{sir} \\ $$

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

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

DC=DB=DE=r CF=((CB)/2)=2 ((CF)/(CD))=((CD)/(AC)) CD^2 =CF×AC ⇒r^2 =2×9=18 ⇒r=3(√2)

$${DC}={DB}={DE}={r} \\ $$$${CF}=\frac{{CB}}{\mathrm{2}}=\mathrm{2} \\ $$$$\frac{{CF}}{{CD}}=\frac{{CD}}{{AC}} \\ $$$${CD}^{\mathrm{2}} ={CF}×{AC} \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\mathrm{2}×\mathrm{9}=\mathrm{18} \\ $$$$\Rightarrow{r}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$

Answered by mr W last updated on 25/Jul/19

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

OC=OP=CP=r=8 ⇒∠OCP=60°=(π/3) ⇒∠PCQ=30°=(π/6) A_(shade) =Cap_(OQ) −Cap_(OP) −Cap_(PQ) =(r^2 /2)((π/2)−sin (π/2))−(r^2 /2)((π/3)−sin (π/3))−(r^2 /2)((π/6)−sin (π/6)) =(r^2 /2)((π/2)−(π/3)−(π/6)−sin (π/2)+sin (π/3)+sin (π/6)) =((64)/2)(−1+(1/2)+((√3)/2)) =16((√3)−1)

$${OC}={OP}={CP}={r}=\mathrm{8} \\ $$$$\Rightarrow\angle{OCP}=\mathrm{60}°=\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow\angle{PCQ}=\mathrm{30}°=\frac{\pi}{\mathrm{6}} \\ $$$${A}_{{shade}} ={Cap}_{{OQ}} −{Cap}_{{OP}} −{Cap}_{{PQ}} \\ $$$$=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\right)−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{3}}−\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\right)−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$$=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{6}}−\mathrm{sin}\:\frac{\pi}{\mathrm{2}}+\mathrm{sin}\:\frac{\pi}{\mathrm{3}}+\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$$=\frac{\mathrm{64}}{\mathrm{2}}\left(−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$=\mathrm{16}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$

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

An other way: A_(shade) =(1/2)×OP×OQ×sin ∠POQ =(1/2)×8×8(√2)×sin (60°−45°) =32(√2)×(((√3)/2)×(1/( (√2)))−(1/2)×(1/( (√2)))) =16((√3)−1)

$${An}\:{other}\:{way}: \\ $$$${A}_{{shade}} =\frac{\mathrm{1}}{\mathrm{2}}×{OP}×{OQ}×\mathrm{sin}\:\angle{POQ} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}×\mathrm{8}\sqrt{\mathrm{2}}×\mathrm{sin}\:\left(\mathrm{60}°−\mathrm{45}°\right) \\ $$$$=\mathrm{32}\sqrt{\mathrm{2}}×\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$=\mathrm{16}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply