Menu Close

Question-64909

Tinku Tara 0 Comments
Pin




Question Number 64909 by Tawa1 last updated on 23/Jul/19
Commented by MJS last updated on 23/Jul/19
the picture shows 2 circles, so the given numbers don′t fit. obviously the right must be greater than the above/below numbers
$$\mathrm{the}\:\mathrm{picture}\:\mathrm{shows}\:\mathrm{2}\:\mathrm{circles},\:\mathrm{so}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{numbers}\:\mathrm{don}'\mathrm{t}\:\mathrm{fit}.\:\mathrm{obviously}\:\mathrm{the}\:\mathrm{right}\:\mathrm{must} \\ $$$$\mathrm{be}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{the}\:\mathrm{above}/\mathrm{below}\:\mathrm{numbers} \\ $$
Commented by Tawa1 last updated on 23/Jul/19
God bless you sir, i get. I will ask you to give me questions on shaded area when am ready done with your solution and sir mrW solution and sir Ajfour
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{get}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{ask}\:\mathrm{you}\:\mathrm{to}\:\mathrm{give}\:\mathrm{me}\:\mathrm{questions}\:\mathrm{on}\:\mathrm{shaded}\:\mathrm{area}\:\mathrm{when}\:\mathrm{am}\: \\ $$$$\mathrm{ready}\:\mathrm{done}\:\mathrm{with}\:\mathrm{your}\:\mathrm{solution}\:\mathrm{and}\:\mathrm{sir}\:\mathrm{mrW}\:\mathrm{solution}\:\mathrm{and}\:\mathrm{sir}\:\mathrm{Ajfour} \\ $$
Answered by MJS last updated on 23/Jul/19
if we exchange 10 ⇌ 18 the big circle x^2 +y^2 =r^2 ⇒ y=±(√(r^2 −x^2 )) the small circle (x+9)^2 +y^2 =(r−9)^2 ⇒ y=±(√((r−9)^2 −(x+9)^2 )) let′s use the upper half circles y_1 =(√(r^2 −x^2 )) y_2 =(√((r−9)^2 −(x+9)^2 )) at x=0: y_1 −y_2 =10 ⇒ y_2 =y_1 −10 y_1 =r y_2 =(√(r(r−18))) (√(r(r−18)))=r−10 r(r−18)=(r−10)^2 2r−100=0 r=50 ⇒ the small circle has radius 50−9=41 ⇒ the shaded area = 50^2 π−41^2 π=819π
$$\mathrm{if}\:\mathrm{we}\:\mathrm{exchange}\:\mathrm{10}\:\rightleftharpoons\:\mathrm{18} \\ $$$$\mathrm{the}\:\mathrm{big}\:\mathrm{circle} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \:\Rightarrow\:{y}=\pm\sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$$\mathrm{the}\:\mathrm{small}\:\mathrm{circle} \\ $$$$\left({x}+\mathrm{9}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left({r}−\mathrm{9}\right)^{\mathrm{2}} \:\Rightarrow\:{y}=\pm\sqrt{\left({r}−\mathrm{9}\right)^{\mathrm{2}} −\left({x}+\mathrm{9}\right)^{\mathrm{2}} } \\ $$$$\mathrm{let}'\mathrm{s}\:\mathrm{use}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{half}\:\mathrm{circles} \\ $$$${y}_{\mathrm{1}} =\sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$${y}_{\mathrm{2}} =\sqrt{\left({r}−\mathrm{9}\right)^{\mathrm{2}} −\left({x}+\mathrm{9}\right)^{\mathrm{2}} } \\ $$$$\mathrm{at}\:{x}=\mathrm{0}:\:{y}_{\mathrm{1}} −{y}_{\mathrm{2}} =\mathrm{10}\:\Rightarrow\:{y}_{\mathrm{2}} ={y}_{\mathrm{1}} −\mathrm{10} \\ $$$${y}_{\mathrm{1}} ={r} \\ $$$${y}_{\mathrm{2}} =\sqrt{{r}\left({r}−\mathrm{18}\right)} \\ $$$$\sqrt{{r}\left({r}−\mathrm{18}\right)}={r}−\mathrm{10} \\ $$$${r}\left({r}−\mathrm{18}\right)=\left({r}−\mathrm{10}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{r}−\mathrm{100}=\mathrm{0} \\ $$$${r}=\mathrm{50} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{small}\:\mathrm{circle}\:\mathrm{has}\:\mathrm{radius}\:\mathrm{50}−\mathrm{9}=\mathrm{41} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}\:=\:\mathrm{50}^{\mathrm{2}} \pi−\mathrm{41}^{\mathrm{2}} \pi=\mathrm{819}\pi \\ $$
Commented by Tawa1 last updated on 23/Jul/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *