Question-71698

Question Number 71698 by TawaTawa last updated on 18/Oct/19

Commented by MJS last updated on 19/Oct/19

we had a similar example a few weeks ago

$$\mathrm{we}\:\mathrm{had}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{example}\:\mathrm{a}\:\mathrm{few}\:\mathrm{weeks}\:\mathrm{ago} \\ $$

Commented by TawaTawa last updated on 19/Oct/19

$$\mathrm{Help}\:\mathrm{sir} \\ $$

Answered by MJS last updated on 19/Oct/19

general solution let the right handed difference be p and the difference on bottom and top be q ⇒ the radius of the big circle C is R the radius of the small circle c is r=R−(p/2) C: x^2 +y^2 −R^2 =0 c: (x−(p/2))^2 +y^2 −(R−(p/2))^2 =0 ⇔ ⇔ x^2 +y^2 −px−R^2 +pR=0 we know that ((x),(y) ) = ((0),((R−q)) ) ∈c put x=0∧y=R−q into c ⇒ (p−2q)R+q^2 =0 ⇒ R=(q^2 /(2q−p)) ⇒ r=(((p−q)^2 +q^2 )/(2(2q−p))) ⇒ green area = area (C) −area (c) = =((p((p−q)^2 +3q^2 ))/(4(2q−p)))π in our case p=18 q=10 R=50 r=41 green area = 819π

$$\mathrm{general}\:\mathrm{solution} \\ $$$$\mathrm{let}\:\mathrm{the}\:\mathrm{right}\:\mathrm{handed}\:\mathrm{difference}\:\mathrm{be}\:{p}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{difference}\:\mathrm{on}\:\mathrm{bottom}\:\mathrm{and}\:\mathrm{top}\:\mathrm{be}\:{q} \\ $$$$\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{big}\:\mathrm{circle}\:{C}\:\mathrm{is}\:{R} \\ $$$$\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{small}\:\mathrm{circle}\:{c}\:\mathrm{is}\:{r}={R}−\frac{{p}}{\mathrm{2}} \\ $$$${C}:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{R}^{\mathrm{2}} =\mathrm{0} \\ $$$${c}:\:\left({x}−\frac{{p}}{\mathrm{2}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} −\left({R}−\frac{{p}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0}\:\Leftrightarrow \\ $$$$\Leftrightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{px}−{R}^{\mathrm{2}} +{pR}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}}\\{{R}−{q}}\end{pmatrix}\:\in{c} \\ $$$$\mathrm{put}\:{x}=\mathrm{0}\wedge{y}={R}−{q}\:\mathrm{into}\:{c}\:\Rightarrow \\ $$$$\left({p}−\mathrm{2}{q}\right){R}+{q}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{R}=\frac{{q}^{\mathrm{2}} }{\mathrm{2}{q}−{p}}\:\Rightarrow\:{r}=\frac{\left({p}−{q}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{2}{q}−{p}\right)} \\ $$$$\Rightarrow\:\mathrm{green}\:\mathrm{area}\:=\:\mathrm{area}\:\left({C}\right)\:−\mathrm{area}\:\left({c}\right)\:= \\ $$$$=\frac{{p}\left(\left({p}−{q}\right)^{\mathrm{2}} +\mathrm{3}{q}^{\mathrm{2}} \right)}{\mathrm{4}\left(\mathrm{2}{q}−{p}\right)}\pi \\ $$$$ \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case} \\ $$$${p}=\mathrm{18}\:\:{q}=\mathrm{10} \\ $$$${R}=\mathrm{50}\:\:{r}=\mathrm{41} \\ $$$$\mathrm{green}\:\mathrm{area}\:=\:\mathrm{819}\pi \\ $$

Commented by TawaTawa last updated on 19/Oct/19

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

Commented by TawaTawa last updated on 21/Oct/19

$$\mathrm{Sir},\:\mathrm{help}\:\mathrm{me}\:\mathrm{check}\:\mathrm{question}\:\:\mathrm{71837} \\ $$

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