Question-59058

Question Number 59058 by necx1 last updated on 04/May/19

Commented by MJS last updated on 04/May/19

$$\mathrm{same}\:\mathrm{as}\:\mathrm{qu}.\:\mathrm{58377} \\ $$

Answered by tanmay last updated on 04/May/19

eqn circle (x−3)^2 +(y−r)^2 =r^2 which passes through (0,6)and(6,6),(0,a) ,(6,a) (0−3)^2 +(6−r)^2 =r^2 9+36−12r+r^2 =r^2 r=((45)/(12))=((15)/4) (x−3)^2 +(y−((15)/4))^2 =((225)/(16)) (0−3)^2 +(a−((15)/(4 )))^2 =((225)/(16)) 9+a^2 −a×((15)/2)+((225)/(16))=((225)/(16)) 18+2a^2 −15a=0 2a^2 −12a−3a+18=0 2a(a−6)−3(a−6)=0 (a−6)(2a−3)=0 a=(3/2) now area of shaded =6×(6−(3/2)) =6×(9/2)=27 swuare unit... check... (6−3)^2 +(6−((15)/4))^2 =9+((81)/(16))→((144+81)/(16))=((225)/(16))→(((15)/4))^2

$${eqn}\:{circle}\:\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${which}\:{passes}\:{through}\:\left(\mathrm{0},\mathrm{6}\right){and}\left(\mathrm{6},\mathrm{6}\right),\left(\mathrm{0},{a}\right)\:,\left(\mathrm{6},{a}\right) \\ $$$$\left(\mathrm{0}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{6}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{9}+\mathrm{36}−\mathrm{12}{r}+{r}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${r}=\frac{\mathrm{45}}{\mathrm{12}}=\frac{\mathrm{15}}{\mathrm{4}} \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}−\frac{\mathrm{15}}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{225}}{\mathrm{16}} \\ $$$$\left(\mathrm{0}−\mathrm{3}\right)^{\mathrm{2}} +\left({a}−\frac{\mathrm{15}}{\mathrm{4}\:}\right)^{\mathrm{2}} =\frac{\mathrm{225}}{\mathrm{16}} \\ $$$$\mathrm{9}+{a}^{\mathrm{2}} −{a}×\frac{\mathrm{15}}{\mathrm{2}}+\frac{\mathrm{225}}{\mathrm{16}}=\frac{\mathrm{225}}{\mathrm{16}} \\ $$$$\mathrm{18}+\mathrm{2}{a}^{\mathrm{2}} −\mathrm{15}{a}=\mathrm{0} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} −\mathrm{12}{a}−\mathrm{3}{a}+\mathrm{18}=\mathrm{0} \\ $$$$\mathrm{2}{a}\left({a}−\mathrm{6}\right)−\mathrm{3}\left({a}−\mathrm{6}\right)=\mathrm{0} \\ $$$$\left({a}−\mathrm{6}\right)\left(\mathrm{2}{a}−\mathrm{3}\right)=\mathrm{0} \\ $$$${a}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${now}\:{area}\:{of}\:{shaded}\:=\mathrm{6}×\left(\mathrm{6}−\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\mathrm{6}×\frac{\mathrm{9}}{\mathrm{2}}=\mathrm{27}\:{swuare}\:{unit}… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${check}… \\ $$$$\left(\mathrm{6}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{6}−\frac{\mathrm{15}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{9}+\frac{\mathrm{81}}{\mathrm{16}}\rightarrow\frac{\mathrm{144}+\mathrm{81}}{\mathrm{16}}=\frac{\mathrm{225}}{\mathrm{16}}\rightarrow\left(\frac{\mathrm{15}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$

Answered by mr W last updated on 04/May/19

Commented by mr W last updated on 04/May/19

let CD=h OA=(h/2) AD=(6/2)=3 OD=(√(3^2 +((h/2))^2 ))=(√(9+(h^2 /4))) OB=OD=(√(9+(h^2 /4))) AB=OA+OB=(h/2)+(√(9+(h^2 /4)))=6 ⇒h+(√(36+h^2 ))=12 ⇒(√(36+h^2 ))=12−h ⇒36+h^2 =144−24h+h^2 ⇒0=9−2h ⇒h=(9/2) Area=6×h=6×(9/2)=27

$${let}\:{CD}={h} \\ $$$${OA}=\frac{{h}}{\mathrm{2}} \\ $$$${AD}=\frac{\mathrm{6}}{\mathrm{2}}=\mathrm{3} \\ $$$${OD}=\sqrt{\mathrm{3}^{\mathrm{2}} +\left(\frac{{h}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{9}+\frac{{h}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${OB}={OD}=\sqrt{\mathrm{9}+\frac{{h}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${AB}={OA}+{OB}=\frac{{h}}{\mathrm{2}}+\sqrt{\mathrm{9}+\frac{{h}^{\mathrm{2}} }{\mathrm{4}}}=\mathrm{6} \\ $$$$\Rightarrow{h}+\sqrt{\mathrm{36}+{h}^{\mathrm{2}} }=\mathrm{12} \\ $$$$\Rightarrow\sqrt{\mathrm{36}+{h}^{\mathrm{2}} }=\mathrm{12}−{h} \\ $$$$\Rightarrow\mathrm{36}+{h}^{\mathrm{2}} =\mathrm{144}−\mathrm{24}{h}+{h}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{0}=\mathrm{9}−\mathrm{2}{h} \\ $$$$\Rightarrow{h}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${Area}=\mathrm{6}×{h}=\mathrm{6}×\frac{\mathrm{9}}{\mathrm{2}}=\mathrm{27} \\ $$

Commented by necx1 last updated on 04/May/19

$${yes}…{Thanks}\:{for}\:{this}\:{style}.{Its}\:{clear} \\ $$$${now} \\ $$

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