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

Commented by MJS last updated on 04/May/19

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

Answered by tanmay last updated on 04/May/19

$$eqncircle{(x-3)}^2+{(y-r)}^2=r^2$$$$whichpassesthrough(0,6)and(6,6),(0,a),(6,a)$$$${(0-3)}^2+{(6-r)}^2=r^2$$$$9+36-12r+r^2=r^2$$$$r=\frac{45}{12}=\frac{15}{4}$$$${(x-3)}^2+{(y-\frac{15}{4})}^2=\frac{225}{16}$$$${(0-3)}^2+{(a-\frac{15}{4})}^2=\frac{225}{16}$$$$9+a^2-a\times \frac{15}{2}+\frac{225}{16}=\frac{225}{16}$$$$18+2a^2-15a=0$$$$2a^2-12a-3a+18=0$$$$2a(a-6)-3(a-6)=0$$$$(a-6)(2a-3)=0$$$$a=\frac{3}{2}$$$$nowareaofshaded=6\times (6-\frac{3}{2})$$$$=6\times \frac{9}{2}=27swuareunit...$$$$$$$$$$$$$$$$$$$$$$$$$$$$check...$$$${(6-3)}^2+{(6-\frac{15}{4})}^2$$$$=9+\frac{81}{16}\to \frac{144+81}{16}=\frac{225}{16}\to {\left(\frac{15}{4}\right)}^2$$

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

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

$$letCD=h$$$$OA=\frac{h}{2}$$$$AD=\frac{6}{2}=3$$$$OD=\sqrt{3^2+{\left(\frac{h}{2}\right)}^2}=\sqrt{9+\frac{h^2}{4}}$$$$OB=OD=\sqrt{9+\frac{h^2}{4}}$$$$AB=OA+OB=\frac{h}{2}+\sqrt{9+\frac{h^2}{4}}=6$$$$\Rightarrow h+\sqrt{36+h^2}=12$$$$\Rightarrow \sqrt{36+h^2}=12-h$$$$\Rightarrow 36+h^2=144-24h+h^2$$$$\Rightarrow 0=9-2h$$$$\Rightarrow h=\frac{9}{2}$$$$Area=6\times h=6\times \frac{9}{2}=27$$

Commented by necx1 last updated on 04/May/19

$$yes...Thanksforthisstyle.Itsclear$$$$now$$

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