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Question Number 181658 by a.lgnaoui last updated on 28/Nov/22

Area of Circle ?

$${Area}\:{of}\:{Circle}\:? \\ $$

Commented by a.lgnaoui last updated on 28/Nov/22

Commented by BaliramKumar last updated on 28/Nov/22

Commented by BaliramKumar last updated on 28/Nov/22

BF×FC = EF×FG  4×4 = 8×(d−8)  2 = d−8  d = 10  2r = 10  r = 5  Area = πr^2  = 25π  Second method  OB^2  = OF^2  + BF^2   r^2  = (8−r)^2  + 4^2   r^2  = 8^2 −2×8×r+r^2  + 16  16r = 80  r = 5  Area = πr^2  = π∙5^2  = 25π

$${BF}×{FC}\:=\:{EF}×{FG} \\ $$$$\mathrm{4}×\mathrm{4}\:=\:\mathrm{8}×\left({d}−\mathrm{8}\right) \\ $$$$\mathrm{2}\:=\:{d}−\mathrm{8} \\ $$$${d}\:=\:\mathrm{10} \\ $$$$\mathrm{2}{r}\:=\:\mathrm{10} \\ $$$${r}\:=\:\mathrm{5} \\ $$$${Area}\:=\:\pi{r}^{\mathrm{2}} \:=\:\mathrm{25}\pi \\ $$$${Second}\:{method} \\ $$$${OB}^{\mathrm{2}} \:=\:{OF}^{\mathrm{2}} \:+\:{BF}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} \:=\:\left(\mathrm{8}−{r}\right)^{\mathrm{2}} \:+\:\mathrm{4}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} \:=\:\mathrm{8}^{\mathrm{2}} −\mathrm{2}×\mathrm{8}×{r}+{r}^{\mathrm{2}} \:+\:\mathrm{16} \\ $$$$\mathrm{16}{r}\:=\:\mathrm{80} \\ $$$${r}\:=\:\mathrm{5} \\ $$$${Area}\:=\:\pi{r}^{\mathrm{2}} \:=\:\pi\centerdot\mathrm{5}^{\mathrm{2}} \:=\:\mathrm{25}\pi \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 28/Nov/22

thank you

$${thank}\:{you} \\ $$

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