Category: Coordinate Geometry

Question Number 190370 by cortano12 last updated on 02/Apr/23 Answered by som(math1967) last updated on 02/Apr/23 Commented by som(math1967) last updated on 02/Apr/23 $${ar}\:{of}\bigtriangleup{FDC}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\mathrm{5}=\mathrm{25}{cm}^{\mathrm{2}} \\…

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Question Number 59058 by necx1 last updated on 04/May/19 Commented by MJS last updated on 04/May/19 $$\mathrm{same}\:\mathrm{as}\:\mathrm{qu}.\:\mathrm{58377} \\ $$ Answered by tanmay last updated on…

Question Number 59000 by Learner last updated on 03/May/19 Answered by MJS last updated on 03/May/19 $$\mathrm{the}\:\mathrm{centers}\:\mathrm{of}\:\mathrm{the}\:\mathrm{small}\:\mathrm{circles}\:\mathrm{form}\:\mathrm{a}\:\mathrm{square} \\ $$$$\mathrm{with}\:\mathrm{side}\:\mathrm{length}\:\mathrm{2}{r}\:\Rightarrow\:\mathrm{its}\:\mathrm{diagonal}\:=\mathrm{2}\sqrt{\mathrm{2}}{r}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{big}\:\mathrm{circle}\:=\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){r}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}\:=\left[\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){r}\right]^{\mathrm{2}} \pi−\mathrm{4}{r}^{\mathrm{2}} \pi=…

Question Number 190019 by ajfour last updated on 26/Mar/23 Commented by ajfour last updated on 26/Mar/23 $${Given}\:{p}\:{and}\:{q}.\:{Find}\:{r}.\: \\ $$$${Parabola}\:{is}\:{y}={x}^{\mathrm{2}} . \\ $$ Commented by mr…