Question Number 105992 by Algoritm last updated on 02/Aug/20 Answered by mr W last updated on 02/Aug/20 $${let}\:{A}={Radius}\:{of}\:{big}\:{semicircle} \\ $$$$\sqrt{\left({A}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }+\sqrt{\left({A}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }=\sqrt{\left(\mathrm{3}+\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{3}−\mathrm{2}\right)^{\mathrm{2}}…
Question Number 171528 by cortano1 last updated on 17/Jun/22 Answered by som(math1967) last updated on 17/Jun/22 $$\left({x}+{y}+{z}\right)^{\mathrm{2}} =\mathrm{162} \\ $$$$\Rightarrow\left({x}+{y}+{z}\right)=\sqrt{\mathrm{162}}=\mathrm{9}\sqrt{\mathrm{2}} \\ $$$$\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{58}…
Question Number 105985 by Algoritm last updated on 02/Aug/20 Answered by mr W last updated on 02/Aug/20 $${r}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{r}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{3}}} \\ $$$${r}=\frac{\mathrm{6}}{\mathrm{2}+\sqrt{\mathrm{3}}}=\mathrm{6}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$${S}=\pi{r}^{\mathrm{2}} =\mathrm{36}\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)\pi \\ $$…
Question Number 105952 by Study last updated on 01/Aug/20 Answered by Her_Majesty last updated on 01/Aug/20 $${green}={square}−\mathrm{2}×\left({square}−{quarter}\right. \\ $$$$\left.{circle}\right)=\mathrm{2}×{quarter}\:{circle}−{square}={half} \\ $$$${circle}−{square}=\mathrm{2}\pi−\mathrm{4} \\ $$ Answered by…
Question Number 105945 by Algoritm last updated on 01/Aug/20 Answered by 1549442205PVT last updated on 02/Aug/20 Commented by 1549442205PVT last updated on 02/Aug/20 $$\mathrm{2R}+\mathrm{BC}=\mathrm{AB}+\mathrm{AC}\Rightarrow\mathrm{R}=\frac{\mathrm{6}+\mathrm{8}−\mathrm{10}}{\mathrm{2}}=\mathrm{2} \\…
Question Number 40255 by ajfour last updated on 17/Jul/18 Commented by MJS last updated on 17/Jul/18 $$\mathrm{are}\:\mathrm{the}\:\mathrm{trapezoids}\:\mathrm{on}\:\mathrm{the}\:\mathrm{right}\:\mathrm{similar}? \\ $$ Commented by MJS last updated on…
Question Number 105769 by ajfour last updated on 31/Jul/20 Commented by ajfour last updated on 31/Jul/20 $${ACDB}\:{is}\:{a}\:{straight}\:{line}.\:{The} \\ $$$${three}\:{coloured}\:{regions}\:{have}\:{equal} \\ $$$${areas}.\:{Given}\:{OA}={p},\:{OB}={q}. \\ $$$${Find}\:{coordinates}\:{of}\:{A}\:{and}\:{B}. \\ $$…
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Question Number 171255 by mnjuly1970 last updated on 11/Jun/22 Commented by mr W last updated on 11/Jun/22 $${S}_{\Delta{APB}} =\frac{{AB}×{AP}\:\mathrm{cos}\:\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{AD}×{AP}\:\mathrm{cos}\:\theta}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{AP}^{\mathrm{2}} }{\mathrm{4}} \\…
Question Number 40170 by ajfour last updated on 16/Jul/18 Commented by ajfour last updated on 16/Jul/18 Commented by ajfour last updated on 16/Jul/18 $${p}=\frac{\mathrm{1}}{\mathrm{2sin}\:\alpha}\:\:\:\:\:\:{q}=\frac{\mathrm{1}}{\mathrm{2sin}\:\beta} \\…