Question Number 40615 by behi83417@gmail.com last updated on 24/Jul/18 Commented by behi83417@gmail.com last updated on 24/Jul/18 $${A}\overset{} {{C}B}=\mathrm{90}^{\bullet} ,{A}\overset{} {{C}E}={E}\overset{} {{C}M}={M}\overset{} {{C}B} \\ $$$${AC}={CM}=\mathrm{5},{CB}=\mathrm{12},{MN}\bot{AB}. \\…
Question Number 40610 by ajfour last updated on 24/Jul/18 Answered by MJS last updated on 25/Jul/18 $$\mathrm{easy}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{with}\:\mathrm{4}+\mathrm{1}\:\mathrm{spheres} \\ $$$$\mathrm{let}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{spheres}\:=\:\mathrm{1} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cone}\:\mathrm{is}\:\mathrm{45}° \\ $$$$\mathrm{its}\:\mathrm{radius}=\mathrm{height}\:\mathrm{is}\:\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{cut}\:\mathrm{through}\:\mathrm{the}\:\mathrm{centers}\:\mathrm{of}\:\mathrm{2}\:\mathrm{opposite}\:\mathrm{bottom}\right.…
Question Number 171677 by cherokeesay last updated on 19/Jun/22 Answered by mr W last updated on 19/Jun/22 $${R}=\mathrm{2}{r} \\ $$$${rectangle}\:{a}×{b}\:{with}\:{b}=\mathrm{2}{r} \\ $$$${a}=\sqrt{\left({r}+{R}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{2}}{r} \\…
Question Number 40594 by ajfour last updated on 24/Jul/18 Commented by ajfour last updated on 24/Jul/18 $${Solution}\:{to}\:{Q}.\mathrm{40581} \\ $$ Answered by ajfour last updated on…
Question Number 40581 by ajfour last updated on 24/Jul/18 Commented by ajfour last updated on 24/Jul/18 $${Find}\:{radius}\:{of}\:{inscribed}\:{sphere} \\ $$$${which}\:{is}\:{tangent}\:{to}\:{all}\:{faces}\:{of} \\ $$$${the}\:{triangular}\:{pyramid}. \\ $$ Commented by…
Question Number 40489 by ajfour last updated on 22/Jul/18 Commented by ajfour last updated on 22/Jul/18 $${In}\:{terms}\:{of}\:{radius}\:{of}\:{circle}\:{R}, \\ $$$${find}\:{a},\:{and}\:{b}. \\ $$ Commented by behi83417@gmail.com last…
Question Number 40457 by behi83417@gmail.com last updated on 22/Jul/18 Answered by MrW3 last updated on 28/Jul/18 $${Q}\mathrm{1}: \\ $$$${the}\:{inscribed}\:{triangle}\:{with}\:{minimal} \\ $$$${perimeter}\:{in}\:{a}\:{given}\:{acute}\:{triangle}\:{is} \\ $$$${that}\:{on}\:{which}\:{links}\:{the}\:{feet}\:{of}\:{the} \\ $$$${altitudes}\:{of}\:{the}\:{given}\:{triangle}.…
Question Number 105992 by Algoritm last updated on 02/Aug/20 Answered by mr W last updated on 02/Aug/20 $${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}}} \\ $$$${r}=\frac{\mathrm{6}}{\mathrm{2}+\sqrt{\mathrm{3}}}=\mathrm{6}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$${S}=\pi{r}^{\mathrm{2}} =\mathrm{36}\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)\pi \\ $$…