Question Number 83411 by ajfour last updated on 02/Mar/20 Commented by ajfour last updated on 02/Mar/20 $${Given}\:{a}\:{regular}\:{triangular} \\ $$$${pyramid}\:{with}\:{base}\:{sides}\:\boldsymbol{{a}}\:{and} \\ $$$${lateral}\:{edges}\:\boldsymbol{{a}}\sqrt{\mathrm{2}}.\:{A}\:{sphere} \\ $$$${passes}\:{through}\:{A}\:{and}\:{is}\:{tangent} \\ $$$${to}\:{the}\:{lateral}\:{edges}\:{SB}\:{and}\:{SC}…
Question Number 148932 by vvvv last updated on 01/Aug/21 $$\frac{\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} }{\boldsymbol{{bc}}}\boldsymbol{{l}}_{\boldsymbol{{a}}} ^{\mathrm{2}} +\frac{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)^{\mathrm{2}} }{\boldsymbol{{ab}}}\boldsymbol{{l}}_{\boldsymbol{{c}}} ^{\mathrm{2}} +\frac{\left(\boldsymbol{{a}}+\boldsymbol{{c}}\right)^{\mathrm{2}} }{\boldsymbol{{ac}}}\boldsymbol{{l}}_{\boldsymbol{{b}}} ^{\mathrm{2}} =\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{l}}_{\boldsymbol{{b}}} ,\boldsymbol{{l}}_{\boldsymbol{{a}}} ,\boldsymbol{{l}}_{\boldsymbol{{c}}} −\boldsymbol{{bissekterissa}} \\…
Question Number 83369 by Power last updated on 01/Mar/20 Commented by mr W last updated on 01/Mar/20 $$\angle{A}=\mathrm{90}° \\ $$$$\Rightarrow{AD}={BD}={DC}=\mathrm{5} \\ $$ Commented by mathmax…
Question Number 83354 by TawaTawa1 last updated on 01/Mar/20 Commented by jagoll last updated on 01/Mar/20 $$\mathrm{NC}\:=\:\mathrm{MC}\:=\:\mathrm{x} \\ $$$$\mathrm{2x}^{\mathrm{2}} \:=\:\mathrm{25}\:\Rightarrow\mathrm{x}^{\mathrm{2}} \:=\:\frac{\mathrm{25}}{\mathrm{2}} \\ $$$$\mathrm{area}\:\mathrm{triangle}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \:=\:\frac{\mathrm{25}}{\mathrm{4}}\: \\…
Question Number 148886 by DELETED last updated on 01/Aug/21 Answered by Rasheed.Sindhi last updated on 01/Aug/21 $$\frac{{AC}}{\mathrm{sin}\:{B}}=\frac{{BC}}{\mathrm{sin}\:{A}} \\ $$$$\frac{\mathrm{8}}{\mathrm{sin}\:\mathrm{60}}=\frac{\mathrm{6}}{\mathrm{sin}\:\theta} \\ $$$$\frac{\mathrm{8}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\mathrm{6}}{\mathrm{sin}\:\theta} \\ $$$$\frac{\mathrm{6}}{\mathrm{sin}\:\theta}=\frac{\mathrm{16}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{3}}}{\mathrm{16}}×\mathrm{6}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}…
Question Number 148880 by DELETED last updated on 01/Aug/21 Answered by DELETED last updated on 01/Aug/21 $$\mid\mathrm{AC}\mid=\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} }=\mathrm{10}\sqrt{\mathrm{2}}\:\mathrm{cm} \\ $$$$\mid\mathrm{CG}\mid=\mathrm{10}\:\mathrm{cm} \\ $$$$\rightarrow\mathrm{Luas}\:\Delta\mathrm{ACG}: \\ $$$$\:\:\:\:=\:\:\frac{\mathrm{10}×\mathrm{10}\sqrt{\mathrm{2}}\:}{\mathrm{2}}=\mathrm{50}\sqrt{\mathrm{2}}\:\mathrm{cm}^{\mathrm{2}}…
Question Number 148881 by DELETED last updated on 01/Aug/21 Answered by Rasheed.Sindhi last updated on 01/Aug/21 $$\blacktriangle{ABC}=\sqrt{\mathrm{s}\left(\mathrm{s}−\mathrm{a}\right)\left(\mathrm{s}−\mathrm{b}\right)\left(\mathrm{s}−\mathrm{c}\right)} \\ $$$$\:\:\:\:\:\:\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{2}}=\frac{\mathrm{7}+\mathrm{6}+\mathrm{8}}{\mathrm{2}}=\mathrm{10}.\mathrm{5} \\ $$$$\blacktriangle{ABC}=\sqrt{\mathrm{10}.\mathrm{5}\left(\mathrm{10}.\mathrm{5}−\mathrm{6}\right)\left(\mathrm{10}.\mathrm{5}−\mathrm{8}\right)\left(\mathrm{10}.\mathrm{5}−\mathrm{7}\right)} \\ $$$$\:\:\:\:\:\:\:=\sqrt{\mathrm{10}.\mathrm{5}×\mathrm{4}.\mathrm{5}×\mathrm{2}.\mathrm{5}×\mathrm{3}.\mathrm{5}}\approx\mathrm{20}.\mathrm{33}\:\mathrm{cm}^{\mathrm{2}} \\ $$…
Question Number 148868 by DELETED last updated on 01/Aug/21 Answered by DELETED last updated on 01/Aug/21 $$\mathrm{given}\:\mathrm{that},\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{AB}=\mathrm{AC}=\mathrm{BC}=\mathrm{6}\:\mathrm{cm}, \\ $$$$\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{10}\:\mathrm{cm}\:\mathrm{hight},\:\mathrm{calculate}\:\mathrm{it}'\mathrm{s}\:\mathrm{volume}. \\ $$ Commented by DELETED last…
Question Number 148858 by Tawa11 last updated on 31/Jul/21 Answered by EDWIN88 last updated on 01/Aug/21 Commented by EDWIN88 last updated on 01/Aug/21 $${shaded}\:{area}\:=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{2}^{\mathrm{2}} \right)\:+\mathrm{16}−\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{4}^{\mathrm{2}}…
Question Number 17771 by b.e.h.i.8.3.417@gmail.com last updated on 10/Jul/17 Commented by b.e.h.i.8.3.417@gmail.com last updated on 10/Jul/17 $${in}\:{X}\overset{\Delta} {{Y}Z}:\: \\ $$$$\left.\:\:\:\:\mathrm{1}\right){XY}={XZ} \\ $$$$\left.\:\:\:\:\mathrm{2}\right){XS}={ST}={TZ}={ZY}\:\: \\ $$$${find}\:{any}\:{possible}\:{angles}\:{in}\:{this}\: \\…