Question Number 47195 by MrW3 last updated on 06/Nov/18 $${Find}\:{the}\:{volume}\:{of}\:{the}\:{pyramid}\:{which} \\ $$$${is}\:{folded}\:{from}\:{a}\:{trangular}\:{paper}\:{with} \\ $$$${sides}\:\boldsymbol{{a}},\:\boldsymbol{{b}}\:{and}\:\boldsymbol{{c}}. \\ $$ Answered by MJS last updated on 06/Nov/18 $$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{use}\:\mathrm{Euler}'\mathrm{s}\:\mathrm{formula} \\…
Question Number 47135 by Tawa1 last updated on 05/Nov/18 Commented by Meritguide1234 last updated on 05/Nov/18 $${ans}\:\left\{\mathrm{2},\mathrm{1}\right\}\:{and}\:\left\{\frac{\mathrm{2}}{\mathrm{5}},−\frac{\mathrm{1}}{\mathrm{5}}\right\} \\ $$ Commented by behi83417@gmail.com last updated on…
Question Number 47113 by MrW3 last updated on 04/Nov/18 Commented by MrW3 last updated on 04/Nov/18 $${Find}\:{the}\:{volume}\:{of}\:{the}\:{triangular} \\ $$$${pyramid}\:{with}\:{given}\:{edge}\:{lengthes}. \\ $$ Commented by MJS last…
Question Number 47001 by behi83417@gmail.com last updated on 03/Nov/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 112500 by I want to learn more last updated on 08/Sep/20 Answered by 1549442205PVT last updated on 08/Sep/20 $$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{smallesr}\:\mathrm{yellow}\:\:\mathrm{part} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{1}}{\mathrm{3}}\mathrm{S}\left(\mathrm{S}=\mathrm{6}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{yellow}\:\mathrm{middle}\:\mathrm{part}\:\mathrm{equal}…
Question Number 178001 by Tawa11 last updated on 11/Oct/22 Commented by Tawa11 last updated on 11/Oct/22 $$\left(\mathrm{b}\right)\:\:\:\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{part}. \\ $$ Answered by aleks041103 last updated on…
Question Number 46919 by behi83417@gmail.com last updated on 02/Nov/18 Commented by ajfour last updated on 02/Nov/18 Commented by ajfour last updated on 02/Nov/18 $${let}\:{CF}\:={h}\:;\:\:{FM}={FB}\:=\:\sqrt{\mathrm{4}−{h}^{\mathrm{2}} }…
Question Number 46906 by behi83417@gmail.com last updated on 02/Nov/18 Commented by behi83417@gmail.com last updated on 03/Nov/18 $${in}\:{A}\overset{\bigtriangleup} {{B}C}:\left({a},{b},{c},\:{as}\:{sides}\right) \\ $$$${D}:\:{midpoint}\:{of}\:{AC}, \\ $$$${BE}:\:{bisect}\:{of}\:\measuredangle{B},{and}\measuredangle{DBD}'. \\ $$$$\boldsymbol{{wanted}}: \\…
Question Number 177978 by mr W last updated on 11/Oct/22 Answered by mr W last updated on 11/Oct/22 $$\boldsymbol{{method}}\:\mathrm{1} \\ $$$${AB}={AF}={DC}=\mathrm{1} \\ $$$${BD}=\mathrm{2}\:\mathrm{sin}\:\mathrm{40}° \\ $$$$\frac{\mathrm{sin}\:\left({C}+\mathrm{110}°\right)}{\mathrm{sin}\:{C}}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{40}°}…
Question Number 46889 by ajfour last updated on 02/Nov/18 Commented by ajfour last updated on 02/Nov/18 $${Q}.\mathrm{46611}\:\left({Alternate}\:{approach}\right) \\ $$ Answered by ajfour last updated on…