Question Number 5319 by Rasheed Soomro last updated on 07/May/16 $$\mathrm{A}\:\boldsymbol{\mathrm{sphere}}\:\mathrm{of}\:\mathrm{radius}\:\boldsymbol{\mathrm{r}}\:\mathrm{contains}\:\mathrm{a}\:\boldsymbol{\mathrm{cube}} \\ $$$$\mathrm{inside}\:\mathrm{it}.\:\mathrm{All}\:\mathrm{the}\:\boldsymbol{\mathrm{vertices}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cube} \\ $$$$\mathrm{touch}\:\mathrm{the}\:\boldsymbol{\mathrm{surface}}\:\mathrm{of}\:\mathrm{sphere}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\boldsymbol{\mathrm{volume}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cube}? \\ $$ Answered by Rasheed Soomro last updated…
Question Number 136377 by MJS_new last updated on 21/Mar/21 Commented by MJS_new last updated on 21/Mar/21 $$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:“\mathrm{square}''\:\mathrm{is}\:\left(\frac{\pi}{\mathrm{3}}+\mathrm{1}−\sqrt{\mathrm{3}}\right){a}^{\mathrm{2}} \:\mathrm{but} \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{this}\:{without}\:\mathrm{integration}? \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{afraid}\:\mathrm{I}'\mathrm{m}\:\mathrm{a}\:\mathrm{blockhead}… \\ $$ Answered…
Question Number 5308 by Rasheed Soomro last updated on 06/May/16 $$\mathrm{3}\:\mathrm{points}\:\mathrm{with}\:\mathrm{the}\:\mathrm{restriction}\:\mathrm{that} \\ $$$$\mathrm{they}\:\mathrm{should}\:\mathrm{be}\:\mathrm{non}-\mathrm{collinear}\:\mathrm{determine} \\ $$$$\mathrm{circle}. \\ $$$$\mathrm{What}\:\mathrm{number}\:\mathrm{of}\:\mathrm{points}\:\mathrm{with}\:\mathrm{what} \\ $$$$\mathrm{restriction}\:\mathrm{determine}\:\mathrm{sphere}? \\ $$ Terms of Service Privacy…
Question Number 5291 by Rasheed Soomro last updated on 05/May/16 $$\mathrm{A}\:\mathrm{circle}\:\mathrm{has}\:\mathrm{been}\:\mathrm{drawn}\:\mathrm{with}\:\mathrm{one} \\ $$$$\mathrm{unit}\:\mathrm{opened}\:\mathrm{compass}\:\mathrm{on}\:\:\mathrm{surface} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{sphere}\:\mathrm{of}\:\mathrm{one}\:\mathrm{unit}\:\mathrm{radius}.\: \\ $$$$\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}? \\ $$ Commented by Yozzii last updated on…
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Question Number 70672 by ajfour last updated on 06/Oct/19 Commented by ajfour last updated on 06/Oct/19 $${Find}\:{sides}\:{of}\:\bigtriangleup{ABC}.\:{Also}\:{find} \\ $$$$\:{the}\:{same}\:{if}\:{instead}\:{of}\:{EF}\:, \\ $$$$\:{DF}=\mathrm{2}\:. \\ $$ Answered by…
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Question Number 70592 by behi83417@gmail.com last updated on 05/Oct/19 Commented by behi83417@gmail.com last updated on 05/Oct/19 $$\mathrm{convex}\:\mathrm{ABCD},\mathrm{is}\:\mathrm{given}\:\mathrm{with}: \\ $$$$\mathrm{S}\left(\mathrm{DCE}\right)=\mathrm{9},\mathrm{S}\left(\mathrm{CBE}\right)=\mathrm{10},\mathrm{S}\left(\mathrm{BAE}\right)=\mathrm{13}\:. \\ $$$$\:\:\Rightarrow\:\mathrm{S}\left(\mathrm{ADE}\right)=? \\ $$ Answered by…
Question Number 70590 by behi83417@gmail.com last updated on 05/Oct/19 Commented by behi83417@gmail.com last updated on 05/Oct/19 $$\mathrm{convex}\:\mathrm{ABCD},\mathrm{is}\:\mathrm{given}\:\mathrm{with}: \\ $$$$\boldsymbol{\mathrm{S}}_{\mathrm{ABCD}} =\boldsymbol{\mathrm{p}}\:\:\:,\:\:\:\:\:\mathrm{AB}+\mathrm{BD}+\mathrm{DC}=\boldsymbol{\mathrm{q}}\:\:\:. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{maximum}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{for}}:\:\boldsymbol{\mathrm{AC}},\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{terms}} \\ $$$$\boldsymbol{\mathrm{of}}:\:\:\boldsymbol{\mathrm{p}}\:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{q}}. \\…
Question Number 70587 by behi83417@gmail.com last updated on 05/Oct/19 Commented by behi83417@gmail.com last updated on 05/Oct/19 $$\mathrm{ABCD},\mathrm{is}\:\mathrm{given}. \\ $$$$\mathrm{AB}=\boldsymbol{\mathrm{a}},\mathrm{BC}=\boldsymbol{\mathrm{b}},\mathrm{CD}=\boldsymbol{\mathrm{c}},\mathrm{DA}=\boldsymbol{\mathrm{d}},\mathrm{AC}=\boldsymbol{\mathrm{m}},\mathrm{BD}=\boldsymbol{\mathrm{n}}. \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}: \\ $$$$\:\:\:\:\:\left(\boldsymbol{\mathrm{mn}}\right)^{\mathrm{2}} =\left(\boldsymbol{\mathrm{ac}}\right)^{\mathrm{2}} +\left(\boldsymbol{\mathrm{bd}}\right)^{\mathrm{2}}…