Question Number 191832 by Mingma last updated on 01/May/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 60647 by ajfour last updated on 23/May/19 Commented by ajfour last updated on 23/May/19 $${Find}\:{maximum}\:{side}\:{length}\:{of}\:{a} \\ $$$${square}\:{circumscribing}\:{a}\:{rectangle} \\ $$$${with}\:{sides}\:{a}\:{and}\:{b}. \\ $$ Commented by…
Question Number 60588 by behi83417@gmail.com last updated on 22/May/19 Commented by behi83417@gmail.com last updated on 22/May/19 $${AD}\:\&{CE}\:{are}\:{angular}\:{bisector}\:{of} \\ $$$$\measuredangle{A}\:\&\measuredangle{C}. \\ $$$${DG}\bot{AB},{EF}\bot{BC},{BH}\bot{DE},{DE}=\frac{\mathrm{1}}{\mathrm{2}}{AC}. \\ $$$$\Rightarrow\frac{{BH}}{{DG}+{EF}}=? \\ $$…
Question Number 126080 by ajfour last updated on 17/Dec/20 Commented by ajfour last updated on 18/Dec/20 $${Cylinder}\:{and}\:{sphere}\:{have}\:{the} \\ $$$${same}\:{radius}.\:{An}\:{equilateral} \\ $$$${triangular}\:{plate}\:{rests}\:{on}\:{sphere} \\ $$$${with}\:{two}\:{vertices}\:{against}\:{curved} \\ $$$${surface}\:{of}\:{cylinder}\:{and}\:{top}\:{vertex}…
Question Number 191602 by ajfour last updated on 26/Apr/23 Commented by ajfour last updated on 26/Apr/23 $${Load}\:{is}\:{to}\:{be}\:{taken}\:{from}\:{A}\:{to}\:{B}. \\ $$$${If}\:{s}={h}\:,\:{find}\:{h}\:\:{in}\:{terms}\:{of}\:{a},{b}. \\ $$ Answered by mr W…
Question Number 60527 by ajfour last updated on 21/May/19 Commented by ajfour last updated on 21/May/19 $$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{of}\:\mathrm{quadrilateral} \\ $$$$\mathrm{OAPB}.\:\mathrm{The}\:\mathrm{ellipse}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{usual}\:\mathrm{one}. \\ $$ Answered by…
Question Number 126052 by fajri last updated on 16/Dec/20 $$ \\ $$$$\mathrm{on}\:\mathrm{any}\:\mathrm{trapezoid}\:\mathrm{ABCD}\:\mathrm{points}\:\mathrm{E}\:\mathrm{and}\: \\ $$$$\mathrm{F}\:\mathrm{are}\:\mathrm{located}\:\mathrm{on}\:\mathrm{CD}\:\mathrm{so}\:\mathrm{that}\:\mathrm{AD}\:\mathrm{is} \\ $$$$\mathrm{paral}{l}\mathrm{el}\:\mathrm{to}\:\mathrm{BE}\:\mathrm{and}\:\mathrm{AF}\:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\: \\ $$$$\mathrm{BC}.\mathrm{Point}\:\mathrm{H}\:\mathrm{is}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{of}\:\mathrm{AF}\: \\ $$$$\mathrm{an}{d}\:\mathrm{BE}\:\mathrm{point}\:\mathrm{G}\:\mathrm{is}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{of}\: \\ $$$$\mathrm{AC}\:\mathrm{and}\:\mathrm{BE}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{AB}\:\mathrm{is}\:\mathrm{4}\:\mathrm{cm} \\ $$$${an}\mathrm{d}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{CD}\:\mathrm{is}\:\mathrm{10}\:\mathrm{cm}\:\mathrm{what}\:\mathrm{is} \\…
Question Number 126049 by fajri last updated on 16/Dec/20 $$ \\ $$$$\mathrm{if}\:\mathrm{in}\:\mathrm{triangle}\:\mathrm{AB}{C},\:\mathrm{AD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{bisector}\:\mathrm{of} \\ $$$$\mathrm{an}{g}\mathrm{le}\:\mathrm{A}\:\mathrm{then}\:\mathrm{BD}\::\:\mathrm{DC}\:=\:\:\:\mathrm{AB}\::\:\mathrm{AC}\: \\ $$$$\mathrm{investigate},\:\mathrm{is}\:\mathrm{the}\:\mathrm{statement}\:\mathrm{true}\:\mathrm{ore} \\ $$$$\mathrm{fals}{e}? \\ $$ Terms of Service Privacy Policy…
Question Number 191577 by TUN last updated on 26/Apr/23 $${Give}\:{A}'{B}'//{AB},\:{B}'{C}'//{BC}\:{and}\:{A}'{C}'//{AC} \\ $$$${Prove}\:{that}:\:\bigtriangleup{ABC}\backsim\bigtriangleup{A}'{B}'{C}' \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191455 by Mingma last updated on 24/Apr/23 Answered by a.lgnaoui last updated on 25/Apr/23 $$\:\bigtriangleup\mathrm{ANB}\:\:\:\mathrm{BM}\bot\mathrm{AN}\:\:\mathrm{et}\:\measuredangle\mathrm{MBA}=\measuredangle\mathrm{MBN} \\ $$$$\:\Rightarrow\mathrm{AB}=\mathrm{BN}=\frac{\boldsymbol{\mathrm{x}}}{\mathrm{cos}\:\boldsymbol{\theta}}\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\:\bigtriangleup\mathrm{PNC}\:\:\:\:\:\:\mathrm{PC}=\mathrm{3cm} \\ $$$$\:\mathrm{BM}\mid\mid\:\mathrm{NP}\:\:\Rightarrow\:\measuredangle\:\mathrm{PNC}=\measuredangle\mathrm{MBN}=\theta=\measuredangle\mathrm{PCN} \\ $$$$\:\Rightarrow\mathrm{NP}=\mathrm{PC}=\mathrm{3}…