Question Number 25930 by Tinkutara last updated on 16/Dec/17 $$\mathrm{A}\:\mathrm{line}\:\mathrm{passes}\:\mathrm{through}\:{A}\left(−\mathrm{3},\:\mathrm{0}\right)\:\mathrm{and} \\ $$$${B}\left(\mathrm{0},\:−\mathrm{4}\right).\:\mathrm{A}\:\mathrm{variable}\:\mathrm{line}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:{AB}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{to}\:\mathrm{cut}\:{x}\:\mathrm{and}\:{y}-\mathrm{axes}\:\mathrm{at} \\ $$$${M}\:\mathrm{and}\:{N}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{intersection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lines}\:{AN}\:\mathrm{and}\:{BM}. \\ $$ Answered by ajfour last updated…
Question Number 156940 by amin96 last updated on 17/Oct/21 Answered by mr W last updated on 17/Oct/21 Commented by mr W last updated on 17/Oct/21…
Question Number 91275 by I want to learn more last updated on 29/Apr/20 Commented by mr W last updated on 29/Apr/20 $$\left({A}\right) \\ $$ Commented…
Question Number 25609 by behi.8.3.4.17@gmail.com last updated on 12/Dec/17 Commented by behi.8.3.4.17@gmail.com last updated on 12/Dec/17 $$\boldsymbol{\mathrm{BE}}=\boldsymbol{\mathrm{EC}},\boldsymbol{\mathrm{AB}}=\mathrm{12},\boldsymbol{\mathrm{AC}}=\mathrm{10} \\ $$$$\boldsymbol{\mathrm{parallel}}\:\boldsymbol{\mathrm{lines}}\:\boldsymbol{\mathrm{to}}:\boldsymbol{\mathrm{AE}},\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{equal}}\:\boldsymbol{\mathrm{distance}} \\ $$$$\boldsymbol{\mathrm{from}}:\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{B}},\boldsymbol{\mathrm{toward}}\:\boldsymbol{\mathrm{C}},\boldsymbol{\mathrm{divide}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{area}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{A}}\overset{\bigtriangleup} {\boldsymbol{\mathrm{B}C}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{ratio}}:\:\mathrm{1}:\mathrm{4}:\mathrm{2}:\mathrm{3}\:. \\…
Question Number 25605 by behi.8.3.4.17@gmail.com last updated on 11/Dec/17 Commented by behi.8.3.4.17@gmail.com last updated on 12/Dec/17 $$\mathrm{from}\:\mathrm{midpoints}\:\mathrm{of}\:\mathrm{sides}\:\mathrm{draw}\:\mathrm{perpendicular} \\ $$$$\mathrm{lines}\:\mathrm{to}\:\mathrm{opposite}\:\mathrm{sides}.\mathrm{find}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{inner}\:\mathrm{hexagon}\:\mathrm{to}\:\mathrm{area}\:\mathrm{of}:\mathrm{A}\overset{\bigtriangleup} {\mathrm{B}C}. \\ $$ Commented…
Question Number 25602 by behi.8.3.4.17@gmail.com last updated on 11/Dec/17 Commented by behi.8.3.4.17@gmail.com last updated on 11/Dec/17 $$\boldsymbol{{A}}\:\boldsymbol{{circle}}\:\boldsymbol{{passes}}\:\boldsymbol{{through}}\: \\ $$$$\boldsymbol{{mid}}\:\boldsymbol{{points}}\:\boldsymbol{{of}}\:\boldsymbol{{sides}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{triangle}}\:\boldsymbol{{A}}\overset{\bigtriangleup} {\boldsymbol{{B}C}}\:. \\ $$$$\mathrm{find}: \\…
Question Number 25479 by AdolfHitler last updated on 11/Dec/17 Answered by jota^ last updated on 11/Dec/17 $${r}\:=\:{radius}\:{circle} \\ $$$${l}\:=\:{side}\:{square} \\ $$$${r}={l}\sqrt{\mathrm{2}} \\ $$$${The}\:{triangle}\:\bigtriangleup\mathrm{APQ} \\ $$$${A}\left(−\mathrm{r},\:\mathrm{0}\right)…
Question Number 90948 by tw000001 last updated on 27/Apr/20 Commented by tw000001 last updated on 27/Apr/20 $$\mathrm{Find}\:\mid\mathrm{PQ}\mid. \\ $$ Commented by tw000001 last updated on…
Question Number 25375 by soomropakistan last updated on 09/Dec/17 $${what}\:{is}\:{HCF}\:\:{of}\frac{\mathrm{1}}{\mathrm{3}\:\:}\:\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\mathrm{1}}{\mathrm{4}}\:? \\ $$ Answered by Rasheed.Sindhi last updated on 09/Dec/17 $$\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{12}} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{8}}{\mathrm{12}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{12}} \\…
Question Number 25350 by Rasheed.Sindhi last updated on 08/Dec/17 $$\mathrm{If}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{two}\:\mathrm{points}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{plane}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{having} \\ $$$$\mathrm{radius}\:\mathrm{r}\:\mathrm{and}\:\mathrm{mAB}>\mathrm{2r}\:,\mathrm{prove}\:\mathrm{that}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one} \\ $$$$\mathrm{of}\:\mathrm{A}\:\mathrm{or}\:\mathrm{B}\:\mathrm{is}\:\mathrm{outside}\:\mathrm{the}\:\mathrm{circle}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com