Question Number 169009 by mr W last updated on 23/Apr/22 Commented by mr W last updated on 23/Apr/22 $${Q}\mathrm{168830}\:{reposted}\:{with}\:{changed}\:{value}. \\ $$ Commented by infinityaction last…
Question Number 168946 by bagjagugum123 last updated on 22/Apr/22 Commented by bagjagugum123 last updated on 22/Apr/22 $${Prove}\:{that}\::\:{w}=\sqrt{\left({x}+{y}+{z}\right)^{\mathrm{2}} −\mathrm{4}{xy}} \\ $$ Answered by infinityaction last updated…
Question Number 168926 by mr W last updated on 22/Apr/22 Commented by mr W last updated on 22/Apr/22 $${M}={midpoint}\:{of}\:{Aa} \\ $$$${L}={midpoint}\:{of}\:{Bb} \\ $$$${K}={midpoint}\:{of}\:{Cc} \\ $$$${prove}\:…
Question Number 103347 by I want to learn more last updated on 14/Jul/20 Commented by I want to learn more last updated on 14/Jul/20 Commented…
Question Number 37804 by ajfour last updated on 17/Jun/18 Commented by ajfour last updated on 17/Jun/18 $${Find}\:{maximum}\:{area}\:{of} \\ $$$${quadrilateral}\:{MNOP}\:{in}\:{terms} \\ $$$${of}\:{radius}\:\boldsymbol{{R}}\:{of}\:{circle}. \\ $$$$\left({O}\:{is}\:{the}\:{center}\:{of}\:{circle}\right). \\ $$…
Question Number 103338 by I want to learn more last updated on 14/Jul/20 Commented by som(math1967) last updated on 14/Jul/20 $$\mathrm{OC}=\mathrm{14}−\mathrm{r} \\ $$$$\mathrm{again}\:\mathrm{OC}=\mathrm{r}\sqrt{\mathrm{2}} \\ $$$$\therefore\mathrm{r}\sqrt{\mathrm{2}}=\mathrm{14}−\mathrm{r}…
Question Number 103318 by I want to learn more last updated on 14/Jul/20 Commented by som(math1967) last updated on 14/Jul/20 $$\frac{\mathrm{AT}}{\mathrm{AB}}=\mathrm{cos}\angle\mathrm{BAT}=\mathrm{cos30}° \\ $$$$\mathrm{AT}=\mathrm{15}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cm} \\ $$$$\therefore\mathrm{BP}=\mathrm{QD}=\left(\mathrm{15}−\frac{\mathrm{15}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\mathrm{cm}…
Question Number 168830 by amin96 last updated on 18/Apr/22 Commented by mr W last updated on 22/Apr/22 $${seems}\:{wrong}! \\ $$$${PB}={PD}=\mathrm{3} \\ $$$${AB}={AP}+{PB}=\mathrm{9}+\mathrm{3}=\mathrm{12} \\ $$$$\frac{{DB}}{{DP}}=\frac{{AB}}{{DB}}\: \\…
Question Number 37754 by tanmay.chaudhury50@gmail.com last updated on 17/Jun/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 37751 by ajfour last updated on 17/Jun/18 Commented by ajfour last updated on 17/Jun/18 $${Find}\:{area}\:{of}\:{quadrilateral}\:{PQRS} \\ $$$${in}\:{terms}\:{of}\:\boldsymbol{{a}},\:\boldsymbol{{b}}. \\ $$$${ABCD}\:{is}\:{a}\:{square}. \\ $$ Commented by…