Question Number 38079 by ajfour last updated on 21/Jun/18 Commented by ajfour last updated on 21/Jun/18 $${Find}\:{coloured}\:{area}\:{ABCD}\:{in} \\ $$$${terms}\:{of}\:{a},{b},{c},{d}\:. \\ $$$${It}\:{is}\:{the}\:{area}\:{common}\:{to}\:{two} \\ $$$${squares}\:{of}\:{side}\:{length}\:{a}. \\ $$…
Question Number 38062 by ajfour last updated on 21/Jun/18 Commented by ajfour last updated on 21/Jun/18 $${how}\:{many}\:{triangles}\:{with}\:{sides} \\ $$$${drawn}\:{in}\:{diagram}\:? \\ $$ Commented by Rasheed.Sindhi last…
Question Number 169088 by cortano1 last updated on 24/Apr/22 Answered by mr W last updated on 24/Apr/22 $${a}=\mathrm{18} \\ $$$${b}=\mathrm{15} \\ $$$${c}=\mathrm{12} \\ $$$${s}=\left(\mathrm{18}+\mathrm{15}+\mathrm{12}\right)/\mathrm{2}=\mathrm{22}.\mathrm{5} \\…
Question Number 38015 by ajfour last updated on 20/Jun/18 Commented by ajfour last updated on 20/Jun/18 $${Find}\:\theta\:{in}\:{terms}\:{of}\:\boldsymbol{{d}},\:\boldsymbol{{R}}\:{if}\:{circle} \\ $$$${has}\:{all}\:{three}\:{coloured}\:{areas}\:{equal}. \\ $$ Commented by MrW3 last…
Question Number 37989 by ajfour last updated on 20/Jun/18 Commented by ajfour last updated on 20/Jun/18 $$\left({i}\right){Find}\:{area}\:{of}\:{blue}\:{square}\:{and}\: \\ $$$${red}\:{square}. \\ $$$$\left({ii}\right){Find}\:\theta\:{for}\:{which}\:{corners}\:{of} \\ $$$${red}\:{square}\:{lies}\:{on}\:{sides}\:{of}\:{blue} \\ $$$${square}.…
Question Number 168979 by MikeH last updated on 22/Apr/22 $${f}\left({x},{y},{z}\right)\:=\:\left(\mathrm{3}{x}^{\mathrm{2}} {y},{x}^{\mathrm{3}} +{y}^{\mathrm{3}} ,\:\mathrm{2}{z}\right) \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{has}\:\mathrm{a}\:\mathrm{potential} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{determined}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 168906 by cortano1 last updated on 21/Apr/22 Answered by mr W last updated on 21/Apr/22 $$\mathrm{tan}\:\frac{{B}}{\mathrm{2}}=\frac{\frac{{b}}{{c}}}{\mathrm{1}+\frac{{a}}{{c}}}=\frac{{b}}{{a}+{c}}=\frac{{b}}{{a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\mathrm{tan}\:\frac{{A}}{\mathrm{2}}=\frac{\frac{{a}}{{c}}}{\mathrm{1}+\frac{{b}}{{c}}}=\frac{{a}}{{b}+{c}}=\frac{{a}}{{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\…
Question Number 103171 by bemath last updated on 13/Jul/20 Commented by bobhans last updated on 13/Jul/20 $$\left(\mathrm{2}{a}\right)\:{a}\:{sector}\:{of}\:{a}\:{circle}\:{with}\:{radius}\:{r}\:{has} \\ $$$${perimeter}\:\mathrm{20}\:{units}\:{and}\:{area}\:\mathrm{21}\:{square} \\ $$$${units}\:.\:{Find}\:{all}\:{possible}\:{of}\:{r}\:{and}\:{the}\: \\ $$$${corresponding}\:{value}\:{of}\:\theta\:. \\ $$…
Question Number 103147 by ajfour last updated on 13/Jul/20 Answered by mr W last updated on 13/Jul/20 $${A}\left({a},\mathrm{0}\right) \\ $$$${B}\left(\mathrm{0},\sqrt{{p}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right) \\ $$$${C}\left(\mathrm{0},{c}\right) \\…
Question Number 168558 by ajfour last updated on 13/Apr/22 Commented by ajfour last updated on 13/Apr/22 $${If}\:{the}\:{shaded}\:{area}\:{be}\:{A}.\:{Find}\:{max} \\ $$$${of}\:{y}_{{P}} \:\left({take}\:{curve}\:{to}\:{be}\:{y}={x}^{\mathrm{2}} \right). \\ $$ Answered by…