Question Number 45366 by behi83417@gmail.com last updated on 12/Oct/18 Commented by behi83417@gmail.com last updated on 12/Oct/18 $${shaded}\:{area}=? \\ $$ Commented by ajfour last updated on…
Question Number 176437 by cherokeesay last updated on 19/Sep/22 Answered by HeferH last updated on 19/Sep/22 $$\:{by}\:{similar}\:{triangles}: \\ $$$$\:{b}\:={AD}\:=\:\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{5}}}\:\:=\:\frac{\left(\mathrm{4}\sqrt{\mathrm{5}}\right)}{\left(\mathrm{1}\:+\:\sqrt{\mathrm{5}}\right)} \\ $$$$\:{h}\:=\:\mathrm{4} \\ $$$$\:{A}_{\bigtriangleup{ABD}} \:=\:\frac{{bh}}{\mathrm{2}}\:=\:\frac{\mathrm{8}\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}\:+\:\mathrm{1}}\:=\:\mathrm{10}\:−\:\mathrm{2}\sqrt{\mathrm{5}\:} \\…
Question Number 110868 by igb last updated on 31/Aug/20 $${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={z} \\ $$$${Level}\:{sets}\:{and}\:{surface}\:{plot} \\ $$$${using}\:{Geogebra} \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 110848 by ajfour last updated on 31/Aug/20 Commented by ajfour last updated on 31/Aug/20 $${Find}\:{c}\:{in}\:{terms}\:{of}\:{a}\:{and}\:{b}. \\ $$ Commented by bemath last updated on…
Question Number 176367 by mr W last updated on 17/Sep/22 Commented by mr W last updated on 17/Sep/22 $${find}\:{the}\:{sum}\:{of}\:{areas}\:{of}\:{all}\:{inscribed} \\ $$$${circles}\:{in}\:{a}\:{triangle}\:{with}\:{side}\:{lengthes} \\ $$$${a},{b},{c}\:{as}\:{shown}. \\ $$…
Question Number 110782 by Aina Samuel Temidayo last updated on 30/Aug/20 $$\mathrm{A}\:\mathrm{triangle}\:\mathrm{has}\:\mathrm{area}\:\mathrm{15}\:\mathrm{and} \\ $$$$\mathrm{circumradius}\:\mathrm{12}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of} \\ $$$$\mathrm{its}\:\mathrm{heights}. \\ $$ Answered by som(math1967) last updated on 30/Aug/20…
Question Number 110781 by Aina Samuel Temidayo last updated on 30/Aug/20 $$\mathrm{Between}\:\mathrm{a}\:\mathrm{square},\mathrm{a}\:\mathrm{triangle}\:\mathrm{and}\:\mathrm{a} \\ $$$$\mathrm{circle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{perimeter},\:\mathrm{which} \\ $$$$\mathrm{shape}\:\mathrm{has}\:\mathrm{the}\:\mathrm{least}\:\mathrm{area}? \\ $$ Commented by mr W last updated on…
Question Number 45187 by MrW3 last updated on 10/Oct/18 Commented by MrW3 last updated on 12/Oct/18 $${Solution}\:{to}\:{Q}\mathrm{45122}. \\ $$$${Find}\:{the}\:{length}\:{of}\:{path}\:{from}\:{A}\:{to}\:{B} \\ $$$${which}\:{has}\:{a}\:{constant}\:{slope}. \\ $$$$ \\ $$$${A}\:{point}\:{P}\:{on}\:{the}\:{path}\:{can}\:{be}\:{described}…
Question Number 176192 by adhigenz last updated on 14/Sep/22 $$\mathrm{Suppose}\:\mathrm{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{rectangle}.\:\mathrm{X}\:\mathrm{and}\:\mathrm{Y}\:\mathrm{are}\:\mathrm{points}\:\mathrm{on}\:\mathrm{BC}\:\mathrm{and}\:\mathrm{CD}\:\mathrm{respectively}, \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{ABX},\:\mathrm{CXY},\:\mathrm{and}\:\mathrm{AYD}\:\mathrm{are}\:\mathrm{3}\:\mathrm{cm}^{\mathrm{2}} ,\:\mathrm{4}\:\mathrm{cm}^{\mathrm{2}} ,\:\mathrm{and}\:\mathrm{5}\:\mathrm{cm}^{\mathrm{2}} \:\mathrm{respectively}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{AXY}. \\ $$ Answered by som(math1967) last updated on…
Question Number 176195 by adhigenz last updated on 14/Sep/22 Answered by behi834171 last updated on 15/Sep/22 $${BC}={BP}={PC}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}},{AB}={AR}={BR}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${PQ}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}−\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}×\mathrm{1}×{cos}\left(\mathrm{150}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}+\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\Rightarrow{PQ}=\sqrt{\frac{\mathrm{7}}{\mathrm{3}}} \\ $$$${cos}\measuredangle{CPQ}=\frac{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{7}}{\mathrm{3}}}\right)^{\mathrm{2}}…