Question Number 15919 by CHANDANA last updated on 15/Jun/17 $${multiply}\:\mathrm{3}{x}+\mathrm{4}{y}+\mathrm{5}{x}−\mathrm{8}{y} \\ $$ Commented by RasheedSoomro last updated on 17/Jun/17 $${Simplify}\:\:\mathrm{3}{x}+\mathrm{4}{y}+\mathrm{5}{x}−\mathrm{8}{y} \\ $$$$\:\:\:\:\:\:\:\:\:{Or} \\ $$$${multiply}\:\left(\mathrm{3}{x}+\mathrm{4}{y}\right)×\left(\mathrm{5}{x}−\mathrm{8}{y}\right) \\…
Question Number 15917 by mrW1 last updated on 15/Jun/17 Commented by mrW1 last updated on 15/Jun/17 $$\mathrm{E},\mathrm{F},\mathrm{G},\mathrm{H}\:\mathrm{are}\:\mathrm{mid}−\mathrm{points}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{quadrilateral}\:\mathrm{ABCD}. \\ $$$$\mathrm{L},\mathrm{K}\:\mathrm{are}\:\mathrm{mid}−\mathrm{points}\:\mathrm{of}\:\mathrm{the}\:\mathrm{diagonals}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{EG},\mathrm{FH},\mathrm{LK}\:\mathrm{intersect}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{point}\:\mathrm{J}\:\:\mathrm{and}\:\mathrm{J}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mid}−\mathrm{point}\:\mathrm{of}\:\:\mathrm{them}.…
Question Number 15908 by mrW1 last updated on 15/Jun/17 Commented by mrW1 last updated on 15/Jun/17 $$\mathrm{On}\:\mathrm{each}\:\mathrm{side}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{constructed}.\:\mathrm{The}\:\mathrm{centroids} \\ $$$$\mathrm{of}\:\mathrm{these}\:\mathrm{triangles}\:\mathrm{form}\:\mathrm{a}\:\mathrm{new}\:\mathrm{triangle}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{new}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{equilateral}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{its}\:\mathrm{sides}\:\mathrm{if}\:\mathrm{the}…
Question Number 15904 by mrW1 last updated on 15/Jun/17 Commented by mrW1 last updated on 15/Jun/17 $$\mathrm{A}\:\mathrm{convex}\:\mathrm{quadrilateral}\:\mathrm{ABCD}\:\mathrm{is} \\ $$$$\mathrm{divided}\:\mathrm{into}\:\mathrm{4}\:\mathrm{triangle}\:\mathrm{by}\:\mathrm{its}\:\mathrm{two} \\ $$$$\mathrm{diagonals}.\:\mathrm{The}\:\mathrm{centroids}\:\mathrm{of}\:\mathrm{these} \\ $$$$\mathrm{triangles}\:\mathrm{form}\:\mathrm{a}\:\mathrm{new}\:\mathrm{quadrilaterial}. \\ $$$$\mathrm{Find}\:\mathrm{out}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{new}\:\mathrm{quadri}−…
Question Number 81331 by mr W last updated on 11/Feb/20 Commented by mr W last updated on 11/Feb/20 $${Given}:\:{the}\:{distances}\:{from}\:{a}\:{point}\:{to} \\ $$$${the}\:{midpoints}\:{of}\:{the}\:{sides}\:{of}\:{a}\:{triangle} \\ $$$${are}\:{p},{q},{r}. \\ $$$${Find}:\:{the}\:{side}\:{lengthes}\:{of}\:{the}\:{triangle}.…
Question Number 81322 by ajfour last updated on 11/Feb/20 Commented by ajfour last updated on 11/Feb/20 $${Find}\:{radius}\:{of}\:{quatercircle} \\ $$$${in}\:{terms}\:{of}\:{a},{b},{c}. \\ $$ Answered by mr W…
Question Number 15784 by mrW1 last updated on 13/Jun/17 Commented by mrW1 last updated on 14/Jun/17 $$\mathrm{ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{quadrilateral}.\: \\ $$$$\mathrm{On}\:\mathrm{each}\:\mathrm{side}\:\mathrm{of}\:\mathrm{it}\:\mathrm{a}\:\mathrm{square}\:\mathrm{is}\:\mathrm{constructed}. \\ $$$$\mathrm{E},\mathrm{F},\mathrm{G},\mathrm{H}\:\mathrm{are}\:\mathrm{center}\:\mathrm{points}\:\mathrm{of}\:\mathrm{these} \\ $$$$\mathrm{squares}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{EG}=\mathrm{FH}\:\mathrm{and} \\ $$$$\mathrm{EG}\bot\mathrm{FH}.…
Question Number 15782 by mrW1 last updated on 13/Jun/17 Commented by mrW1 last updated on 13/Jun/17 $$\mathrm{E}\:\mathrm{and}\:\mathrm{F}\:\mathrm{are}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{AB}\:\mathrm{and}\:\mathrm{CD}.\: \\ $$$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{area}\:\mathrm{yellow}=\mathrm{area}\:\mathrm{green} \\ $$ Answered by…
Question Number 81303 by mr W last updated on 11/Feb/20 Commented by mr W last updated on 11/Feb/20 $${Find}\:{x}=? \\ $$ Answered by mr W…
Question Number 81289 by 37996827 last updated on 11/Feb/20 $${discussthesymmetryofthefollowingcurves} \\ $$$${squareofx}+{squareofy}=\mathrm{1} \\ $$ Commented by mr W last updated on 11/Feb/20 $${itisacirclewithradiusof}\mathrm{1}{andcenterat} \\ $$$${theoriginwhichissymmetricaboutevery}…