Question Number 148821 by vvvv last updated on 31/Jul/21 $$\boldsymbol{{S}}=\frac{\mathrm{4}}{\mathrm{3}}\sqrt{\boldsymbol{{m}}\left(\boldsymbol{{m}}−\boldsymbol{{m}}_{\boldsymbol{{a}}} \right)\left(\boldsymbol{{m}}−\boldsymbol{{m}}_{\boldsymbol{{b}}} \right)\left(\boldsymbol{{m}}−\boldsymbol{{m}}_{\boldsymbol{{c}}} \right)} \\ $$$$\boldsymbol{{m}}=\frac{\boldsymbol{{m}}_{\boldsymbol{{a}}} +\boldsymbol{{m}}_{\boldsymbol{{b}}} +\boldsymbol{{m}}_{\boldsymbol{{c}}} }{\mathrm{2}} \\ $$$$\boldsymbol{{m}}_{\boldsymbol{{a}}} ;\boldsymbol{{m}}_{\boldsymbol{{b}}} ;\boldsymbol{{m}}_{\boldsymbol{{c}}} −\boldsymbol{{mediani}} \\ $$$$\boldsymbol{{prove}}…
Question Number 17743 by b.e.h.i.8.3.417@gmail.com last updated on 10/Jul/17 Answered by mrW1 last updated on 10/Jul/17 $$\mathrm{let}\:\mathrm{k}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{l} \\ $$$$\mathrm{A}\left(\mathrm{0},\mathrm{k},\mathrm{0}\right) \\ $$$$\mathrm{B}\left(\mathrm{k},\mathrm{0},\mathrm{0}\right) \\ $$$$\mathrm{C}\left(\mathrm{0},−\mathrm{k},\mathrm{0}\right) \\ $$$$\mathrm{D}\left(−\mathrm{k},\mathrm{0},\mathrm{0}\right)…
Question Number 148797 by kameda last updated on 31/Jul/21 Answered by puissant last updated on 31/Jul/21 $$\left.\mathrm{1}\right) \\ $$$$\mathrm{sin}\alpha=\frac{\mathrm{AB}}{\mathrm{OA}}\:\Rightarrow\:\mathrm{sin}\alpha=\frac{\mathrm{AB}}{\mathrm{1}}\:\mathrm{car}\:\mathrm{c}'\mathrm{est}\:\mathrm{un} \\ $$$$\mathrm{cercle}\:\mathrm{de}\:\mathrm{rayon}\:\mathrm{1}.. \\ $$$$\Rightarrow\mathrm{AB}=\mathrm{sin}\left(\mathrm{x}\right)\:\mathrm{car}\:\alpha\:\mathrm{est}\:\mathrm{represent}\acute {\mathrm{e}}\:\mathrm{par}\:\mathrm{x}. \\…
Question Number 148775 by Jonathanwaweh last updated on 31/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 17692 by b.e.h.i.8.3.417@gmail.com last updated on 09/Jul/17 Answered by Tinkutara last updated on 09/Jul/17 Commented by b.e.h.i.8.3.417@gmail.com last updated on 09/Jul/17 $${thanks}\:{sir}.{nice}\:{and}\:{smart}. \\…
Question Number 148739 by Tawa11 last updated on 30/Jul/21 Commented by Tawa11 last updated on 30/Jul/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}\:\:\:\:\:\:\:\mathrm{JSMT} \\ $$ Commented by Tawa11 last updated on…
Question Number 17653 by 1kanika# last updated on 09/Jul/17 $$\mathrm{A}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{with}\:\mathrm{its}\:\mathrm{end}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{coordinate} \\ $$$$\mathrm{axes}\:\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{length} \\ $$$$\mathrm{of}\:\mathrm{its}\:\mathrm{intersect}\:\mathrm{on}\:\mathrm{the}\:\mathrm{coordinate}\: \\ $$$$\mathrm{axes}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{C}\:. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{mid}\:\mathrm{points}\:\mathrm{of} \\ $$$$\mathrm{this}\:\mathrm{segment}\:. \\ $$$$\mathrm{Ans}.\:\mathrm{is}\:\:\:\mathrm{8}\left(\mid\mathrm{x}\mid^{\mathrm{3}} +\mid\mathrm{y}\mid^{\mathrm{3}}…
Question Number 17645 by Tinkutara last updated on 09/Jul/17 $$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{the}\:\mathrm{point}\:{M}\:\mathrm{lying}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{interior}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parallelogram}\:{ABCD}, \\ $$$$\mathrm{two}\:\mathrm{parallels}\:\mathrm{to}\:{AB}\:\mathrm{and}\:{AD}\:\mathrm{are}\:\mathrm{drawn}, \\ $$$$\mathrm{intersecting}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:{ABCD}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{points}\:{P},\:{Q},\:{R},\:{S}\:\left(\mathrm{See}\:\mathrm{Figure}\right).\:\mathrm{Prove} \\ $$$$\mathrm{that}\:{M}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{diagonal}\:{AC}\:\mathrm{if}\:\mathrm{and} \\ $$$$\mathrm{only}\:\mathrm{if}\:\left[{MRDS}\right]\:=\:\left[{MPBQ}\right]. \\ $$ Commented…
Question Number 17614 by Tinkutara last updated on 08/Jul/17 $$\mathrm{The}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{CA}\:=\:\mathrm{CB}.\:\mathrm{P}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{between}\:\mathrm{A} \\ $$$$\mathrm{and}\:\mathrm{B}\:\left(\mathrm{and}\:\mathrm{on}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{line}\:\mathrm{AB}\:\mathrm{to}\:\mathrm{C}\right).\:\mathrm{D}\:\mathrm{is}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{perpendicular}\:\mathrm{from}\:\mathrm{C}\:\mathrm{to}\:\mathrm{PB}.\:\mathrm{Show}\:\mathrm{that} \\ $$$$\mathrm{PA}\:+\:\mathrm{PB}\:=\:\mathrm{2}\centerdot\mathrm{PD}. \\ $$ Commented by b.e.h.i.8.3.417@gmail.com…
Question Number 17580 by b.e.h.i.8.3.417@gmail.com last updated on 08/Jul/17 Commented by b.e.h.i.8.3.417@gmail.com last updated on 08/Jul/17 $${triangle}\:{ABC},{is}\:{given}\:{with}: \\ $$$${AB}={c},{BC}={a},{AC}={b} \\ $$$${red}\:\:{lines},{are}\:{perpendicular}\:{bisectors} \\ $$$${of}\:{sides}. \\ $$$$\left.\mathrm{1}\right){find}\:{radius}\:{of}\:{a}\:{circle}\:{that}\:{passes}…