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Category: Geometry

Question-77420

Question Number 77420 by BK last updated on 06/Jan/20 Commented by mr W last updated on 06/Jan/20 $${the}\:{three}\:{triangles}\:{are}\:{similar}. \\ $$$${the}\:{radii}\:{of}\:{their}\:{incircles}\:{are}\:{in} \\ $$$${the}\:{same}\:{ratio}\:{as}\:{their}\:{side}\:{lengthes}, \\ $$$${therefore}\:{x}=\mathrm{5}. \\…

S-ABCD-3-2-2-BAO-MAO-22-5-BCM-DCM-S-AOB-

Question Number 11880 by @ANTARES_VY last updated on 03/Apr/17 $$\boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{ABCD}}} =\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\angle\boldsymbol{\mathrm{BAO}}=\angle\boldsymbol{\mathrm{MAO}}=\mathrm{22},\mathrm{5}° \\ $$$$\angle\boldsymbol{\mathrm{BCM}}=\angle\boldsymbol{\mathrm{DCM}} \\ $$$$\boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{AOB}}} =? \\ $$ Terms of Service Privacy Policy…

x-2-dcosx-

Question Number 11827 by uni last updated on 01/Apr/17 $$\int\mathrm{x}^{\mathrm{2}} \mathrm{dcosx}=? \\ $$ Answered by mrW1 last updated on 01/Apr/17 $$={x}^{\mathrm{2}} \mathrm{cos}\:{x}−\int\mathrm{cos}\:{xd}\left({x}^{\mathrm{2}} \right) \\ $$$$={x}^{\mathrm{2}}…

sin-2-7cos-2-6-find-

Question Number 11781 by tawa last updated on 31/Mar/17 $$\mathrm{sin}\left(\mathrm{2}\theta\right)\:+\:\mathrm{7cos}\left(\mathrm{2}\theta\right)\:=\:\mathrm{6} \\ $$$$\mathrm{find}\:\theta \\ $$ Answered by sma3l2996 last updated on 31/Mar/17 $$\mathrm{2}{sin}\left(\theta\right){cos}\left(\theta\right)+\mathrm{7}{cos}^{\mathrm{2}} \left(\theta\right)−\mathrm{7}{sin}^{\mathrm{2}} \left(\theta\right)=\mathrm{6} \\…

A-quadrilateral-with-consecutive-sides-of-lenght-7-15-15-and-d-is-inscribed-in-a-circle-with-its-diameter-d-Find-the-radius-of-circle-

Question Number 11759 by Joel576 last updated on 31/Mar/17 $$\mathrm{A}\:\mathrm{quadrilateral}\:\mathrm{with}\:\mathrm{consecutive}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{lenght}\:\mathrm{7},\:\mathrm{15},\:\mathrm{15},\:\mathrm{and}\:{d} \\ $$$$\mathrm{is}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{its}\:\mathrm{diameter}\:{d}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{circle} \\ $$ Answered by ajfour last updated on 31/Mar/17 Commented by…