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Question Number 58876 by mr W last updated on 01/May/19

Commented by math1967 last updated on 01/May/19

Is it (1/3) part sir?

Isit13partsir?

Commented by mr W last updated on 01/May/19

correct sir!

correctsir!

Commented by math1967 last updated on 01/May/19

△AOB∼△EOD ∴((BO)/(DO))=((AB)/(DE))=(2/1)  Again ((Ar△AOB)/(Ar△AOD))=((BO)/(DO))=(2/1)  ∴△AOB:△DAB=2:3  ∴△AOB=(2/3)△DAB=(2/3)×(1/2)ABCD                                                  =(1/3)ABCD

AOBEODBODO=ABDE=21AgainArAOBArAOD=BODO=21AOB:DAB=2:3AOB=23DAB=23×12ABCD=13ABCD

Commented by math1967 last updated on 01/May/19

Answered by tanmay last updated on 01/May/19

tanθ=((a/2)/a)=(1/2)  the eq of st lines which enclose the shaded area  (1) x axis   (2) (x/a)+(y/a)=1(3)  y=tan(90−θ)x  y=cotθx→y=2x  now solve y=2x and x+y=a  3x=a  x=(a/3) and y=((2a)/3)  the shaded triangle  say ABC  point A(0,0)  B(a,0)  C((a/3),((2a)/3))  area of △ABC=(1/2)×a×((2a)/3)=(a^2 /3)  area of shaded triangle=(1/3)(area of square)

tanθ=a2a=12theeqofstlineswhichenclosetheshadedarea(1)xaxis(2)xa+ya=1(3)y=tan(90θ)xy=cotθxy=2xnowsolvey=2xandx+y=a3x=ax=a3andy=2a3theshadedtrianglesayABCpointA(0,0)B(a,0)C(a3,2a3)areaofABC=12×a×2a3=a23areaofshadedtriangle=13(areaofsquare)

Commented by tanmay last updated on 01/May/19

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