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Question-16785




Question Number 16785 by RasheedSoomro last updated on 26/Jun/17
Commented by RasheedSoomro last updated on 26/Jun/17
Without using area-formula find out  the ratio of shaded part to the whole.
$$\mathrm{Without}\:\mathrm{using}\:\mathrm{area}-\mathrm{formula}\:\mathrm{find}\:\mathrm{out} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{part}\:\mathrm{to}\:\mathrm{the}\:\mathrm{whole}. \\ $$
Commented by mrW1 last updated on 26/Jun/17
(((√3)/2))^2 =(3/4)
$$\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Answered by sandy_suhendra last updated on 27/Jun/17
AB=BC=CD=DE=EF=FA=x  JK=KM=MN=NS=SP=(√((1/4)x^2 +(1/4)x^2 −2((1/2)x)^2 cos 120°)) = (1/2)x(√3)       A_(JKMNSP)  : A_(ABCDEF)  = ((1/2)x(√3))^2 : x^2  = (3/4)
$$\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DE}=\mathrm{EF}=\mathrm{FA}=\mathrm{x} \\ $$$$\mathrm{JK}=\mathrm{KM}=\mathrm{MN}=\mathrm{NS}=\mathrm{SP}=\sqrt{\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\right)^{\mathrm{2}} \mathrm{cos}\:\mathrm{120}°}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\sqrt{\mathrm{3}}\:\:\:\:\: \\ $$$$\mathrm{A}_{\mathrm{JKMNSP}} \::\:\mathrm{A}_{\mathrm{ABCDEF}} \:=\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} :\:\mathrm{x}^{\mathrm{2}} \:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by RasheedSoomro last updated on 28/Jun/17
Thanks SANDY!
$$\mathcal{T}{hanks}\:\mathcal{SANDY}! \\ $$
Answered by RasheedSoomro last updated on 28/Jun/17
Commented by RasheedSoomro last updated on 28/Jun/17
Ratio of shaded part to whole               =  ((small triangles in shaded part)/(small triangles in whole))              =((18)/(24))=(3/4)
$$\mathrm{Ratio}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{part}\:\mathrm{to}\:\mathrm{whole} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\frac{\mathrm{small}\:\mathrm{triangles}\:\mathrm{in}\:\mathrm{shaded}\:\mathrm{part}}{\mathrm{small}\:\mathrm{triangles}\:\mathrm{in}\:\mathrm{whole}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{18}}{\mathrm{24}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$
Commented by mrW1 last updated on 28/Jun/17
very fine!
$$\mathrm{very}\:\mathrm{fine}! \\ $$

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