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Question Number 85950 by TawaTawa1 last updated on 26/Mar/20

Commented by TawaTawa1 last updated on 26/Mar/20

Which colour area is bigger. please show workings.

$$\mathrm{Which}\:\mathrm{colour}\:\mathrm{area}\:\mathrm{is}\:\mathrm{bigger}. \\ $$$$\mathrm{please}\:\mathrm{show}\:\mathrm{workings}. \\ $$

Answered by jagoll last updated on 26/Mar/20

Commented by jagoll last updated on 26/Mar/20

blue area = 2×((1/4)πR^2 −(1/2)R^2 ) = R^2 ((1/2)π−1) = (1/2)R^2 (π−2)

$$\mathrm{blue}\:\mathrm{area}\:=\:\mathrm{2}×\left(\frac{\mathrm{1}}{\mathrm{4}}\pi\mathrm{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{R}^{\mathrm{2}} \right) \\ $$$$=\:\mathrm{R}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\pi−\mathrm{1}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{R}^{\mathrm{2}} \left(\pi−\mathrm{2}\right) \\ $$

Commented by jagoll last updated on 26/Mar/20

black area = (1/4)π(2R)^2 −((1/2)πR^2 +R^2 ) = πR^2 −(1/2)πR^2 −R^2 = (1/2)πR^2 −R^2 = (1/2)R^2 (π−2)

$$\mathrm{black}\:\mathrm{area}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{2R}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\pi\mathrm{R}^{\mathrm{2}} +\mathrm{R}^{\mathrm{2}} \right) \\ $$$$=\:\pi\mathrm{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\pi\mathrm{R}^{\mathrm{2}} −\mathrm{R}^{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\pi\mathrm{R}^{\mathrm{2}} −\mathrm{R}^{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{R}^{\mathrm{2}} \left(\pi−\mathrm{2}\right) \\ $$

Commented by TawaTawa1 last updated on 26/Mar/20

God bless you sir. I appreciate your time.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}. \\ $$

Commented by jagoll last updated on 26/Mar/20

blue area = black area.

$$\mathrm{blue}\:\mathrm{area}\:=\:\mathrm{black}\:\mathrm{area}. \\ $$

Commented by TawaTawa1 last updated on 26/Mar/20

Sir, please explain the blue area why it is 2 × ((1/4)πR^2 − (1/2)R^2 )

$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{explain}\:\mathrm{the}\:\mathrm{blue}\:\mathrm{area}\:\mathrm{why}\:\mathrm{it}\:\mathrm{is}\:\:\mathrm{2}\:×\:\left(\frac{\mathrm{1}}{\mathrm{4}}\pi\mathrm{R}^{\mathrm{2}} \:\:−\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{R}^{\mathrm{2}} \right) \\ $$

Commented by TawaTawa1 last updated on 26/Mar/20

Also black area same thing.

$$\mathrm{Also}\:\mathrm{black}\:\mathrm{area}\:\mathrm{same}\:\mathrm{thing}. \\ $$

Commented by jagoll last updated on 26/Mar/20

the blue area is twice the width of the circle

$$\mathrm{the}\:\mathrm{blue}\:\mathrm{area}\:\mathrm{is}\:\mathrm{twice}\:\mathrm{the}\:\mathrm{width} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{circle} \\ $$

Commented by jagoll last updated on 26/Mar/20

Commented by jagoll last updated on 26/Mar/20

look at picture

$$\mathrm{look}\:\mathrm{at}\:\mathrm{picture} \\ $$

Commented by TawaTawa1 last updated on 26/Mar/20

Alright sir, what of (1/4)πR^2 . and (1/2)πR^2 Sorry for two many question sir

$$\mathrm{Alright}\:\mathrm{sir},\:\:\mathrm{what}\:\mathrm{of}\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\pi\mathrm{R}^{\mathrm{2}} .\:\:\:\mathrm{and}\:\:\frac{\mathrm{1}}{\mathrm{2}}\pi\mathrm{R}^{\mathrm{2}} \\ $$$$\mathrm{Sorry}\:\mathrm{for}\:\mathrm{two}\:\mathrm{many}\:\mathrm{question}\:\mathrm{sir} \\ $$

Commented by jagoll last updated on 26/Mar/20

in blue are, the area of one tile is a quarter of the area of a circle minus the area of a triangle

$$\mathrm{in}\:\mathrm{blue}\:\mathrm{are},\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{one}\:\mathrm{tile} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{quarter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle} \\ $$$$\mathrm{minus}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$ \\ $$

Commented by TawaTawa1 last updated on 26/Mar/20

God bless you sir. I appreciate.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by jagoll last updated on 26/Mar/20

thank you

$$\mathrm{thank}\:\mathrm{you}\: \\ $$

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