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Question Number 112467 by Aina Samuel Temidayo last updated on 08/Sep/20

Answered by 1549442205PVT last updated on 08/Sep/20

Side of the square equal to (a/( (√2))) Width of rectangle equal to ((2a)/( (√2)))=a(√2) The area of rectangle is S_1 =2a.a(√2)=2a^2 (√2) (1) The area of semicircle equal to (1/2)π((a/(2(√2))))^2 =((πa^2 )/(16))(2).Side of the right triangle equal to ((a(√2))/( (√2)))=a−this is also side of two squares The area of two squares and triangle equal to 2a^2 +(1/2)a^2 =(5/2).a^2 =(3) From (1)(2)(3) we infer grey area equal to 2a^2 (√2) +((πa^2 )/(16))+(5/2)a^2 −(a^2 /2) =((a^2 (𝛑+32+32(√2)))/(16))

$$\mathrm{Side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{a}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{Width}\:\mathrm{of}\:\mathrm{rectangle}\:\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{2a}}{\:\sqrt{\mathrm{2}}}=\mathrm{a}\sqrt{\mathrm{2}} \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{rectangle}\:\mathrm{is}\:\mathrm{S}_{\mathrm{1}} =\mathrm{2a}.\mathrm{a}\sqrt{\mathrm{2}}=\mathrm{2a}^{\mathrm{2}} \sqrt{\mathrm{2}}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{semicircle}\:\mathrm{equal}\:\mathrm{to}\:\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\frac{\mathrm{a}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$=\frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{16}}\left(\mathrm{2}\right).\mathrm{Side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{right}\:\mathrm{triangle}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\frac{\mathrm{a}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}=\mathrm{a}−\mathrm{this}\:\mathrm{is}\:\mathrm{also}\:\mathrm{side}\:\mathrm{of}\:\mathrm{two}\:\mathrm{squares} \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{two}\:\mathrm{squares}\:\mathrm{and} \\ $$$$\mathrm{triangle}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{2a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{a}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}.\mathrm{a}^{\mathrm{2}} =\left(\mathrm{3}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right)\left(\mathrm{2}\right)\left(\mathrm{3}\right)\:\mathrm{we}\:\mathrm{infer}\:\mathrm{grey}\:\mathrm{area}\: \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{2a}^{\mathrm{2}} \sqrt{\mathrm{2}}\:+\frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{16}}+\frac{\mathrm{5}}{\mathrm{2}}\mathrm{a}^{\mathrm{2}} −\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{\boldsymbol{\mathrm{a}}^{\mathrm{2}} \left(\boldsymbol{\pi}+\mathrm{32}+\mathrm{32}\sqrt{\mathrm{2}}\right)}{\mathrm{16}} \\ $$

Commented by Aina Samuel Temidayo last updated on 08/Sep/20

Ok. Thanks.

$$\mathrm{Ok}.\:\mathrm{Thanks}. \\ $$

Commented by 1549442205PVT last updated on 13/Sep/20

You are welcome

$$\mathrm{You}\:\mathrm{are}\:\mathrm{welcome} \\ $$

Commented by Aina Samuel Temidayo last updated on 13/Sep/20

Is your User ID ′...′ ?

$$\mathrm{Is}\:\mathrm{your}\:\mathrm{User}\:\mathrm{ID}\:'...'\:? \\ $$

Commented by Rasheed.Sindhi last updated on 21/Sep/20

width of the rectangle =diagonal of the square=a≠a(√2)

$${width}\:{of}\:{the}\:{rectangle} \\ $$$$={diagonal}\:{of}\:{the}\:{square}={a}\neq\mathrm{a}\sqrt{\mathrm{2}} \\ $$

Commented by 1549442205PVT last updated on 21/Sep/20

In question given side of square equal to a⇒diagonal of sq.=a(√2)

$$\mathrm{In}\:\mathrm{question}\:\mathrm{given}\:\mathrm{side}\:\mathrm{of}\:\mathrm{square}\:\mathrm{equal} \\ $$$$\mathrm{to}\:\mathrm{a}\Rightarrow\mathrm{diagonal}\:\mathrm{of}\:\mathrm{sq}.=\mathrm{a}\sqrt{\mathrm{2}} \\ $$

Commented by Rasheed.Sindhi last updated on 21/Sep/20

Then the diagram is incorrect.

$${Then}\:{the}\:{diagram}\:{is}\:{incorrect}. \\ $$

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