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Question Number 180423 by mr W last updated on 12/Nov/22

Commented by Rasheed.Sindhi last updated on 12/Nov/22

I think sir that the area depends upon mO_1 O_2 ^(−)

$$\mathcal{I}\:{think}\:{sir}\:{that}\:{the}\:{area}\:{depends} \\ $$$${upon}\:{m}\overline {\mathrm{O}_{\mathrm{1}} \mathrm{O}_{\mathrm{2}} } \\ $$

Commented by mr W last updated on 12/Nov/22

O_1 and O_2 are clearly defined: the center of one circle lies on the other circle.

$${O}_{\mathrm{1}} \:{and}\:{O}_{\mathrm{2}} \:{are}\:{clearly}\:{defined}:\: \\ $$$${the}\:{center}\:{of}\:{one}\:{circle}\:{lies}\:{on}\:{the}\: \\ $$$${other}\:{circle}. \\ $$

Commented by mr W last updated on 12/Nov/22

find the area of the square.

$${find}\:{the}\:{area}\:{of}\:{the}\:{square}. \\ $$

Commented by Rasheed.Sindhi last updated on 12/Nov/22

Right sir!

$$\mathcal{R}{ight}\:\boldsymbol{{sir}}! \\ $$

Answered by JDamian last updated on 12/Nov/22

((((√7)−1)/2))^2 ?

$$\left(\frac{\sqrt{\mathrm{7}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ? \\ $$

Commented by mr W last updated on 12/Nov/22

yes

$${yes} \\ $$

Answered by mr W last updated on 12/Nov/22

1^2 =((1/2))^2 +((a/( (√2))))^2 +2×(1/2)×(a/( (√2)))×((√2)/2) 2a^2 +2a−3=0 a=((−1+(√7))/2) ⇒area a^2 =2−((√7)/2)

$$\mathrm{1}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{a}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{\:\sqrt{\mathrm{2}}}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{3}=\mathrm{0} \\ $$$${a}=\frac{−\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\Rightarrow{area}\:{a}^{\mathrm{2}} =\mathrm{2}−\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$

Commented by mr W last updated on 12/Nov/22

Commented by mr W last updated on 12/Nov/22

((1/2)+(a/2))^2 +((a/2))^2 =1^2 2a^2 +2a−3=0 a=((−1+(√7))/2)

$$\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{3}=\mathrm{0} \\ $$$${a}=\frac{−\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$

Commented by ajfour last updated on 12/Nov/22

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Commented by ajfour last updated on 12/Nov/22

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$${i}\:{posted}\:\mathrm{4500}\:{songs}'\:{cover}\:\:{in}\: \\ $$$$\mathrm{10}\:{months}!\:\left({Hindi}\:\&\:{English}\right) \\ $$

Commented by mr W last updated on 12/Nov/22

thanks sir! nice to see that you are back to here!

$${thanks}\:{sir}! \\ $$$${nice}\:{to}\:{see}\:{that}\:{you}\:{are}\:{back}\:{to}\:{here}! \\ $$

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