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Question Number 35968 by RoachDN last updated on 26/May/18

Commented by Joel579 last updated on 26/May/18

$$\mathrm{Would}\mathrm{you}\mathrm{mind}\mathrm{to}\mathrm{translate}\mathrm{the}\mathrm{question}\mathrm{into}\mathrm{english}?$$

Answered by tanmay.chaudhury50@gmail.com last updated on 28/May/18

$$letsideofbigsquare=A$$$$sideofsmallsquare=a$$$$distancefromcentreofcircletosmallswuare$$$$is(\frac{A}{2}+a)$$$${(\frac{A}{2}+a)}^2+{\left(\frac{a}{2}\right)}^2=r^2$$$$again2r=A\sqrt{2}$$$${(\frac{A}{2}+a)}^2+{\left(\frac{a}{2}\right)}^2=\frac{A^2}{2}$$$$\frac{A^2}{4}+Aa+a^2+\frac{a^2}{4}=\frac{A^2}{2}$$$$A^2+4Aa+4a^2+a^2=2A^2$$$$5a^2+4Aa-A^2=0$$$$5a^2+5Aa-Aa-A^2=0$$$$5a(a+A)-A(a+A)=0$$$$(5a-A)(a+A)=0$$$$soa=\frac{A}{5}$$$$a^2=\frac{A^2}{25}$$$$A\sqrt{2}=2r$$$$A=\frac{2r}{\sqrt{2}}$$$$A=\frac{2\times 14\sqrt{2}}{\sqrt{2}}=28$$$$a^2=\frac{28\times 28}{25}=31.36$$$$so4a^2=4\times 31.36=125.44$$

Answered by MJS last updated on 27/May/18

$$PQ=a=28;WX=x$$$$\mathrm{center}\mathrm{of}\mathrm{circle}=I,\mathrm{radius}=r=14\sqrt{2}$$$$\frac{W+X}{2}=J$$$$\mathrm{triangle}IJX$$$$\mid IJ{\mid }^2+\mid JX{\mid }^2=\mid IX{\mid }^2$$$${(14+x)}^2+{\left(\frac{x}{2}\right)}^2={\left(14\sqrt{2}\right)}^2$$$$\frac{5}{4}{x}^2+28x+196=392$$$$x^2+\frac{112}{5}x-\frac{784}{5}=0$$$$x_1=-28\Rightarrow \mathrm{not}\mathrm{useable}$$$$x_2=\frac{28}{5}=x$$$$x^2=\frac{784}{25}=31.36=1\mathrm{square}$$$$4x^2=125.44=4\mathrm{squares}$$$$\mathrm{there}\mathrm{might}\mathrm{be}\mathrm{an}\mathrm{error}\mathrm{of}\mathrm{translation}...$$

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