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Question Number 18799 by tawa tawa last updated on 29/Jul/17

Answered by Tinkutara last updated on 30/Jul/17

$$\mathrm{Diagonal}\mathrm{of}{1}^{\mathrm{st}}\mathrm{square}=\frac{12\sqrt{2}}{2}$$$$\mathrm{Diagonal}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{square}=\frac{12\sqrt{2}}{4}$$$$\mathrm{Diagonal}\mathrm{of}{3}^{\mathrm{rd}}\mathrm{square}=\frac{12\sqrt{2}}{8}$$$$\mathrm{Area}\mathrm{of}{1}^{\mathrm{st}}\mathrm{square}={\left(\frac{12\sqrt{2}}{2}\right)}^2\mathbf{÷}2=36$$$$\mathrm{Area}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{square}={\left(\frac{12\sqrt{2}}{4}\right)}^2\mathbf{÷}2=9$$$$\mathrm{Area}\mathrm{of}{3}^{\mathrm{rd}}\mathrm{square}={\left(\frac{12\sqrt{2}}{8}\right)}^2\mathbf{÷}2=\frac{9}{4}$$$$\mathrm{This}\mathrm{is}\mathrm{a}\mathrm{GP}\mathrm{with}a=36\mathrm{and}r=\frac{1}{4}.$$$$\mathrm{Sum}\mathrm{of}10\mathrm{terms}=\frac{36}{1-\frac{1}{4}}=48$$$$\mathrm{Hence}\mathrm{sum}\mathrm{of}\mathrm{areas}\mathrm{of}\mathrm{all}\mathrm{these}\mathrm{squares}$$$$\mathrm{is}48\mathrm{sq}.\mathrm{units}.$$$$\mathrm{Area}\mathrm{remaining}=144-48=96\mathrm{sq}.\mathrm{units}$$

Commented by tawa tawa last updated on 30/Jul/17

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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