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  2. Sum of the areas of two squares is 544 m2. If the difference of their perimeters is 32 m, find the sides of the two squares.     
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Sum of the areas of two squares is 544 m2. If the difference of their perimeters is 32 m, find the sides of the two squares. 

 

 

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\\ \text { Let 'a' and 'b' be the sides of two squares }\\ &\text { According to the question } \\ a^{2}+b^{2}=544 ....(i) \\ 4 a-4 b=32 \\ a-b=8 ....(ii) \\ \text{Put value of a from equation (i) in equation (ii)} \\ (b+8)^2 + b^2 = 544 \\ b^{2}+16 b +64+b^2 = 544 \\ b^{2}+8 b-240=0 \\ b^{2}+20b - 12b -240=0 \\ (b+20)(b-12)=0 \\ b=12 \mathrm{m} \\ \mathrm{a}=12+8=20 \mathrm{m} \\

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