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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.     

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. 

 

 

Answers (1)

\\ \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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