sum of areas of two squares is 468msq . if the difference of their perimeters is 24m find the sides of the two squares .....
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a^2 + b^2 = 468
4a - 4b = 24
4(a-b) = 24
a-b=6
a=b+6
a^2=b^2 + 12b + 36
b^2=b^2
b^2 + b^2 + 12b=468-36
2(b^2) +12b=432
b^2 + 12b=216
b(b+12)=216
4a - 4b = 24
4(a-b) = 24
a-b=6
a=b+6
a^2=b^2 + 12b + 36
b^2=b^2
b^2 + b^2 + 12b=468-36
2(b^2) +12b=432
b^2 + 12b=216
b(b+12)=216
b^2 + 6b - 216 = 0
b = 12
a = 18
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