Sum of areas of two squares is 468msq . if the difference of their per

1.	Sum of areas of two squares is 468msq . if the difference of their perimeters is 24m find the sides of the two squares .....need step by step ans
Answer» 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 b^2 + 6b - 216 = 0 b = 12 a = 18

Sum of areas of two squares is 468msq . if the difference of their perimeters is 24m find the sides of the two squares .....need step by step ans

Answer»

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

b^2 + 6b - 216 = 0

b = 12

a = 18

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Sum of areas of two squares is 468msq . if the difference of their perimeters is 24m find the sides of the two squares .....need step by step ans

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