sum of the areas of two squares is 468m sq .if. the difference between their perimeters is 24 m.find the sides of the two squares
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a^2+b^2=468
4a-4b=24
a-b = 6-------(1)
(a-b)^2=6^2
a^2+b^2-2ab=36
468-2ab=36
2ab=432
ab =216
(a+b)^2=a^2+b^2+2ab
(a+b)^2=468+432
(a+b)^2=900
a+b=30-------(2)
add(2)and(1)
2a=36
a=18
b=12
4a-4b=24
a-b = 6-------(1)
(a-b)^2=6^2
a^2+b^2-2ab=36
468-2ab=36
2ab=432
ab =216
(a+b)^2=a^2+b^2+2ab
(a+b)^2=468+432
(a+b)^2=900
a+b=30-------(2)
add(2)and(1)
2a=36
a=18
b=12
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