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Question

Sum of area of two squares is 468m2. If the difference of their permeter is 24 m, and the sides of the two squares.

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Solution

Sum of the areas of two squares =468m2
Let a and b be the sides of the two squares.
a2+b2=468 .....(1)
Also given that,
the difference of their perimeters =24m
4a4b=24
ab=6
a=b+6 .......(2)
We need to find the sides of the two squares.
Substituting the value of a from equation (2) in equation (1), we have,
(b+6)2+b2=468
b2+12b+36+b2=468
2b2+12b+36468=0
2b2+12b432=0
b2+6b216=0
b2+18b12b216=0
b(b+18)12(b+18)=0
(b+18)(b12)=0
b=18 or 12
Since side cannot be negative.Hence b=12m
a=b+6=12+6=18m


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