sum of the areas of two squares is 468 m2. if difference of their perimeters is 24 m. find tha sides of the two squares
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Let us say that the sides of the two squares are 'a' and 'b'
Sum of their areas = a^2 + b^2 = 468
Difference of their perimeters = 4a - 4b = 24
=> a - b = 6
=> a = b + 6
So, we get the equation
(b + 6)^2 + b^2 = 468
=> 2b^2 + 12b + 36 = 468
=> b^2 + 6b - 216 = 0
=> b = 12
=> a = 18
The sides of the two squares are 12 and 18
Sum of their areas = a^2 + b^2 = 468
Difference of their perimeters = 4a - 4b = 24
=> a - b = 6
=> a = b + 6
So, we get the equation
(b + 6)^2 + b^2 = 468
=> 2b^2 + 12b + 36 = 468
=> b^2 + 6b - 216 = 0
=> b = 12
=> a = 18
The sides of the two squares are 12 and 18
HARISHKABILAN:
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Here is ur answer hope this helps if so mark as brainliest.
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