Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Answers
so, their perimeter will be 4x and 4yrespectively and their areas will be x² and y²respectively.
[as you know perimeter of square = 4 × side length and area of square = (side length)²]
It is given that 4x − 4y = 24 [Difference of perimeter]
or x − y = 6
x = y + 6.........(1)
Also, x² + y² = 468 [sum of squares is 468]
=> (6+y)² +y² = 468 [ put eq (1) ]
=> 36 + y²+12y + y² = 468
=> 2y² + 12y – 432 = 0
=> y² + 6y – 216 = 0
=> y² +18y – 12y – 216 = 0
=> y(y + 18)(y – 12) = 0
=> y = - 18 or 12.
However, side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m
SOLUTION:
Let the length of each side of a square be x . Then its perimeter = 4x
[Perimeter of a square = 4×side]
Given: difference of the perimeters of two squares = 24 m
Perimeter of second square - perimeter of first square = 24
Perimeter of second square - 4x = 24
Perimeter of second square = 24 +4x
Length of each side of second square = (24 +4x)/4 = 4(6+x)/4= (6+x) m
Given : Sum of the area of two squares = 468 m²
Area of first square + Area of second square = 468 m²
x² + (6+x)² = 468
[Area of a square = side²]
x² + (6)² + x² + 2×6×x= 468
[(a+b)²= a²+b²+2ab]
2x² + 36 + 12x= 468
2x² + 12x + 36 - 468= 0
2x² + 12x + 36 - 432 = 0
2(x² +6x - 216)= 0
x² +6x - 216 = 0
x² +18x -12x -216 =0
[By factorization]
x(x +18) -12(x +18)=0
(x - 12)(x + 18) = 0
(x - 12)= 0 or (x + 18) = 0
x = 12 or x = -18
[Side can't be negative]
so , x = 12
Side of first square= (x) = 12 m
Side of second square = 6 +x = 6+ 12= 18 m
Hence, the side of a square is 12 m and side of second square is 18 m.
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