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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.
Let the side of the first square be x and the side of the second square be y.
Area of a square = side × side
Perimeter of a square = 4 × side
Therefore, the area of the first and second square is x2 and y2 respectively.
Also, the perimeters of the first and second square are 4x and 4y respectively.
Applying the known conditions:
(i) x2 + y2 = 468 ...(1)
(ii) 4x - 4y = 24 ....(2) considering x > y
4(x - y) = 24
x - y = 6
x = 6 + y
Substitute x = y + 6 in equation (1)
(y + 6)2 + y2 = 468
y2 +12y + 36 + y2 = 468
2y2 + 12y + 36 = 468
2(y2 + 6y + 18) = 468
y2 + 6y + 18 = 234
y2 + 6y + 18 - 234 = 0
y2 + 6y - 216 = 0
Solving by factorization method,
y2 + 18y - 12y - 216 = 0
y (y + 18) - 12 (y + 18) = 0
(y + 18) (y - 12) = 0
y + 18 = 0 and y - 12 = 0
y = -18 and y = 12
y cannot be a negative value as it represents the side of the square.
Side of the first square x = y + 6 = 12 + 6 = 18 m
Side of the second square = 12 m.
Sum of the areas of two squares 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Class 10 Maths NCERT Solutions Chapter 4 Exercise 4.3 Question 11
Sum of the areas of two squares 468 m², if the difference of perimeters is 24 m, then the side of the first square is 18 m and the side of the second square is 12 m.
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