The sum of the areas of two squares is 640 m². If the difference in their perimeters is 64 m, find the sides of the two squares.
Answers
Given :
- Sum of areas of two squares is 640m²
- And their perimeters is 64 m
To find :
- Sides of the two squares.
Solution :
Let length and breadth of the both squares are x and y
Sum of Y squares
X² + y² = 640 - - - - (1)
And difference in their perimeters
4x - 4y = 64 - - - - - (2)
==> x - y = 64/4
==> x - y = 16
Putting value of x we get
x² + y² = 640
- ➾ (16 +y) ² + y² = 640
- ➾356 + 32y + y² +y² = 640
- ➾356 + 32y + 2y² = 640
- ➾ 192 - 16y + y² = 0
- ➾192 - (24-8)y² + y² = 0
- ➾192 - 24y - 8y + y² = 0
- ➾8 (y+24) - y (y+24) = 0
- ➾ (y+24) (y - 8) = 0
- ➾ y = 24 or y = - 8
➾ y = –8 rejected as the side of a square cannot be negative.
Hence, side of the first square,
➾ y = 24 m
and side of the second square,
➾X = (24 – 16) m ...[Using (ii)]
= 8 m.
Given :
- Sum of the areas of two squares = 640 m²
- Difference in perimeters of two squares = 64 m
To Find :
- Sides of two squares = ?
Solution :
Let side of one square be "x" and other square be "y".
As, we are given :
Sum of the areas of two squares = 640 m²
- Area of square = side²
=> x² + y² = 640 m² - (i)
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Now, we are given :
Difference in perimeters of two squares = 64 m
- Perimeter of square = 4 × side
=> 4x - 4y = 64 m
=> 4(x - y) = 64 m
=> x - y = 64/4
=> x - y = 16 m - (ii)
From equation (ii), we have :
=> x = 16 + y
Substitute it in equation (i) :
=> (16 + y)² + y² = 640
=> 256 + y² + 32y + y² = 640
=> 2y² + 32y - 640 + 256 = 0
=> 2y² + 32y - 384 = 0
=> 2(y² + 16y - 192) = 0
=> y² + 16y - 192 = 0
=> y² + 24y - 8y - 192 = 0
=> y(y + 24) - 8(y + 24) = 0
=> (y - 8) (y + 24) = 0
=> y - 8 = 0 ; y + 24 = 0
=> y = 8 ; y = - 24
y = - 24 will be rejected because side can never be negative.
Hence, side of one square is 8 m.
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Now, by putting y = 8 in equation (ii) :
=> x - 8 = 16
=> x = 16 + 8
=> x = 24 m
Hence, side of second square is 24 m.
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Hence, side of both squares are 8 m and 24 m.
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