**Problem 1 :**

The side of a square exceeds the side of another square by 4
cm and the sum of the area of two squares is 400 sq.cm. Find the dimensions of
the squares.

**Solution :**

Let x be side length of one square

The side of a square exceeds the side of another square by 4 cm

So, the side length of another square = x + 4

Area of one square with side length x = x^{2}

Area of one square with side length x + 4 is (x + 4)²

Sum of the area of two squares = 400 sq.cm

x^{2} + (x + 4)^{2} = 400

x^{2} + x^{2} + 2 ⋅ x ⋅ 4 + 4^{2} = 400

2x^{2} + 8x + 16 - 400 = 0

2x^{2} + 8x - 384 = 0

By dividing the entire equation by 2, we get

x^{2} + 4x - 192 = 0

(x + 16) (x - 12) = 0

x + 16 = 0 x = -16 |
x - 12 = 0 x = 12 |

Therefore sides of one square is 12 cm.

Side length of another square = (12 + 4) = 16 cm.

**Problem 2 :**

The length of the rectangle exceeds its width by 2 cm and the area of the rectangle is 195 sq.cm. Find the dimensions of the rectangle.

**Solution :**

Let x and y be the width and length of rectangle respectively

The length of the rectangle exceeds its width by 2 cm

So, length (y) = x + 2

Area of the rectangle = 195 sq.cm

Length ⋅ width = 195

x⋅(x + 2) = 195

x^{2} + 2x - 195 = 0

(x + 15) (x - 13) = 0

x + 15 = 0 x = -15 |
x - 13 = 0 x = 13 |

Here x represents width of the rectangle. So, the negative value is not possible.

To find the value of y we have to apply the value of x in the equation y = x + 2

y = 13 + 2

y = 15 cm

Therefore length of rectangle is 15 cm and width of the rectangle is 13 cm.

**Problem 3 :**

The footpath of uniform width runs all around a rectangular field 28 meters long and 22 meters wide. If the path occupies 600 m² area, find the width of the path.

**Solution :**

Let x be the width of the path

Length of the rectangular field = 28 m

Width of the rectangular field = 22 m

Area of the path = 600 m^{2}

Length of the larger rectangle :

= 28 + x + x

= 28 + 2x

Width of the larger rectangle :

= 22 + x + x

= 22 + 2x

Area of the path = Area of larger rectangle - Area of smaller rectangle

600 = (28 + 2x) (22 + 2x) - 28 ⋅ 22

600 = 616 + 56x + 44x + 4x^{2} - 616

600 = 56x + 44x + 4x^{2}

600 = 4x^{2} + 100 x

4x² + 100x = 600

Dividing the entire equation by 4, we get

x^{2 }+ 25x = 150

x² + 25x - 150 = 0

(x - 5) (x + 30) = 0

x - 5 = 0 x = 5 |
x + 30 = 0 x = -30 |

Negative value is not possible. Because x represents width of the path.

Therefore width of the path is 5 m

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