Geometry Problems on Squares
Problems on area and perimeter of squares with detailed solutions are presented.
Area and Perimeter of a SquarePerimeter of a Square
Perimeter = 4 S , S is the side length of the square.
Area of a Square
Area = S^{ 2} , S is the side length of the square.
Problems with Deatiled SolutionsProblem 1
When the sides of a square are each increased by 2 feet its area increases by 44 feet^{ 2}. Find the side length S before the increase.
Solution to Problem 1:

Let S be the side length before the increase, the area A1 is given by
A1 = S^{ 2}

Let S + 2 the side after the increase, the area A2 is given by
A2 = (S + 2)^{ 2}

But A2 = A1 + 44, hence
A1 + 44 = (S + 2)^{ 2}

Substitute A1 by S^{ 2}
in the above equation.
S^{ 2} + 44 = (S + 2)^{ 2}

Expand, group like terms and rewrite the equation as follows.
4 S = 40

Solve for S.
S = 10 feet.

As an exercise, find areas for S = 10 and for S = 12 and check the the difference is 44 feet^{ 2}.
Problem 2
Find the area and perimeter of a square with diagonal of 200 meters.
Solution to Problem 2:

Use Pythagora's theorem to write
S^{ 2} + S^{ 2} = 200^{ 2}

Solve for S^{ 2} to find the area
S^{ 2} = 20000 m^{ 2}

We need to find S to find the perimeter
S = 100 sqrt(2)
Perimeter = 4 S = 400 sqrt(2) m.
Problem 3
What happens to the area of a square if we double its side?
Solution to Problem 3:

The area A1 of a square of side length S is given by.
A1 = S^{ 2}

Double the side to 2S and find the new area A2.
A2 = (2 S)^{ 2} = 4 S^{ 2}

The area is multiplied by 4.
Problem 4
A square garden (green) of 400 m^{ 2} is to be surrouned by a walkway (yellow) of constant width x. The total area of the walkway has to be 500 m^{ 2}. Find the width x of the walkway.
Solution to Problem 4:

The total area A of the garden and the walkway is given by.
A = 400 + 500 = 900 m^{ 2}

The side S of the garden is given by.
S = sqrt(400) = 20 m

The outside of the walkway is a square of side
S + x + x = S + 2x = 20 + 2x.

The total area of the large square is equal to 900 m^{ 2}, hence the equation:
(20 + 2x)^{ 2} = 900

We now solve for x
x = 5 and x = 25

x is a measure of length and has to be positive, hence
x = 5 meters.

As an exercise, find the side of the larger square and its area and check with the total value of the area 900 m.^{ 2}
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