**Perimeter and Area of Square Worksheet :**

Worksheet given in this section is much useful to the students who would like to practice problems on perimeter and area of square.

Before look at the worksheet, if you would like to learn the stuff perimeter and area of square,

**Problem 1 :**

If the length of each side of a square is 8.5 cm, then find its perimeter.

**Problem 2 :**

The length of each diagonal of a square is 2√2 cm. Find its perimeter.

**Problem 3 :**

If a square has the side length of 7.5 cm, then find its area.

**Problem 4 :**

The length of each side of a square is 3√5 cm. Find its area.

**Problem 5 :**

The area of a square is 32 square inches. Find the length of its diagonal.

**Problem 6 :**

The square has side length 36 inches. Find its area in square feet.

**Problem 7 : **

The lengths of each side of two squares are 4 cm and 5 cm. Find the ratio of their perimeters.

**Problem 8 : **

The lengths of each side of two squares are 4 cm and 5 cm. Find the ratio of their areas.

**Problem 9 :**

AB is one of the sides of the square ABCD and the side AB is defined by A(0, 2) and B(6, 9). Find the perimeter of the square ABCD.

**Problem 10 :**

PR is one of the diagonals of the square PQRS and the diagonal PQ is defined by P(1, 4) and Q(4, 8). Find the area of the square PQRS.

**Problem 1 :**

If the length of each side of a square is 8.5 cm, then find its perimeter.

**Solution:**

Formula for perimeter of a square :

= 4s^{ }

Substitute 14 for s.

= 4(8.5)

= 34

So, the perimeter of the square is 34 cm.

**Problem 2 :**

The length of each diagonal of a square is 2√2 cm. Find its perimeter.

**Solution:**

To find the perimeter of a square, first we have to know the length of each side.

Let s be the length of each side of the square.

Draw a sketch.

In the figure shown below, consider the right triangle ABC.

By Pythagorean Theorem, we have

AB^{2} + BC^{2} = AC^{2}

Substitute.

s^{2} + s^{2} = (2√2)^{2}

Simplify and solve for s.

2s^{2} = 2^{2} ⋅(√2)^{2}

2s^{2} = 4 ⋅(2)

2s^{2} = 8

Divide each side by 2.

s^{2} = 4

Find positive square root on both sides.

√s^{2} = √4

√s^{2} = √(2 ⋅ 2)

s = 2

Formula for perimeter of a square.

Perimeter = 4s

Substitute 2 for s.

= 4(2)

= 8

So, the perimeter of the the square is 8 cm.

**Problem 3 :**

If a square has the side length of 7.5 cm, then find its area.

**Solution:**

When the length of a side is given, formula for area of a square :

= s^{2 }

Substitute 24 for s.

= (7.5)^{2}

= 56.25

So, area of the square is 56.25 square cm.

**Problem 4 :**

The length of each side of a square is 3√5 cm. Find its area.

**Solution:**

When the length of a side is given, formula for area of a square :

= s^{2}

Substitute 3√5 for s.

= (3√5)^{2}

Simplify.

= 3^{2} ⋅ (√5)^{2}

= 9 ⋅ 5

= 45

So, the area of the square is 45 square cm.

**Problem 5 :**

The area of a square is 32 square inches. Find the length of its diagonal.

**Solution:**

Area of the square = 32 in^{2}

1/2 ⋅ d^{2} = 32

Multiply each side by 2.

d^{2} = 64

Find positive square root on both sides.

√d^{2} = √(8 ⋅ 8)

d = 8

So, the length of diagonal is 8 inches.

**Problem 6 :**

The square has side length 36 inches. Find its area in square feet.

**Solution:**

When the length of a side is given, formula for area of a square :

= s^{2 }

Substitute 12 for s.

= 36^{2}

= 1296 in^{2} -----(1)

We know

12 inches = 1 ft

Square both sides.

(12 inches)^{2} = (1 ft)^{2}

12^{2} in^{2} = 1^{2} ft^{2}

144 in^{2} = 1 ft^{2}

Therefore, to convert square inches into meter square feet, we have to divide by 144.

(1)-----> Area of the square = 1296 in^{2}

Divide the right side by 144 to convert in^{2} into ft^{2}.

Area of the square = (1296 / 144) ft^{2}

= 9 ft^{2}

So, the area of the square is 9 square feet.

**Problem 7 : **

The lengths of each side of two squares are 4 cm and 5 cm. Find the ratio of their perimeters.

**Solution:**

Formula for perimeter of a square :

= 4s

= 4(4) = 16 cm |
= 4(5) = 20 cm |

Ratio of the perimeters :

= 16 : 20

Divide each term by 4.

= 4 : 5

So, the ratio of the perimeters of two squares is 4 : 5.

**Problem 8 : **

The lengths of each side of two squares are 4 cm and 5 cm. Find the ratio of their areas.

**Solution:**

Formula for area of a square :

= s^{2}

= 4 = 16 cm |
= 5 = 25 cm |

Ratio of the areas :

= 16 : 25

So, the ratio of the areas of two squares is 16 : 25.

**Problem 9 :**

AB is one of the sides of the square ABCD and the side AB is defined by A(0, 2) and B(6, 9). Find the perimeter of the square ABCD.

**Solution:**

**Distance between the two points (x _{1}, y_{1}) and (x_{2}, y_{2}) is **

**= √[(x _{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]**

**To find the distance between A and B, substitute **

**(x _{1}, y_{1}) = (0, 2)**

**(x _{2}, y_{2}) = (6, 9)**

**in the above formula.**

**Distance between A and B : **

**= √[(6-0) ^{2 }+ (9-2)^{2}]**

**= √[6 ^{2 }+ 7^{2}]**

**= ****√[36 ^{ }+ 49]**

**= ****√85**

**Therefore, the length of one of the sides is ****√85**** units.**

Formula for perimeter of a square :

= 4s

Substitute s = **√8**5.

= 4**√8**5

Use calculator and simplify.

≈ 36.88

So, the perimeter of the square ABCD is about 36.88 units.

**Problem 10 :**

PR is one of the diagonals of the square PQRS and the diagonal PQ is defined by P(1, 4) and Q(4, 8). Find the area of the square PQRS.

**Solution:**

**Distance between the two points (x _{1}, y_{1}) and (x_{2}, y_{2}) is **

**= √[(x _{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]**

**To find the distance between P and R, substitute **

**(x _{1}, y_{1}) = (1, 4)**

**(x _{2}, y_{2}) = (4, 8)**

**in the above formula.**

**Distance between P and R : **

**= √[(4-1) ^{2 }+ (8-4)^{2}]**

**= √[3 ^{2 }+ 4^{2}]**

**= ****√[9 ^{ }+ 16]**

**= ****√25**

**= 5**

**Therefore, the length of the diagonal PR is 5**** units.**

When the length of a diagonal is given, formula for area of a square :

= 1/2 ⋅ d^{2}

Substitute d = 5.

= 1/2 ⋅ 5^{2}

Simplify.

= 1/2 ⋅ 25

= 12.5

So, the area of the square PQRS is 12.5 square units.

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