## AREA AND PERIMETER OF A SQUARE WORD PROBLEMS

Area and perimeter of a square word problems are much useful to kids who would like to practice problems in mensuration.

## Area and perimeter of a square word problems

(1) Find the area of the square having side length 24 cm.

(2) A square is of area 64 cm². Then find its side length.

(3) The square having side length 25 cm. Find the area in meter.

(4) Find the area of the square whose diagonal is measuring  4cm.

(5) The diagonals of two squares are in the ratio 2:5. Find the ratio of their area.

(6) Find the perimeter of the square having side length 24 cm

(7) Find the perimeter of the square having side length 15 cm

(8) A square is of area 64 cm². What is its perimeter?

(9) Find the perimeter of the square whose diagonal is measuring 4cm.

(10) The perimeter of two squares are 40 cm and 32 cm. Find the perimeter of third square whose area is equal to the difference of the area of two squares.

(11) The perimeter of five squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square in area to the sum of the areas of those squares is :

## Area and perimeter of a square word problems - answers

Question 1 :

Find the area of the square having side length 24 cm.

Solution:

Area of the square = a² = 24²

= 24 x 24

= 576 cm²

Hence, area of square is 576 cm²

Let us see the solution of next question on "Area and perimeter of a square word problems".

Question 2 :

A square is of area 64 cm². Then find its side length.

Solution :

Area of the square = 64 cm²

a²  =  64 cm²

a  =  √64

=   √8 x 8 ==>  8 cm

Hence, side length of square is 8 cm

Let us see the solution of next question on "Area and perimeter of a square word problems".

Question 3 :

The square having side length 25 cm. Find the area in meter.

Solution :

Area of the square  = a²    =  25²  ==> 25 x 25 ==> 625 cm²

100 cm = 1 m

=  625/100  ==> 6.25 m²

Let us see the solution of next question on "Area and perimeter of a square word problems".

Question 4 :

Find the area of the square whose diagonal is measuring  4 cm.

Solution :

The diagonal AC divides the square into two right triangles.Δ ACB and Δ ADC. In triangle ACB right angle is at B.

So the side which is opposite to right angle is called as hypotenuse.

By using Pythagorean theorem

AC²  =  AB² + BC²

4²  =  x² + x²

16  =  2x²

8  =  x²

√8  =  x

√2 x 2 x 2  =  x

2√2  =  x

Therefore, the length of each side is 2√2 cm

Area of the square  =  a²

=  (2√2)²

=  2²(√2)²

=  4 (2)

=  8 cm²

Let us see the solution of next question on "Area and perimeter of a square word problems".

Question 5 :

The diagonals of two squares are in the ratio 2:5. Find the ratio of their area.

Solution :

Let the diagonals of two squares be 2x and 5x respectively.

Area of a square when diagonal is given = (1/2) x d²

Area of first square  =  (1/2) (2x)²

=  (1/2)  4x²   ==> 2x²

Area of second square  =  (1/2) (5x)²

=  (1/2)  25x²   ==> 25x² / 2

Ratio of their areas ==> 2x²  :  (25x² / 2)

=   4 :  25

Let us see the solution of next question on "Area and perimeter of a square word problems".

Question 6 :

Find the perimeter of the square having side length 24 cm

Solution :

Perimeter of a square = 4a

here a  =  24 cm

=  4 ( 24 )

=  96 cm

Hence, perimeter of the square is 96 cm

Let us see the solution of next question on "Area and perimeter of a square word problems".

Question 7 :

Find the perimeter of the square having side length 15 cm

Solution :

Perimeter of a square = 4a

here a = 15 cm

= 4 ( 15 )

= 60 cm

Therefore perimeter of the square is 60 cm

Question 8:

A square is of area 64 cm². What is its perimeter?

Solution :

Area of a square = 64 cm²

a² = 64 cm²

a = √ 64

a = √8 x 8

a = 8 cm

Now we have to find the perimeter

Perimeter of the square = 4a

= 4 (8)

= 32 cm

Therefore perimeter of the square is 32 cm

Question 9 :

Find the perimeter of the square whose diagonal is measuring 4 cm.

Solution:

In the above square we have two right triangles. Those are ⊿ ACB and ⊿ ADC. In ⊿ ACB right angled at B. The side which is opposite to this angle is called hypotenuse side. We can find the other sides using Pythagorean theorem.

AC²  = AB² + BC²

Since all sides are equal in square the sides AB and BC are equal in length.

Let AB = x so BC = x

AC² = x² + x²  ==>  4² = 2x² ==>  x² = 16/2 ==>  x² = 8

x = √8

x = √2 x 2 x 2

x = 2 √2 cm

Hence, length of all sides = 2 √2 cm

Perimeter of square = 4a

= 4 (2 √2)

= 8 √2 cm

Hence, perimeter of new square is 8 √2 cm.

Question 10 :

The perimeter of two squares are 40 cm and 32 cm. Find the perimeter of third square whose area is equal to the difference of the area of two squares.

Solution :

Let "a" and "b" are side length of first and second squares respectively.

Perimeter of first square = 40 cm

4 a  =  40 ==> a  =  10

Perimeter of second square = 32 cm

4 b  =  32 ==> b  =  8

Area of third square  =  10 ²   -  8²  ==> 36 cm²

Side length of third square  =  √36 ==> 6 cm

Perimeter of third square = 4 (6) = 24 cm

Hence, perimeter of third square is 24 cm

Question 11 :

The perimeter of five squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square in area to the sum of the areas of those squares is:

Solution :

Area of new square = Sum of the areas of given squares

Perimeter of 1st square = 24 cm ==> 4a = 24 ==> a = 6

Side length of 1st square is 6 cm

Perimeter of 2nd square = 32 cm ==> 4b = 32 ==> b = 8

Side length of 2nd square is 8 cm

Perimeter of 3rd square = 40 cm ==> 4c = 40 ==> c = 10

Side length of 3rd square is 10 cm

Perimeter of 4th square = 76 cm ==> 4d = 76 ==> d = 19

Side length of 4th square is 19 cm

Perimeter of 5th square = 80 cm ==> 4e = 80 ==> e = 20

Side length of 5th square is 20 cm

The side lengths of given five squares are 6 cm, 8 cm, 10 cm, 18 cm and 20 cm respectively.

Sum of the areas of given squares =  6² + 8² + 10² + 19² + 20²

Area of new square = 961

Side length of new square = √961 = 31

Perimeter of new square = 4 (31) = 124 cm

## More shapes

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