Area and perimeter word problems

AREA AND PERIMETER WORD PROBLEMS

About "Area and perimeter word problems"

Area and perimeter word problems are much useful to the students who would like to practice problems on area and perimeter of different shapes.

Before we look at area and perimeter word problems, let us look at some basic stuff about area and perimeter.

Difference between area and perimeter

Area

The region occupied by the closed figure is known as area.

Perimeter

The continuous line forming the boundary of a closed figure is known as perimeter.

Measuring unit of area

Square unit.(sq) Measures two dimensions.

Example : 36 in² or 36 inches squared

Measuring unit of perimeter

Linear unit measures one dimension.

Example : 36 in. or 36 inches

Formula for area and perimeter

Square

Area of square = a²

Perimeter of square = 4a

Here "a" stands for length of each side of the square

Rectangle

Area of rectangle = L x W

Perimeter of rectangle = 2(L + w)

Here L and w stand for the length and width of the rectangle.

Triangle

Area of Equilateral triangle = (√3/4)a²

Perimeter of Equilateral triangle = 3a

Here "a" stands for the length of each side.

Scalane triangle

Area of scalene triangle = √s(s-a)(s-b)(s-c)

Here, s = (a + b + c)/2

Perimeter of scalene triangle = a + b + c

Quadrilateral

Area of quadrilateral = (1/2) x d x (h₁ + h₂)

Perimeter of quadrilateral = Sum of all four sides

Parallelogram

Area of parallelogram = b x h

Perimeter of quadrilateral = Sum of all four sides

Rhombus

Area of rhombus = (1/2) x (d₁ x d₂)

Perimeter of rhombus = Sum of all four sides

Trapezium

Area of trapezoid = (1/2) (a + b) x h

Perimeter of trapezoid = Sum of all four sides

Circle

Area of circle = Πr²

Perimeter of circle = 2Πr

Semi circle

Area of Semi circle = (1/2)Πr²

Perimeter of semi circle = (Π+2)r

Quadrant

Area of quadrant = (1/4)Πr²

Perimeter of quadrant = [(Π/2) + 2]r

Sector

Area of the sector = (θ/360)Πr² sq. units

(or) Area of the sector = (1/2)lr sq. units

Length of arc = (θ/360)2Πr

Perimeter of sector = L + 2r

Area and perimeter word problems

Problem 1 :

Find the area of the square whose diagonal is 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 square = a²

= (2√2)²

= 2²(√2)²

= 4(2)

= 8 cm²

Hence, the area of the square is 8 square centimeter.

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Problem 2 :

The area of a square is 64 cm². What is its perimeter?

Solution :

Area of the square = 64 cm²

a² = 64

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

Hence, the perimeter of the square is 32 cm

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Problem 3 :

The length and width of the rectangle are in the ratio 5:2 .If the perimeter of the rectangle is 294 cm, then find the length and width of the rectangle.

Solution :

From the given ratio 5:2,

length = 5x

width = 2x

Given : Perimeter of the rectangle = 294 cm²

2(Length + Width) = 294

Length + Width = 147

5x + 2x = 147

7x = 147

x = 147/7

x = 21

Length = 5(21) = 105 cm

Width = 2(21) = 42 cm

Hence, the length and width of the rectangle are 105 cm and 42 cm respectively.

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Problem 4 :

The perimeter of a rectangle is 50 cm. The length is 15 cm. What is the area?

Solution :

Perimeter of the rectangle = 50 cm

2 (L + B) = 50

L + B = 25

Here the length is 15 cm

15 + B = 25

B = 25 - 15

B = 10 cm

Hence, the width is 10 cm

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Problem 5 :

Find the area of rhombus if the lengths of the diagonals are 15 cm and 10 cm.

Solution :

Here d₁ = 15 cm d₂ = 10 cm

Area of the rhombus = (1/2) x (d₁ x d₂)

= (1/2) x 15 x 10

= 15 x 5

= 75 cm²

Hence, the area of the rhombus is 75 square cm.

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Problem 6 :

In a trapezium, the length of one parallel side is two more than the other parallel side and the height is 4 cm. The area of the trapezium is 64 cm². Find the two parallel sides.

Solution :

Area of the trapezium = 64 cm²

Let "a" be the length one of the parallel sides.

Then, the length of other parallel side b = a + 2

Area of a trapezium = (1/2) (a + b) x h

(1/2) (a + b) x h = 64

(a + a + 2) x 4 = 128

2a + 2 = 32

2a = 30

a = 15

The length of the other parallel side b = 15 + 2 = 17 cm

Hence, the lengths of the two parallel sides are 15 cm and 17 cm.

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Problem 7 :

Find the area of the sector whose radius is 35 cm and perimeter is 147 cm.

Solution :

Radius = 35 cm

Perimeter of sector = 147 cm

L + 2r = 147 cm

Here r = 35 cm

L + 2(35) = 147

L + 70 = 147

L = 147 - 70

L = 77 cm

Area of the sector = (1/2) x Lx r square units

= (1/2) x 77 x 35

= 38.5 x 35

= 1347.5 sq. units

Hence, the area of the sector is 1347.5 sq. units

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Problem 8 :

Find the length of arc if the perimeter of sector is 45 cm and radius is 10 cm.

Solution:

Perimeter of sector = 45 cm

L + 2r = 45

here r = 10 cm

L + 2(10) = 45

L + 20 = 45

L = 45 - 10

L = 35 cm

Hence, the length of the arc is 35 cm.

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Problem 9 :

Find the circumference of the semi-circle whose diameter is 7 cm.

Solution :

r = diameter/2

r = 7/2

r = 3.5

Now we can apply the formula

Circumference of semi circle = (Π+2)r

Plug r = 3.5 and Π = 22/7

= [(22 + 4)/7]x 3.5

= [22+14] x 0.5

= 36 x 0.5

= 18 cm

Hence, the circumference of the semi circle is 18 cm.

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Problem 10 :

The length of a rectangle is 70 cm and width is 30 cm. If the length is increased by 10%, find the area of the rectangle

Solution :

After the length is increased by 10%,

length = 110% of 70 = 1.1 x 70 = 77 cm

There is no change in width. So width = 30 cm

Area of the rectangle = l x w

= 77 x 30

= 2310

Hence, the required area is 2310 square cm.

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Problem 11 :

The length of a rectangle is 80 cm and width is 40 cm. If the length is increased by 20% and width is decreased by 10%, By what percentage will the area be increased or decreased ?

Solution :

When the length = 80 cm and width = 40 cm,

Area of the rectangle = l x w = 80 x 40 = 3200 sq.cm ------(1)

After the length is increased by 20% and width is decreased by 10%,

length = 120% of 80 = 1.2 x 80 = 96 cm

width = 90% of 40 = 0.9 x 40 = 36 cm

Now, area of the rectangle = 96 x 36 = 3456 ------(2)

Comparing (1) and (2), area is increased by 256 sq.cm.

Percentage increase in in area = (256/3200) x 100 % = 8 %

Hence, the area is increased by 8%.

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Problem 12 :

The width of a rectangle is 10 m and the area is 2000000 square cm. Find the length in meters

Solution :

Given : Area of the rectangle = 200000 sq.cm

To convert sq.cm into sq.meters, we have to divide by 10000

Therefore, area of the rectangle = 2000000/10000 = 200 sq.m

Area of the rectangle = 200 sq.meters

l x 10 = 200

l = 20 m

Hence, the length is 20 meters.

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Problem 13 :

If the area of the rectangle is increased from 70 to 84 square cm.,find the increase in percentage

Solution :

Actual increase in the area = 84 - 70 = 14 sq.cm

Percentage increase in area = (14/70) x 100 %

Percentage increase in area = 20%

Hence, the percentage increase in the area is 20%.

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Problem 14 :

The area of a rhombus is 78 square cm and one of the diagonals is 12 cm. Find the length of other diagonal

Solution :

Area of the rhombus = 78 square cm

(1/2) x (d₁ x d₂) = 78

(1/2) x (12 x d₂) = 78

12 x d₂ = 156

d₂ = 12

Hence, the length of other diagonal is 12 cm.

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Problem 15 :

Find the cost of carpeting a rectangle shaped room whose length is 13 m and width is 9 m if the cost of carpeting is $2 per square meter.

Solution :

Area of the room = l x w

Area of the room = 13 x 9

Area of the room = 117 sq.m

Cost of carpeting per sq.m = $2

Cost of carpeting for 117 sq.m = 2 x 117

Cost of carpeting for 117 sq.m = 234

Hence, the cost of carpeting is $234.

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Problem 16 :

The length of a rectangle is two times its width. If the perimeter is 24 cm, find the area of the rectangle.

Solution :

Let "x" be the width of the rectangle.

Then, length = 2x

Given : Perimeter of the rectangle = 24 cm

2(l + w) = 24

2(2x + x) = 24

2(3x) = 24

3x = 12

x = 4 ------> width = 4 cm

Length = 2x = 2(4) = 8 cm

Area of the rectangle = 8 x 4

= 32 sq.cm

Hence, area of the rectangle is 32 sq.cm.

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Problem 17 :

If the ratio of the radii of two circles is is 2 : 3, find the ratio of their area.

Solution :

From the given ratio 2 : 3,

Radius of the first circle = 2x

Radius of the first circle = 3x

Ratio of their area = Π(2x)² : Π(3x)²

= (2x)² : (3x)²

= 4x² : 9x²

= 4 : 9

Hence, ratio of their area is 4 : 9.

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Problem 18 :

David has a square garden whose one of the sides is covered by the wall of his house. The length of each side of the garden is 25 feet. If the cost of fencing is $3 per feet, find the total cost of fencing the garden.

Solution :

Since one of the sides of the square garden is covered by the wall, fencing has to be put for the other three sides.

Since the shape is square, the length of all the sides will be equal.

The length of each side = 25 feet.

Sum of the lengths of 3 sides = 3 x 25 = 75 feet.

So, fencing has to be done for 75 feet.

Cost fencing per feet = $3

Cost fencing for 75 feet = $3 x 75

= 225

Hence, the total cost for fencing the garden is $225.

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Problem 19 :

The radius of a circular path of garden is 6 meters. A circular path of width 2 meters is laid around and outside the garden. Find the area of the path.

Solution :

From the given information and the picture given above,

radius of the outer circle R = 8 m

radius of the inner circle is r = 6 m

Area of the path = Area of the outer circle - Area of the inner circle

Area of the circular path = ΠR² - Πr²

= Π(R² - r²)

Plug Π = 22/7, R = 8 and r = 6

= (22/7) x (8² - 6²)

= (22/7) x (64 - 36)

= (22/7) x 28

= 22 x 4

= 88

Hence, the area of the circular path is 88 square meters.

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Problem 20 :

The poster has a border whose width is 2 cm on each side. If the printed material has the length of 8 cm and width of 6 cm, find the area of the border. .

Solution :