Angle between two planes: A plane in geometry is a flat surface that extends in two dimensions indefinitely but has no thickness. The angle formed...
Angle between Two Planes: Definition, Angle Bisectors of a Plane, Examples
November 10, 2024In our daily life, we can see wide Applications of Perimeter and Area. Perimeter is the length of the boundaries of a closed figure, and the area is the region or space occupied by an object or figure in the geometry. Mensuration deals with the formulas and the applications of the perimeter and the area of geometrical objects or shapes.
The perimeter and the area applications involve rectangular parks, cross paths, square plots, triangular grounds, etc. This topic is essential in mathematics, as we make use of these calculations in real-life scenarios too. In this article, we will discuss in detail the Applications of Perimeter and Area, Definition, Formulas, Solved Examples, etc. Continue reading to know more.
Perimeter is the measurement of the boundaries of the object or shapes. Perimeter is the length of the closed figure’s boundaries or the total distance of the boundaries of any two-dimensional closed figure. We have different types of formulas that are used to find the perimeter of the two-dimensional figure. The perimeter is generally measured using units like \({\text{cm, m,}}\,{\text{km, feet,}}\,{\text{inches,}}\) etc.
The area is the space or region occupied by any figure or shape in the given plane. The space occupied by three-dimensional figures is of two types: lateral surface area or curved surface area and the total surface area. The space or region enclosed by all the faces of the three-dimensional figure is called the total surface area. In general, the area is measured in square units like \({{\text{m}}^2}{\text{, c}}{{\text{m}}^2}{\text{,}}\,{\text{k}}{{\text{m}}^2}{\text{, sq}}{\text{.feet,}}\,{\text{sq}}.{\text{inches}}\) etc.
As we know, that perimeter is the sum of the lengths of the boundaries of any closed figure, and the space or region enclosed by the figure in the two-dimensional plane is called the area.
We have different types of formulas to calculate the area and the perimeter of the figures. The types of perimeter and area are triangular shapes, square and rectangular shapes etc. Some of the popularly used formulas are tabulated below:
Learn About Area and Perimeter
We know the measurements like area and the perimeter are very useful in our daily life. We have a wide range of applications of these. Some of the essential uses of the area and the perimeter are listed below:
As we know that perimeter and area have their applications in various fields. Mainly they have their applications in three-dimensional objects like cylinders, cones, cubes, spheres etc.
A general lateral surface area or the curved surface area of three-dimensional figures can be found by multiplying the perimeter of the base and height of the object. And, also the volume of the three-dimensional figure can be found by multiplying the area of the base of the object with the height of the given object.
We know that a quadrilateral with equal lengths of all four sides is called the square, and the quadrilateral with two pairs of opposite sides is called the rectangle.
Square is a quadrilateral with all angles equal and all sides of equal length.
1. perimeter \( = 4 \times {\rm{side}} = 4a\)
2. Area \( = {\rm{side}} \times {\rm{side}} = {a^2}\)
Consider a rectangle with length \(\left( l \right)\) and breadth \(\left( b \right)\):
1. Perimeter \( = 2\left( {{\rm{length}} + {\rm{breadth}}} \right) = 2\left( {l + b} \right)\)
2. Area \( = {\rm{length}} \times {\rm{breadth}} = l \times b\)
As we know, perimeter and area have many applications, such as finding the area of the paths, roads etc. To solve the problem related to perimeter and area, follow some steps to solve it easily.
Q.1. A constructor has his land in the shape of the polygon shown below. Find the area and the perimeter of the given figure.
Ans: The perimeter of the closed figure is nothing but the sum of the lengths of the boundaries. The perimeter of the given figure is given below:
\(6~{\text{cm}} + 18~{\text{cm}} + 6~{\text{cm}} + 3~{\text{cm}} + 11~{\text{cm}} + 9.5~{\text{cm}} + 6~{\text{cm}} = 59.5~{\text{cm}}.\)
Dividing the given polygon into two regions as shown below, such that region \(A\) represents the rectangle and region \(B\) represents the scalene triangle.
The area of the given figure is the sum of the areas of the rectangle and the scalene triangle.
Area of region \(A\) (Rectangle):
From the figure, length \( = 18\,{\text{cm}},\) breadth \( = 6\,{\text{cm}}.\)
The area of the rectangle is \({\rm{length}} \times {\rm{breadth}}\)
So, the area of the region \(A = 18 \times 6 = 108\,{\text{c}}{{\text{m}}^2}\)
Area of region \(B\) (Scalene triangle):
From the figure, the length of the triangle’s base is \(9\,{\text{cm}},\) and the triangle’s height is \(9\,{\text{cm}}.\)
We know that area of the triangle is \(\frac{1}{2} \times {\rm{base}} \times {\rm{height}} = \frac{1}{2} \times 9 \times 9 = 40.5\,{\rm{c}}{{\rm{m}}^2}\)
Thus, total area\( = \)area of the region \(A + \) area of the region \(B\)
\( = 108 + 40.5 = 148.5~{\text{c}}{{\text{m}}^2}\)
Therefore, the area of the given land is \(148.5~{\text{c}}{{\text{m}}^2}.\)
Q.2. Keerthi is planting a garden with the dimensions as shown below. She wants to cover the entire surface of the garden with a thin layer of mulch, which costs \(Rs.3\,{\text{per}}\,{\text{sq}}.{\text{ft}}.\) Find the total amount she has to spend on the entire mulch.
Ans: Divide the given figure into two regions as shown below, such that region \(A\) represents the rectangle and region \(B\) represents the trapezoid.Area of region \(A\) (Rectangle):
From the figure, length \(= 8\,{\text{ft}},\) breadth \(= 4\,{\text{ft}},\)
\({\rm{length}} \times {\rm{breadth}}\) gives the area of the rectangle.
So, the area of the region \(A = 8 \times 4 = 32\,{\text{sq}}.{\text{ft}}\)
Area of region \(B\) (Trapezoid):
We know that area of the trapezium \( = \frac{1}{2} \times \left( {{\rm{sum}}\,{\rm{of}}\,{\rm{parallel}}\,{\rm{sides}}} \right) \times {\rm{height}}{\rm{.}}\)
\( = \frac{1}{2} \times \left({8 + 14} \right) \times 4 = 44\,{\text{sq}}.{\text{ft}}\)
Thus, total area \( = \) area of the region \(A + \) area of the region \(B\)
\( = 32 + 44 = 76\,{\text{sq}}.{\text{ft}}\)
Given, cost of mulch is \(Rs.\,3\,{\text{per}}\,{\text{sq}}.{\text{ft}}.\)
The total cost of mulch \( = \)cost of mulch \(\,{\text{per}}\,{\text{sq}}.{\text{ft}} \times \) area of the figure
\( = Rs.\,3 \times 76 = Rs.228\)
Hence, she has to pay a total of \(Rs.228\) for th entire mulch.
Q.3. As shown in the figure, two crossroads of width \(5\,{\text{cm}}\) pass through the centre of a rectangular park. Find the area of the roads.
Ans: From the given figure:
\(PQ = 5\,{\rm{m}}\) and \(PS = 45\,{\rm{m}}\)
\(EH = 5\,{\rm{m}}\) and \(EF = 70\,{\rm{m}}\)
\(KL = 5\,{\rm{m}}\) and \(KN= 5\,{\rm{m}}\)
The area of the crossroads is the area of the shaded region.
Area of the crossroads \( = \) area of the rectangle \(PQRS + \) area of the rectangle \(EFGH – \) area of the square \(KLMN.\)
Area of the crossroads \( = PS \times PQ + EF \times EH – KL \times KN\)
Area of the crossroads \( = \left( {45 \times 5 + 70 \times 5 – 5 \times 5} \right){{\rm{m}}^2} = \left( {225 + 350 – 25} \right){{\rm{m}}^2} = 550\,{{\rm{m}}^2}\)
Hence, the area of the crossroads is \(550\,{{\rm{m}}^2}.\)
Q.4. A \(15\,{\text{inches}}\) pizza is served to customers. Calculate its circumference.
Ans: Diameter of pizza \(\left( d \right)15\,{\text{inches}}\)
The circumference of the circle is \(C = \pi d\)
\(C = \pi \times 15 = 15\pi \,{\text{inches}}\)
Hence, the circumference of pizza is \(15\pi \,{\text{inches}}\)
Q.5. The length and width of the rectangular farm are \(80\) yards and \(60\) yards. Find the area of the farm.
Ans: Given: Length of the rectangle \(\left( l \right) = 80\,{\text{yds}}.\)
Width of the farm \(b = 60\,{\text{yds}}.\)
\({\rm{length}} \times {\rm{breadth}}\) gives the area of the rectangle.
So, the area of the farm is \( = 80 \times 60 = 4800\,{\text{sq}}.{\text{yd}}.\)
In this article, we have studied the definitions of the perimeter and the area of the figures. This article also gives some formulas to find the figures’ area and perimeter like triangles (equilateral, isosceles, right-angled, scalene) and quadrilaterals (square, parallelograms, rectangle, rhombus etc.).
This article gives the applications and importance of the perimeter and the area. This article also gives the solved examples that help us to solve the problems easily.
Q.1. What are the applications of the perimeter?
Ans: The perimeter is used to find the total length of the boundary of any shape or figure.
Q.2. What are the applications of the area?
Ans: The area is used to find the space or region occupied by any shape or figure. Examples are finding the space enclosed by the paths, cross-walks etc.
Q.3. What are the applications of perimeter and area in our everyday life?
Ans: Applications of area and perimeter can be seen in everyday life, like finding the area of the floor of the house, the area of the footpath, fencing of the park with a wire, etc.
Q.4. What is the perimeter formula?
Ans: Perimeter formula tells that the perimeter of any figure or shape is the sum of the lengths of the boundaries.
Q.5. What is the distance taken around the circle?
Ans: The distance around the circle is called the circumference or perimeter of the circle.