Let us study the area of closed figures in geometry. We know that area is the space occupied by the object or figure in the geometry. A closed shape can be described as enclosed by a figure whose line segments and curves are connected in geometry. The closed figures start and end at the same point.

We can say that all plane figures (polygons, circles etc.) are closed figures, as they are connected closely with the line segments. The measure of the region bounded by the closed figure in a plane is called the area of the closed figure.

We will discuss the formulas and methods of finding the closed figures’ area in detail.

Define Closed Figures

A closed shape can be described as a shape enclosed by a figure whose line segments and/or curves are connected in geometry. Closed figures have a continuous path that is connected by the line segments or curves.

The starting point of the closed figures coincides with the endpoint. Thus, the closed figures start and end at the same point.

In geometry, we have different types of closed figures, such as triangles, circles, quadrilaterals (square, rectangle, trapezium, parallelogram, rhombus and kite), pentagons, hexagons etc.

The below figures shown are examples of the closed figures.

Area of Closed Figures

The region of space occupied by the object or figure in the geometry is called the area of the figure. The measure of the region bounded by the closed figure in a plane is called the area of the closed figure.
The area of the closed figures can be measured in the squaring units like \({\text{c}}{{\text{m}}^2},~{{\text{m}}^2},~{\text{k}}{{\text{m}}^2},{\text{sq}}.{\text{ft}},\) etc. Some formulas are generally associated with standard closed figures like circles, triangles, and quadrilaterals etc., to find the area.
If any closed figure is divided into two or more regions, then the area of the given closed figure is equal to the sum of the areas of the regions so formed.

Area of the Closed Figures – Circles

A circle is a closed figure that is formed by no sides and no vertices. Thus, the circle is the loci of the points that are at a fixed distance from the same point.

The area of the circle can be found by using the formulas that are listed below:

\({\text{area}} = \pi {r^2} = \frac{{\pi {d^2}}}{4}\)
Here, \(r\) is the radius of the circle, and \(d\) is the diameter of the circle.

Area of the Closed Figures – Triangles

Triangle is a closed figure formed by three line segments and three vertices. The various formulas are associated with the different types of triangles such as equilateral triangle, scalene triangle, isosceles triangle and right-angled triangle.

1. Area \( = \frac{1}{2} \times {\text{base}} \times {\text{height}}\)
2. \( = \sqrt {s\left({s – a} \right)\left({s – b} \right)\left({s – c} \right)} \)

Type of Triangle	Area formula
Equilateral triangle	\(\frac{{\sqrt 3 }}{4} \times {\text{sid}}{{\text{e}}^2}\)
Isosceles triangle	\(\frac{1}{2} \times b \times \sqrt {\left({{a^2} – \frac{{{b^2}}}{4}} \right),} \) Here, \(a\) is the length of the equal side, \(b\) is the length of the unequal side
Right-angled triangle	\(\frac{1}{2} \times {\text{base}} \times {\text{height}}\)/td>
Scalene triangle	\( = \sqrt {s\left({s – a} \right)\left({s – b} \right)\left({s – c} \right)} ,\) Where \(s = \) semiperimeter of the triangle.

Area of the Closed Figures – Quadrilaterals

The closed figure formed with the four-line segments is called the quadrilaterals. We have different types of quadrilaterals such as square, rectangle, trapezium, parallelograms, rhombus and kite. We have a set of standard formulas to find all the areas of these closed figures.

Quadrilateral	Formula
Square	\({\text{side}} \times {\text{side}}\)
Rectangle	\({\text{length}} \times {\text{breadth}}\)
Rhombus	\(\frac{1}{2} \times {\text{diagonal}}\,1 \times {\text{diagonal}}\,2\)
Parallelogram	\({\text{base}} \times {\text{height}}\)
Trapezium	\(\frac{1}{2} \times \left({{\text{sum}}\,{\text{of}}\,{\text{parallel}}\,{\text{sides}}} \right) \times {\text{height}}\)
Kite	\(\frac{1}{2} \times {\text{diagonal}}\,1 \times {\text{diagonal}}\,2\)

In general, the area of the quadrilateral can be found by using the formula:
Area \( = \frac{1}{2} \times \) length of any diagonal \( \times \) (sum of heights from the other vertices on the diagona).

The area of the general quadrilateral can be found by splitting the quadrilateral into two triangles. And, the area of the quadrilateral is equal to the sum of the areas of triangles so formed.

Area of Closed Figures – Pentagons

The closed figures formed by five line segments are called the pentagons. There are two types of pentagons: Regular pentagons or irregular pentagons. A regular pentagon has all its five sides of equal measure, whereas are the irregular pentagon has different sides.

The area of the pentagon means the space occupied by the five sides of the closed figure in it. The area of the regular pentagon can be found by using the formula:

\({\text{Area}} = \frac{1}{4}\left[{\sqrt{5\left({5 + 2\sqrt 5 } \right)} \times {s^2}} \right]\)

Where,
\(s\) is the side of the regular pentagon.

The area of the irregular polygon is found by dividing it into triangles and finding the area of that triangles by standard formulas. Now, the area of the irregular pentagon is equal to the sum of the areas of triangles so formed.

Area of Closed Figures – Hexagon

The closed figures formed by using the six-line segments are called the hexagons. There are two types of hexagons: regular hexagons or irregular hexagons. A regular hexagon has all its six sides of equal measure, whereas are the irregular hexagon has different sides.

The area of the hexagon is the space enclosed between the six sides. The area of the regular hexagon can be found by using the formula:
Area of the regular hexagon \( = \frac{{\sqrt 3 }}{2}{s^2}\)
Here,
\(s\) is the side of the measure of the regular hexagon.
The area of the irregular hexagon is found by dividing the whole hexagon into triangles and/or rectangles. Now, find the area of triangles, rectangles so formed by the standard formulas. Now, the total area is the sum of the areas of triangles and rectangles so formed.

Solved Examples

Q.1. Keerthi is planting a garden (closed figure as shown below) with the dimensions as sown below. She wants to cover the entire surface of the garden with a thin layer of mulch, which costs \({\rm{Rs}}.3\,{\rm{per}}\,{\rm{sq}}\,{\rm{ft}}\) Find the total amount she has to spend on the entire mulch.

Ans: To find the area of the garden (closed figure), divide the given closed 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 \( = 4\,{\text{ft}}.\)
\({\text{Length}} \times {\text{breadth}}\) 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 \( = \frac{1}{2} \times \left({{\text{sum}}\,{\text{of}}\,{\text{parallel}}\,{\text{sides}}} \right) \times {\text{height}}.\)
\( = \frac{1}{2} \times \left({8 + 14} \right) \times 4 = 44\,{\text{sq}}.{\text{ft}}\)
Thus, the total area of the closed figure \( = \) area of the region \(A + \) area of the region \(B\)
\( = 32 + 44 = 76\,{\text{sq}} \cdot {\text{ft}}\)
Given, cost of mulch is \(Rs.\,3\,{\text{per}}\,{\text{sq}}.{\text{ft}}.\)
The total cost of mulch \( = \) cost of mulch per sq. ft \( \times \) area of the figure.
\( = {\rm{Rs}}.3 \times 76 = {\rm{Rs}}.228\)
Hence, she has to pay a total of \(Rs.228\) for th entire mulch.

Q.2. Ramu made the kite as shown below with the given measurements for the kite festival. Find the area of the kite made by Ramu.

Ans: Given Ramu made a kite \(PQRS.\)
The area of the kite \(PQRS\) is equal to the sum of the two parts, as shown in the figure.
In triangle \(PQS\) semi perimeter \(\left( s \right) = \frac{{4 + 4 + 5}}{2} = \frac{{13}}{2} = 6.5\,{\text{units}}.\)
So, \(\operatorname{ar} \left({\Delta PQS} \right) = \sqrt {6.5\left({6.5 – 4} \right)\left({6.5 – 4} \right)\left({6.5 – 5} \right)} = 2.5\sqrt{9.75} \,{\text{sq}}.{\text{units}}.\)
In triangle \(QRS,\) semi perimeter \(\left( s \right) = \frac{{6 + 6 + 5}}{2} = \frac{{17}}{2} = 8.5\,{\text{units}}.\)
So, \(\operatorname{ar} \left({\Delta QRS} \right) = \sqrt {8.5\left({8.5 – 6} \right)\left({8.5 – 6} \right)\left({8.5 – 5} \right)} = 2.5\sqrt {29.75} \,{\text{sq}}.{\text{units}}.\)
The area of the kite \(PQRS = \operatorname{ar} \left({\Delta PQS} \right) + \operatorname{ar} \left({\Delta QRS} \right) = 2.5\,\left({\sqrt {9.75} + \sqrt {29.75} } \right){\text{sq}}.{\text{units}}\)
Hence, the area of the kite made by Ramu is \(2.5\left({\sqrt {9.75} + \sqrt {29.75} } \right)\,{\text{sq}}.{\text{units}}.\)

Q.3. Find the area of the closed figure as shown in the below image of a radius \(9\,{\text{cm}}.\)(Use: \(\pi = \frac{{22}}{7}\))

Ans: Given: Radius of the circle \(\left( r \right) = 9\,{\text{cm}}.\)

The area of the circle with radius \(‘r’\) is \(\pi {r^2}.\)
\( = \frac{{22}}{7} \times {9^2} = 254.57~{\text{c}}{{\text{m}}^2}\)
Hence, the area of the given closed figure is \(254.57~{\text{c}}{{\text{m}}^2}.\)

Q.4. 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 \({\text{length}} \times {\text{breadth}}.\)
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 \(\frac{1}{2} \times {\text{base}} \times {\text{height}} = \frac{1}{2} \times 9 \times 9 = 40.5~{\text{c}}{{\text{m}}^2}\)
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.5. Find the area of the closed figure formed with three line segments of lengths \(3\,{\text{cm}},\,3\,{\text{cm}},\,4\,{\text{cm}}.\)
Ans: We know that the closed figure formed by the three line segments is the triangle.
Given, length of the sides of the triangle is \(3\,{\text{cm}},\,3\,{\text{cm}},\,4\,{\text{cm}}.\)

So,\(a = 3\,{\text{cm}}\) and \(b = 4\,{\text{cm}}\)
We know that area of the isosceles triangle is \(\frac{1}{2} \times b \times \sqrt {\left({{a^2} – \frac{{{b^2}}}{4}} \right)} .\)
\( = \frac{1}{2} \times 4 \times \sqrt {{3^2} – \frac{{{4^2}}}{4}} \)
\( = 2 \times \sqrt{9 – 4} \)
\( = 2\sqrt 5 ~{\text{c}}{{\text{m}}^2}\)
Hence, the area of the given triangle is \( = 2\sqrt 5 ~{\text{c}}{{\text{m}}^2}.\)

Summary

In this article, we have discussed the definitions of closed figures, areas of closed figures. This article also gives the area of the closed figures formed by curves (circles) and the area of the closed figures formed by the line segments such as triangles, quadrilaterals, pentagons hexagons etc.

This article gives the solved examples that help us solve the numerical problems in closed figures easily.

FAQs

Q.1. What is the area of the closed figures?
Ans: The region of space occupied by the object or figure in the geometry is called the area of the figure.

Q.2. How do you find the area of the closed figures formed by the curves?
Ans: The area of the closed figures formed by curves is the circle, and its area is \(\pi {r^2},r\) is the radius.

Q.3. What is the closed shape formed by six sides?
Ans: The closed figure formed by six sides is known as a hexagon.

Q.4. How do you find the area of the closed figures formed by the five-line segments of regular shape?
Ans: The area of the closed figures formed by the five-line segments of a regular shape (a regular pentagon) can be calculated by using \(\frac{1}{4}\left[{\sqrt {5\left({5 + 2\sqrt 5 } \right)} \times {s^2}} \right],\) where \(s\) is side.

Q.5. What are the closed figures?
Ans: Closed figures have a continuous path that is connected by the line segments or curves.