The area is the amount of space occupied by the object or figure in the geometry. We can measure the area of a closed figure by using the formulas. A closed figure can be represented as a shape surrounded by a shape whose line segments and curves are joined in the geometry. Examples of closed figures are triangles, circles, parallelograms, squares, rectangles, etc. In this article, you can find areas of different closed figures, axioms for congruent figures with questions and answers.
Area of a Closed Figure – Definition
The area of a closed figure is the measure of the space or region bounded by the closed figure in a 2D geometry. Any closed figure is divided into two or more regions, the area of a closed figure is equal to the sum of the areas of the regions. The area of the closed figure is measured in the square units.
Measurement of Area
In order to measure the area of the closed figures, we have to know the formulas of areas of different shapes. The area of a closed figure is measured in square units in the region. We will discuss the formulas and methods of finding the area of a closed figure here in detail.
Area of Closed Figure – Triangle
A triangle is a closed figure with three line segments and three vertices. There are three types of triangles equilateral, isosceles, right-angled, and scalene triangles.
Area of triangle = 1/2 × base × height
Area of equilateral triangle = √3/4 × side²
Area of Isosceles triangle = 1/2 × b × √(a² – b²/4), where a is the length of the equal side and b is the length of the unequal side
Area of Right-angled triangle = 1/2 × base × height
Area of Scalene triangle = √s(s – a)(s – b)(s – c), where s is the semi perimeter of the triangle
Area of Closed Figure – Rectangle
A rectangle is a closed planar figure with four sides and four vertices. The area of a rectangle can be measured by multiplying the length and width.
Area of a rectangle = length × width
Area of Closed Figure – Square
A square is a regular polygon with all equal sides. It has four line segments and four sides. The formula to measure the area of the square is the product of square and square.
Area of a square = s × s
Area of Closed Figure – Rhombus
The shape of the Rhombus is similar to the shape of the diamond. It has four equal sides. A rhombus has two diagonals d1 and d2. The area of a rhombus formula is shown below
Area of Rhombus = 1/2 × d1 × d2
Area of Closed Figure – Parallelogram
A parallelogram is a quadrilateral with four line segments and four vertices. The Area of the parallelogram is measured by multiplying the base and height of the given figure.
Area of parallelogram = base × height
Area of Closed Figure – Trapezium
A trapezium is a quadrilateral with at least one pair of parallel sides. It is a closed figure with four line segments. The area of trapezium can be calculated using the lengths of two of its parallel sides and height.
Area of trapezium = 1/2 × sum of parallel sides × height
Area of Closed Figure – Pentagon
A pentagon is a closed figure formed by five line segments and five vertices. A pentagon is of two types they are regular and irregular polygon.
Area = 1/4[√5(5 + 2√5) × s²]
where,
s is the side of the pentagon
Area of Closed Figure – Hexagon
The closed figure formed by using the six-line segments is known as the hexagon. Hexagons are classified into two types regular hexagon and irregular hexagon. A regular hexagon has 6 equal sides and an irregular hexagon has unequal sides. The formula to measure the area of the hexagon is given below.
Area of a hexagon = √3/2 s²
Area of Closed Figure – Circle
A circle has no sides and no vertices. It has a radius and diameter. To measure the area of the circle we have to consider the radius of the circle.
Area of a circle = Πr²
where,
Π = 3.14 or 22/7
Solved Problems on How to find out the Area of a Closed Figure
Example 1.
Find the area of the closed figure formed with three line segments of lengths 3cm, 3cm, 4cm.
Solution:
We know that
The closed figure formed by the three line segments is the triangle.
Given that
The length of the sides of the triangle is 3cm, 3cm, 4cm.
So that a = 3cm and b = 4cm
We know that area of the isosceles triangle is ½ × b × √(a² – b²/4)
= ½ × 4 × √(3² – 4²/4)
= 2 × √9 – 4
2√5 cm²
Therefore the area of the given triangle is 2√5 cm²
Example 2.
Find the area of the closed figure formed with three line segments of lengths 6cm, 6cm, 3cm.
Solution:
We know that
The closed figure formed by the three line segments is the triangle.
Given that
The length of the sides of the triangle is 6cm,6cm,3cm.
So that a = 6cm and b = 3cm
We know that area of the isosceles triangle is ½ × b × √(a² – b²/4)
= ½ × 3 × √(6² – 3²/4)
= 2 × √36 – 9/4
9.75cm²
Therefore the area of the given triangle is 9.75cm²
Example 3.
Find the area of the closed figure as shown in the below image of a radius of 6cm
Solution:
Given that
Radius of the circle ( r ) = 6cm
The area of the circle with radius r is πr²
We know that
π = 22/7
= 22/7 × 6² = 113.4 square cm
Hence the area of the given closed figure is 113.4 square cm
Example 4.
Swetha 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 the sheet, which costs Rs.3 per sqft Find the total amount she has to spend on the entire sheet.
Solution:
To find the area of the garden divide the given closed figure into two regions as shown below
such that,
Region A represents the rectangle
Region B represents the trapezoid
Area of region A (rectangle)
From the figure,
length =8ft, breadth =4ft.
Area of the rectangle = Length × breadth
So that the area of the region A = 8 × 4 = 32 square feet
Area of region B (trapezoid)
We know that
Area of the trapezium = 1/2 × (sum of parallel sides) × height.
= 1/2 × (8 + 14) × 4 = 44 square feet
The total area of the closed figure = area of region A + area of the region B
= 32 + 44 = 76 square feet
Given that
The cost of the sheet is Rs.3 per sq. ft.
The total cost of sheet = cost of mulch per sq. ft × area of the figure.
= Rs.3 × 76 = Rs.228
Therefore, she has to pay a total of Rs.228 for the entire sheet.
Example 5.
A constructor has his land in the shape of the polygon shown below. Find the area and the perimeter of the given figure.
Solution:
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 cm+18 cm+6 cm+3 cm+11 cm+9.5 cm+6 cm=59.5 cm
Dividing the polygon into two regions as shown below
such that,
Region A represents the rectangle
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 cm, breadth =6 cm,
The area of the rectangle = length × breadth
So that the area of the region A = 18 × 6 = 108 cm²
Area of region B (Scalene triangle)
From the figure
The length of the triangle’s base is 9cm, and the triangle’s height is 9cm.
We know that
Area of the triangle = 1/2 × base × height = 1/2 × 9 × 9 = 40.5 cm²
Total area = area of the region A + area of the region B
=108 + 40.5 = 148.5 cm²
Therefore, the area of the given land is 148.5 cm².
FAQs on Area of a Closed Figure
1. What is the closed shape formed by five sides?
The closed figure formed by five line segments is known as the pentagon.
2. What is the closed shape formed by six sides?
The closed figure formed by six line segments is known as a hexagon.
3. What is the area of the closed figures?
The area of the closed figure is measured based on the shape.