# What are Area Formulas for different Geometric Shapes?

There are many geometrical shapes/figures available and we can find the area for each figure using different formulas based on their shape. The area of a shape can be defined as a space enclosed by a closed geometric figure in a two-dimensional plane. In this article, we are finding the area of a few geometric shapes using different formulas.

### Area of Triangle

The area of a triangle is calculated by multiplying one half the base times the height.

Area of Triangle = (1/2) × Base × Height

**Example: Find the area of a triangle with a height of 6cm and a base of 4cm.**

**Solution:**

To find the area of triangle, we had a formula i.e., (1/2) × base × height

So, for this Triangle Base = 4 & Height = 6

Area = (1/2) × 4 × 6

= (1/2) × 24

= 12

So the area of the given triangle is 12cm

^{2}.

### Area of Square

In a square, the length of all sides are the same and to find the area for the square we need to find the square of the length of a side.

Area of Square = Side × Side = Side

^{2}

**Example: Find the area of a** **square with each side of length 4cm.**

**Solution:**

From given figure, We have square in which length of each side is 4cm. We can find area to it by

Area of Square = Side × Side

= 4 × 4

= 16

So for the given square the area is 16cm

^{2}.

### Area of Rectangle

The area of a rectangle can be calculated by multiplying its length and width of it. Because in rectangle opposite faces have the same length(measurements).

Area of Rectangle = length × width

Note:The formula for area of parallelogram is same as of formula for finding area of rectangle.

**Example: What is the area of the rectangle of length 10 cm and width 5cm.**

**Solution:**

From given figure,

length = 10cm

width = 5cm

Area of rectangle = length × width

= 10 × 5

= 50

So, the area of given rectangle is 50cm

^{2}

**Area of Circle**

The area of a circle can be calculated by using the formula πr^{2 }where r is the radius of a circle and π = 22/7

**Example: Find the circle area with a** **radius of 5cm.**

**Solution:**

From given data,

Radius r = 5cm

area of circle = π × radius

^{2}= (22/7) × 5

^{2}= (22/7) × 25

= 78.57 approximately

So the area of circle is 78.57 cm

^{2}

### Area of Trapezoid

In a trapezoid, we have 2 different lengths of bases i.e., top and bottom lengths. To find the area of a trapezoid the formula is half the sum of lengths of two bases times height.

Area of Trapezoid = (1/2) × (base1 + base2) × height

**Example: Find the area of below given Trapezoid**

**Solution:**

From the given figure,

length of Base1 = 7cm

length of Base2 = 5cm

height of trapezium = 4cm

Area of Trapezium = (1/2) × (base1 + base2) × height

= (1/2) × (5 + 7) × 4

= (1/2) × 12 × 4

= 24cm

^{2}So area of given trapezium is 24cm

^{2}.

**Area of Eclipse**

Before knowing the formula for finding the eclipse, we need to know about the eclipse.

In geometry, an eclipse is in the shape of an oval and it has a major axis and minor axis as shown in the figure. To find the area of the eclipse we need to perform multiplication among the length of the semi-major axis, semi-minor axis and π.

Area of eclipse = π × (Semi-major axis length) × (Semi-minor axis length)

Semi-major axis length = length of major axis/2

Semi-minor axis length = length of minor axis/2

**Example: What is the area of the** **eclipse when its length of the** **major axis and the** **minor axis is 6cm and 3cm respectively.**

**Solution:**

From the given data,

length of major axis = 6

length of minor axis = 3

length of semi major axis = length of major axis/2

= 6/2

= 3

length of semi minor axis = length of minor axis/2

= 3/2

=1.5

area of eclipse = π × (length of semi major axis) × (length of semi minor axis)

= (22/7) × 3 × 1.5

= 14.14 approximately

So for the given eclipse area is 14.14cm

^{2}

### Area of Rhombus

The area of the Rhombus is calculated by finding half of the product of the lengths of two diagonals of a rhombus. The formula is given by:

Area of Rhombus = (1/2) × Length(Diagonal

_{1}) × Length(Diagonal_{2})

Pictorial representation of rhombus is given below:

**Example: Find the area of the** **rhombus where the lengths of diagonals are 3cm and 6cm.**

**Solution:**

From the given data,

Length of Diagonal

_{1 }= 3cmLength of Diagonal

_{2 }= 6cmArea of Rhombus = (1/2) × Length(Diagonal

_{1}) × Length(Diagonal_{2})= (1/2) × 3 × 6

= 9

So the area of rhombus is 9cm

^{2}.