Geometry Formulas

Geometry is a branch of Mathematics that revolves around shapes, sizes, and relative positions of properties and figures. The section deals with lengths, volumes, and areas. Geometry is of two kinds: Plane Geometry and Solid Geometry. Plane Geometry deals with area, length, perimeter of different shapes, and Solid Geometry calculates the volume of geometric figures and shapes.

Students must study geometry formulas extensively to understand the topics in-depth. The academic experts at Extramarks have designed a list of all geometry formulas to help the students in their educational courses.

Class-wise Most Important Geometry Formulas

This section present geometry formulas classwise

Class 8

• Perimeter of Square = 4a
• Area of Square = a²
• Perimeter of Rectangle = 2(l + b)
• Area of Rectangle = l x b
• Perimeter of Triangle = Sum of All Sides
• Area of Triangle = 1/2 × b × h
• Area of Trapezoid = 1/2 × (b1 + b2) × h
• Area of Circle = π × r²
• Circumference of Circle = 2πr

Class 9

• Volume of a Cuboid = Base Area × Height = Length × Breadth × Height

• Volume of a Cone = (1 / 3 )πr²h

• Volume of a Sphere = (4/3) π r³

• Volume of a Hemisphere  = (2/3) πr³

• Surface Area of a Cuboid = 2(lb + bh + hl), where ‘l’, ‘b’ and ‘h’ are the length, breadth, and height respectively.

• Curved Surface Area of a Cone = 1 /2 × l × 2πr = πrl, where ‘r’ is its base radius and ‘l’ its slant height

• Surface Area of a Sphere = 4 π r²

• The volume of a Cuboid = Base Area × Height = Length × Breadth × Height

• The volume of a Cube = a3 where ‘a’ is the edge of the cube.

• The volume of a Cylinder = πr²h, where ‘r’ and ‘h’ are radius and height respectively.

• Volume of a Cone = (1 / 3)πr²h

• Volume of a Sphere = (4/3) πr³
• Volume of a Hemisphere = (2/3) πr³

Class 10

• Volume of Prism = B × h

• Lateral Surface Area of Prism (LSA) = p × h 

• The tangent to a circle equation x² + y² = a² for a line y = mx + c is given by the equation y = mx ± a √ [1+ m2].
• The tangent to a circle equation x² + y² = a2 at (a1,b1) is xa1 + yb1 = a²

Class 11

• Slope m = rise/run = Δy/Δx = y2−y1/x2−x1
• Point-Slope Form y−y1 = m (x−x1)

Class 12

• Cartesian equation of a plane: lx + my + nz = d 
• Distance between two points P(x1, y1, z1) and Q(x2, y2, z2): PQ = √ ((x1 – x2)² + (y1 – y2)² + (z1 – z2)²) 

Basic Geometry Formulas

• Perimeter of a Square = P = 4a; Where a = Length of the sides of a Square

• Perimeter of a Rectangle = P = 2(l+b); Where, l = Length, b = Breadth

• Area of a Square = A = a²; Where a = Length of the sides of a Square

• Area of a Rectangle = A = l×b; Where, l = Length, b = Breadth

• Area of a Triangle = A = ½×b×h; Where, b = base of the triangle; h = height of the triangle

• Area of a Trapezoid = A = ½×(b1 + b2)×h; Where b1 & b2 are the bases of the Trapezoid; h = height of the Trapezoid

• Area of a Circle = A = π×r²

• Circumference of a Circle = A = 2πr; Where, r = Radius of the Circle

• Surface Area of a Cube = S = 6a2; Where, a = Length of the sides of a Cube

• The curved surface area of a Cylinder  = 2πrh

• Total surface area of a Cylinder = 2πr(r + h)

• Volume of a Cylinder = V = πr²h; Where, r = Radius of the base of the Cylinder; h = Height of the Cylinder

• Curved surface area of a cone =  πrl

• Total surface area of a cone = πr(r+l) = πr[r+√(h2+r2)]

• Volume of a Cone = V = ⅓×πr²h; Where r = Radius of the base of the Cone, h = Height of the Cone

• Surface Area of a Sphere = S = 4πr²

• Volume of a Sphere = V = 4/3×πr3, Where r = Radius of the Sphere

Type of Geometry Formulas

We can study Geometry formulas by classifying them into 2D geometry formulas and 3D geometry formulas. Let’s learn them

2D Geometry Formulas

• Perimeter of Square = 4a
• Area of Square = a²
• Perimeter of Rectangle = 2(l + b)
• Area of Rectangle = l x b
• Perimeter of Triangle = Sum of All Sides
• Area of Triangle = 1/2 × b × h
• Area of Trapezoid (A) = 1/2 × (b1 + b2) × h
• Area of Circle = π × r²
• Circumference of Circle(C) = 2πr

3D Geometry Formulas

• Curved Surface Area of Cylinder = 2πrh
• Total Surface Area of Cylinder = 2πr(r + h)
• Volume of Cylinder = V = πr²h
• Surface Area of Cube = 6a²
• LSA of Cube = 4a²
• Volume of Cube = a³
• Surface Area of Cuboid = 2(lb + bh + hl)
• LSA of Cuboid = 2(l + b)h
• Volume of Cube = l × b × h
• Curved Surface Area of a Cone = πrl
• Total Surface Area of Cone = πr(r + l) = πr[r + √(h² + r²)]
• Volume of a Cone = V =1/2× πr²h
• Surface Area of a Sphere = 4πr²
• Volume of a Sphere = 4/3 × πr³
• Curved Surface Area of a hemisphere = 2πr²
• Surface Area of a hemisphere = 3πr²
• Volume of a Sphere = 4/3 × πr³

How to Apply Geometry Formulae in Mathematics?

The formulas are used to make out the perimeter, breadth, height, area, surface area, and many more. 

Let us see an example.

Example: Calculate the area of a trapezium with parallel sides of 24cm and 20cm and a distance of 15cm between them.

Given the length of parallel sides 24cm and 20cm, and the distance between parallel sides is 15cm.

We know that area of trapezium = ½ × (Sum of parallel sides) × Height 

= ½ × (24+20) ×15cm²

= 22×15cm² = 330cm²

Hence, the area of the given trapezium is 330cm²

Solved Examples on Geometry Formulas

Example 1: The length and breadth of a wall are 20 ft. and 15 ft. respectively. The owner wants to get it covered with wallpaper. The price of the wallpaper is ₹50 per sq. ft. Find the Cost of covering the wall with the wallpaper?

Solution:

Area of the wall = l × b 

= 20 × 15

= 300 sq. ft.

Cost of wallpaper = Area of the wall × Cost per sq. ft.

= 300 × 50

= ₹ 15000

Example 2: Calculate the area of a triangle with a 25m base and a 14m height.

Solution:

Here, base = 25cm and Height =14cm

Area of a triangle = (1/2 ×base × height) sq. units 

= (1/2×25×14) cm²

= 175cm²

Therefore, the area of the given triangle is 175cm2.

Example 3: Find the volume of a cube whose edge measures 8cm.

Solution:

Given, edge a=8cm,

we know that volume of a cube =a³ cubic units 

= 83= 8×8×8cm³ = 512cm3

Hence, the volume of the given cube is 512cm³

Example 4: Calculate the circumference of a circle with a radius of 10.5cm

Solution:

Here, r=10.5 = 105/10= 21/2cm

we know that circumference of a circle =2πr

=2 × 22/7 × 21/2 cm 

= 66 cm

Therefore, the circumference of a given circle is 66cm.