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Ungrouped Data: Know Formulas, Definition, & Applications
December 11, 2024Geometry Formulas: Geometry is a branch of mathematics that studies the relationships among points, lines, angles, surfaces, measurements, and solid shape features. Geometric formulas find its use at various places. Scientists, engineers, and students use geometric formulas to determine geometric shape dimensions, area, volume, and other metrics.
Shapes with only two dimensions: length and breadth are calculated using the two-dimensional planar geometry formula. Three-dimensional geometry is used to calculate solid shapes with three dimensions, such as a cube, cuboid, sphere, cylinder, or cone, that have length, width, and height or depth. This chapter contains the geometry formulas basic sums to help students with their preparations.
Geometry formulas are used to calculate the dimensions, perimeter, area, surface area, volume, and other properties of two-dimensional and three-dimensional geometric shapes. Flat shapes such as squares, circles, and triangles are examples of two dimensions shapes, while cubes, cuboids, spheres, cylinders, and cones are three-dimensional shapes. You can access the important geometry formulas PDF from Embibe.
Basic geometry formulas are listed below:
The perimeter of a triangle–
The sum of the lengths of sides of a triangle is called its perimeter. If the lengths of the sides of a triangle are (a, b) and (c), then,
The perimeter ((a + b + c), {rm{units}})
The perimeter of a triangle is generally denoted by (2s), where (s) is the semi-perimeter of the triangle.
Thus, (2s=(a + b + c), {rm{units}}).
Area of a triangle (general) formula:
In a two-dimensional plane, the area of a triangle is defined as the total space occupied by its three sides.
({rm{Area}}, {rm{of}}, {rm{Triangle}} = frac{1}{2} times b times h, {rm{sq}}.{rm{units}}) Where (b = {rm{base}}) and (h = {rm{height}})
Area of an equilateral triangle:
An equilateral triangle is a triangle with the same length of each of its three sides in geometry.
({\rm{Area}}, {rm{of}}, {rm{an}}, {rm{equilateral}}, {rm{triangle}}, = frac{{sqrt 3 }}{4}{a^2}, {rm{sq}}, {rm{units}}), where (a) is the side of an equilateral triangle.
Heron’s formula: Heron of Alexandria was the first to discover Heron’s formula. It is used to calculate the area of various triangles, including equilateral, isosceles, and scalene triangles, as well as quadrilaterals.
({rm{Area}}, Delta ABC, = sqrt {s(s – a)(s – b)(s – c)}, {rm{sq}}, {rm{units}}) where s = frac{{{rm{Perimeter}}}}{2}= frac{{a + b + c}}{2},{rm{units}})
The perimeter of a quadrilateral: A quadrilateral’s perimeter is equal to the sum of its side lengths. A quadrilateral is a four-sided polygon that can be regular or irregular.
If the lengths of the sides of a triangle are (a, b,c) and (d) ({rm{units}}) then
\({\rm{Perimeter}}\, = \,(a + b + c + d)\,{\rm{units}}\)
Perimeter of a square: A square is a quadrilateral with all sides equal and all angles measuring the same as a right-angle.
Suppose there is a square with \(x\) units of length on each side.
\({\rm{Perimeter}}\,\,{\rm{of}}\,{\rm{square}} = \,4x\,{\rm{units}}\)
Perimeter of a rectangle: A quadrilateral with opposite sides equal and \({90^ \circ }\) angles on each side.
Consider a rectangle with a length of \(l\) and a width of \(b\).
\({\rm{Perimeter}}\,{\rm{of}}\,{\rm{a}}\,{\rm{rectangle}} = 2\left( {l + b} \right){\rm{units}}\)
Area of a quadrilateral: The quadrilateral area is defined as the region enclosed by the quadrilateral’s sides.
Area formulas of few quadrilaterals are given in the table below:
Shape Name | Figure | Area of formulas |
Square | \({\rm{Area}}\,{\rm{of}}\,{\rm{square}}\, = \,{x^2}\,{\rm{sq}}\,{\rm{units}}\) | |
Rectangle | \({\rm{Area}}\,{\mkern 1mu} {\rm{of}}{\mkern 1mu} \,{\rm{at}}{\mkern 1mu} \,{\rm{rapzium}}{\mkern 1mu} \, = \) \({\mkern 1mu} \frac{1}{2}({\rm{sum}}{\mkern 1mu} {\rm{of}}{\mkern 1mu} {\rm{parallel}}{\mkern 1mu} \,{\rm{sides}}) \times {\rm{Height}}{\mkern 1mu} \,{\rm{sq}}\,{\mkern 1mu} {\rm{units}}\) | |
Rhombus | \({\rm{Area of arhombus = }}\frac{1}{2}\,{\rm{(product }}\,{\rm{of}}\,{\rm{ diagonals)}}\,{\rm{sq}}\,{\rm{units }}\) | |
Parallelogram | \({\rm{Area }}\,\,{\rm{of }}\,{\rm{a}}\,{\rm{ parallelogram = }}\frac{1}{2}{\rm{(}}\,{\rm{base}}\times {\rm{height)\,sq }}\,{\rm{units }}\) | |
Trapezium | \({\rm{Area }}\,\,{\rm{of }}\,{\rm{a}}\,{\rm{ trapezium = }}\frac{1}{2}{\rm{(}}\,{\rm{Sum }}\,{\rm{of }}\,{\rm{parallel\,sides)}}\, \times \,{\rm{Height sq units}}\) | |
Kite | \({\rm{Area}}\,{\rm{of}}\,{\rm{a}}\,{\rm{kite}}\,{\rm{ = }}\,\frac{1}{2}\, \times ({\rm{product}}\,{\rm{of}}\,{\rm{diagonals}})\,{\rm{sq}}\,{\rm{units}}\) |
The set of all points in the plane that are a fixed distance (the radius) from a fixed point (the centre) is called a circle.
Circumference of a circle: The circumference is the boundary of a circle in geometry.
Circumference of a circle ( = 2 times pi times r)
Area of a circle: The space enclosed by a circle boundary is called the area of a circle.
Area of circle ( = pi {r^2}) where (r) is the radius of a circle and (pi = frac{{22}}{7}).
Geometry formulas of three-dimensional figures (solid shapes):
The basic formulas of three–dimensional figures are volume, total surface area and lateral or curved surface area.
The volume of solid figures: The volume of a three-dimensional object is commonly described as the capacity of the thing to hold the matter (solid, liquid, gas, or plasma) or space occupied by the matter (solid, liquid, gas, or plasma) inside the three-dimensional object.
The total surface area of solid figures: The total area of the surface of a three-dimensional object is its total surface area.
The lateral or curved surface area of solid figures: A lateral surface is a surface on all of the object’s sides, excluding its base and top (when they exist).
The geometry formulas sheet for 3–dimensional figures are given in below figure:
Download – Geometry Formulas PDF
Shape Name | Figures | Formulas |
Cube | Lateral surface area = 4{a^2}, {rm{sq}}{rm{.}}, {rm{units}}) Total surface area = 6{a^2}, {rm{sq}}{rm{.}}, {rm{units}}) Volume = {a^3}, {rm{cubic}}, {rm{units}}) Diagonal = sqrt 3 a{rm{units}}) | |
Cuboid | Lateral surface area = 2h(l + b), {rm{sq}}, {rm{units}}) Total surface area = 2h(lb + bh + hl), {rm{sq}}, {rm{units}}) Volume = lbh {rm{cubic}}, {rm{units}}) Diagonal = sqrt {{l^2} + {b^2} + {h^2}}, {rm{units}}) | |
Cylinder | Curved surface area = 2pi rh {rm{sq}}, {rm{units}}) Total surface area ( = \,2\pi r(r + h)\,{\rm{sq}}\,{\rm{units}}\) Volume \( = \,\pi {r^2}h\,{\rm{cubic}}\,{\rm{units}}\) | |
Hollow Cylinder | Curved surface area \( = \,2\pi h(R + r)\,{\rm{sq}}\,{\rm{units}}\) Total surface area \( = \,2\pi (Rh + rh + {R^2} – {r^2})\,{\rm{sq}}\,{\rm{units}}\) Volume \( = \pi \left( {{R^2} – {r^2}} \right)h\,{\rm{cubic}}\,{\rm{units}}\) | |
Cone | Curved surface area \( = \,\pi rl\,{\rm{ sq}}\,{\rm{units}}\) Total surface area \( = \,\pi r(r + l)\,{\rm{ sq}}\,{\rm{units}}\) Where, \(\,l = \)slant height | |
Hollow Cone | Volume \( = \frac{1}{3} \times \pi \left( {{R^2} – {r^2}} \right)h\,{\rm{cubic}}\,{\rm{units}}\) | |
Sphere | Surface area \( = 4\pi {r^2}{\rm{sq\,units}}\) Volume \({\rm{ = }}\frac{4}{3}\pi {r^3}\,{\rm{cubic}}\,{\rm{units}}\,\) | |
Hollow Sphere | Volume \({\rm{ = }}\frac{4}{3}\pi ({R^3}{\rm{ – }}\,{{\rm{r}}^3} \,{\rm{)cubic}}\,{\rm{units}}\,\) | |
Hemisphere | Curved surface area \( = 2\pi {r^2}{\rm{sq \,units}}\) Total surface area \( = 3\pi {r^2}{\rm{sq\, units}}\) Volume \({\rm{ = }}\frac{2}{3}\pi {r^3}\,{\rm{cubic}}\,{\rm{units}}\,\) | |
Hollow Hemisphere | Curved surface area \( = 2\pi ({R^2} + {r^2})\,{\rm{sq \,units}}\) Total surface area \( = 2\pi \left( {{R^2} – {r^2}} \right) – \pi \left( {{R^2} – {r^2}} \right)\,{\rm{sq}}\,{\rm{units}}\) Volume \({\rm{ = }}\frac{2}{3}\pi ({R^3} – {r^3})\,{\rm{cubic}}\,{\rm{units}}\,\) | |
Frustrum Cone | Curved surface area \(\, = \pi (R + r)l\,{\rm{sq \,units}}\) Total surface area \(\, = \pi l(R + r) + \pi {R^2} + \pi {r^2}\,{\rm{sq\, units}}\) Where (l\, = \,\sqrt {{h^2} + {{(R – r)}^2}} {\rm{units}}\) Volume = \frac{1}{3}\pi h\left( {{R^2} + {r^2} + Rr} \right)\,{\rm{cubic}}\,{\rm{units}}\) |
Check other important Maths articles:
Geometry formulas are essential for solving all plane and solid geometry mathematical problems.
Types of Geometry Formulas:
Geometry formulas are the most common concern among students in math classes. They are used to work out the length, perimeter, area, and volume of different geometric forms and figures. Numerous geometric formulas deal with height, breadth, length, radius, perimeter, area, surface area or volume, and other topics.
Several coordinate geometric formulas are rather complicated. You may have never seen them before; however, we use some simple formulae in our daily lives to compute length, space, and other things.
Geometry formulas are used in maths to find the perimeter and area of plane figures, surface area and volume of three-dimensional shapes.
Example \(\,1\): Calculate the area of a trapezium with parallel sides of \(24\,{\rm{cm}}\) and \(20\,{\rm{cm}}\) and a distance of \(15\,{\rm{cm}}\) between them.
Solution: Given the length of parallel sides \(24\,{\rm{cm}}\) and \(20\,{\rm{cm}}\), and the distance between parallel sides is \(\,15\,{\rm{cm}}\).
We know that area of trapezium \( = \,\frac{1}{2} \times ({\rm{Sum}}\,{\rm{of}}\,{\rm{parallel}}\,{\rm{sides}}\,) \times {\rm{Height}}\)
\( = \,\frac{1}{2} \times (24\, + \,20\,) \times 15\,{\rm{c}}{{\rm{m}}^2}\)
\( = \,22\,\, \times \,15\,{\rm{c}}{{\rm{m}}^2} = 330\,{\rm{c}}{{\rm{m}}^2}\)
Hence, the area of the given trapezium is \(330\,{\rm{c}}{{\rm{m}}^2}\)
Example \(\,2\): Calculate the volume, total surface area, and lateral surface area of a cuboid that is \(8\) metres long, \(6\) metres wide, and \(3.5\) metres tall.
Solution:
The volume of the cuboid \( = (l \times b \times h)\,{\rm{cubic}}\,{\rm{units}}\)
\( = (8 \times 6 \times 3.5)\,{{\rm{m}}^3} = 168\,{{\rm{m}}^3}\)
The total surface area of the cuboid \( = 2(lb + bh + lh)\,{\rm{sq}}\,{\rm{units}}\)
\({\rm{ = 2(8}} \times {\rm{6}}\,{\rm{ + }}\,{\rm{6}} \times {\rm{3}}{\rm{.5 + 8}} \times {\rm{3}}{\rm{.5)}}{{\rm{m}}^2} = 194\,{{\rm{m}}^2}\)
The lateral surface area of the cuboid \( = 2 \times h(l\, + \,b)\,{\rm{sq}}\,{\rm{units}}\)
\({\rm{ = 2}} \times {\rm{3}}{\rm{.5}} \times {\rm{(}}\,{\rm{8}}\,{\rm{ + }}\,{\rm{6)}}{{\rm{m}}^2}\, = \,98\,{{\rm{m}}^2}\)
Here you will find some geometry formulas with examples curated by our experts.
Q.1. Determine the length of the longest pole that can be placed in a \(10\,{\rm{m}}\) by \(10\,{\rm{m}}\) by \(5\,{\rm{m}}\) room.
Ans. Here \(l\, = 10\,{\rm{m}},\,b\, = \,10\,{\rm{m}}\) and \(h\, = \,5\,{\rm{m}}\)
Length of the longest pole \( = \) Length of the diagonal
\( = \sqrt {{l^2} + {b^2} + {h^2}} \,{\rm{units}}\)
\( = \,\sqrt {{{10}^2} + {{10}^2} + {5^2}} {\rm{m}}\,{\rm{ = }}\,\sqrt {225} \,{\rm{m}}\,{\rm{ = }}\,{\rm{15}}\,{\rm{m}}\)
Hence, the length of the longest pole is \(15\,{\rm{m}}\).
Q.2. Calculate the area of a triangle with a \(25\,{\rm{m}}\) base and a \(14\,{\rm{m}}\) height.
Ans. Here, base \( = \,25\,{\rm{cm }}\) and Height \( = \,14\,{\rm{cm }}\)
Area of a triangle \( = \left( {\frac{1}{2} \times {\rm{base}} \times {\rm{height}}} \right)\,{\rm{sq}}\,{\rm{units}}\)
\( = \left( {\frac{1}{2} \times 25 \times 14} \right)\,{\rm{c}}{{\rm{m}}^2}\,{\rm{ = }}\,{\rm{175}}\,{\rm{c}}{{\rm{m}}^2}\)
Therefore, the area of a given triangle is \({\rm{175}}\,{\rm{c}}{{\rm{m}}^2}\).
Q.3. Calculate the circumference of a circle with a radius of \(10.5\,{\rm{cm}}\)
Ans. Here, \(r = \,10.5\, = \frac{{105}}{{10}} = \,\frac{{21}}{2}\,{\rm{cm}}\)
We know that circumference of a circle \( = \,2\pi r\)
\( = \,2\, \times \,\frac{{22}}{7} \times \frac{{21}}{2}\,{\rm{cm = }}\,{\rm{66\, cm}}\)
Therefore, the circumference of a given circle is \({\rm{66\, cm}}.\)
Q.4. Find the volume of a cube whose edge measures \({\rm{8 \,cm}}.\)
Ans. Given, edge \(a = 8\,{\rm{cm}}\)
We know that volume of a cube \( = {a^3}\,{\rm{cubic}}\,{\rm{units}}\)
\( = {8^3} = 8 \times 8 \times 8\,{\rm{c}}{{\rm{m}}^3} = 512\,{\rm{c}}{{\rm{m}}^3}\)
Hence, the volume of the given cube is \(512\,{\rm{c}}{{\rm{m}}^3}\)
Q.5. The ends of a \(45\,{\rm{cm}}\) high cone’s frustum have radii of \(28\,{\rm{cm}}\) and \(7\,{\rm{cm}}\) respectively. Calculate the volume, curved surface area, and total surface area of this object.
Ans. Given, \(h = 45\,{\rm{cm}},\,{r_1} = 28\,{\rm{cm}}\) and \({r_2}\, = 7\,{\rm{cm}}\)
Volume of a frustrum \( = \frac{1}{3}\pi h({r_1}^2 + {r_2}^2 + {r_1}{r_2})\)
\( = \frac{1}{3} \times \frac{{22}}{7} \times 45[{(28)^2} + {(7)^2} + (28)(7)]\,{\rm{c}}{{\rm{m}}^3} = \,48510\,{\rm{c}}{{\rm{m}}^3}\)
we have \( = \,\sqrt {{h^2} + {{({r_1} – {r_2})}^2}} = \,\sqrt {{{(45)}^2} + {{(28 – 7)}^2}} {\rm{cm}}\)
\( = 3\,\sqrt {{{15}^2} + {7^2}} = 49.65\,{\rm{cm}}\)
So, the curved surface area of the frustum
\( = \pi ({r_1} + {r_2})l = \frac{{22}}{7}(28 + 7)(49.65) = 54615.5\,{\rm{c}}{{\rm{m}}^2}\)
Total surface area of the frustum
\( = \pi l({r_1} + {r_2}) + \pi {r_1}^2 + \pi {r_2}^2 = [5461.5\, + \frac{{22}}{7}({28^2}) + \frac{{22}}{7}{(7)^2}]{\rm{c}}{{\rm{m}}^2} = 8079.5\,{\rm{c}}{{\rm{m}}^2}\)
In this article, we learnt about the definition of geometry formulas, basic geometry formulas, types of geometry formulas, uses of geometry formulas, how Geometry formulas apply in maths, solved examples on geometry formulas and FAQs on geometry formulas.
The learning outcome of this article is how to use geometry formulas in finding the perimeter and area of plane figures and the volume and surface areas of solid figures.
Here are some of the most commonly asked questions on Geometry Formulas:
Q.1: What are the basic geometry formulas?
Ans: The basic geometry formulas are:
The perimeter of a triangle:
Perimeter \( = \,(a + b + c)\,{\rm{units}}\)
The perimeter of a quadrilateral:
Perimeter \( = \,(a + b + c + d)\,{\rm{units}}\)
Perimeter \( = \,4x\,{\rm{units}}\)
Perimeter of a rectangle \( = \,2(l \times b)\,{\rm{units}}\)
Circumference of a circle \( = \,2\pi r\,{\rm{units}}\)
Area of a triangle:
\({\rm{Area}}\,{\rm{of}}\,{\rm{a}}\,{\rm{triangle}} = \,\frac{1}{2} \times b \times h\,{\rm{sq}}{\rm{.}}\,{\rm{units}}\)
\({\rm{Area}}\,{\rm{of}}\,{\rm{an \,equilateral}}\,{\rm{triangle}} = \,\frac{{\sqrt 3 }}{4}{a^2}\,{\rm{sq}}{\rm{.}}\,{\rm{units}}\)
Area of a quadrilateral:
\({\rm{Area}}\,{\rm{of}}\,{\rm{a}}\,{\rm{square}}\, = \,{x^2}\,{\rm{sq}}\,{\rm{units}}\)
\({\rm{Area}}\,{\rm{of}}\,{\rm{rectangle}}\, = \,l \times b\,\,{\rm{sq}}\,{\rm{units}}\)
\({\rm{Area}}\,{\rm{of}}\,{\rm{rhombus}} = \,\frac{1}{2} \times \,({\rm{product}}\,{\rm{of}}\,{\rm{diagonals}})\,{\rm{sq}}\,{\rm{units}}\)
\({\rm{Area}}\,{\rm{of}}\,{\rm{parallelogram}} = \,({\rm{base}}\times{\rm{height}})\,{\rm{sq}}\,{\rm{units}}\)
\({\rm{Area}}\,{\rm{of}}\,{\rm{trapezium}} = \frac{1}{2}\,{\rm{(Sum}}\,{\rm{of}}\,{\rm{parallel\,sides}}) \times {\rm{Height}}\,{\rm{sq}}\,{\rm{units}}\)
\({\rm{Area}}\,{\rm{of}}\,{\rm{a}}\,{\rm{kite}} = \frac{1}{2}\,{\rm{(Product\, of \,diagonals}})\,{\rm{sq}}\,{\rm{units}}\)
Q.2: Are there formulas in geometry?
Ans: Yes, there are many formulas in geometry, in that few formulas are there to compute perimeter and area of a plane figure, and few formulas are there to compute surface area and volume.
Q.3: What is the formula of the cube?
Ans: There are formulas to calculate the surface and volume of a cube.
They are:
The lateral surface area of a cube \( = 4{a^2}{\rm{sq}}{\rm{.units}}\)
The total surface area of a cube \(6{a^2}{\rm{sq}}{\rm{.units}}\)
The volume of a cube \( = {a^3}\,{\rm{cubic}}\,{\rm{units}}\)
The volume of a cube \( = \sqrt 3 a\,{\rm{units}}\)
Q.4: What is the formula of the cylinder?
Ans: There are formulas to calculate the surface and volume of a cube.
There are:
The curved surface area of a cylinder \( = 2\pi rh\,{\rm{sq}}{\rm{.}}\,{\rm{units}}\)
The he total surface area of a cylinder \({\rm{ = }}\,2\pi r(r\, + \,h)\,{\rm{sq}}{\rm{.units }}\)
The volume of a cylinder \( = \pi {r^2}h\,{\rm{cubic}}\,{\rm{units}}\)
Q.5: What is the formula of the cuboid?
Ans: There are formulas to calculate the surface and volume of a cube.
There are:
Lateral surface area of a cuboid \( = 2h(l + b{\rm{)\,sq}}{\rm{.}}\,{\rm{units}}\)
Total surface area of a cuboid \( = 2\left( {lb + bh + hl} \right)\,{\rm{sq}}\,{\rm{units}}\)
Volume of a cuboid \( = l \times b \times h\,{\rm{cubic}}\,{\rm{units}}\)
Diagonal of a cuboid \({\rm{ = }}\,\sqrt {{l^2} + {b^2} +{h^2}} \,{\rm{units}}\)
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