# Mensuration Formulas for Class 8 Maths Chapter 11

## Mensuration Formulas for Class 8 Maths Chapter 11

Are you looking for Mensuration formulas or important points that are required to understand Mensuration for class 8 maths Chapter 11? You are the right place to get all information about Mensuration class 8 maths chapter 11. Mensuration formulas play a vital role in preparing you for the class 8 exam as well as higher studies. Mensuration formulas are very helpful for better scores in the exam.  Check Mensuration formulas according to class 8:

 $\ Area\;of\;Square = l^{2}$ $\ Perimeter\;of\;Square = 4 \times l$Where, l : length of side $\ Area\;of\;Rectangle = l \times w$ $\ Perimeter\;of\;Rectangle = 2 (l+w)$Where, L = Length w = Width $\ Area\;of\;Circle = \pi r^{2}$ $\ Perimeter\;of\;Circle = 2 \pi r$Where, 𝒓 = Radius d = Diameter d = 2𝒓 $\ Area\;of\;Scalene\;Triangle = \sqrt{s(s-a)(s-b)(s-c)}$ $\ Perimeter\;of\;Scalene\;Triangle = a+b+c$Where, a, b, c are Side of Scalene Triangle $Area\;of\;Isoscele\;Triangle =\frac{1}{2}bh$ $Altitude\;of\;an\;Isosceles\;Triangle=\sqrt{a^{2}-\frac{b^{2}}{4}}$ Where, b = Base of the isosceles triangle h = Height of the isosceles triangle & a = length of the two equal sides $\large Perimeter\;of\;Isosceles\;Triangle,P=2\,a+b$ Where, a = length of the two equal sides b = Base of the isosceles triangle $Area \;of \;an \;Right\;Triangle = \frac{\sqrt{1}}{2}bh$ $Perimeter \;of \;an \;Right \;Triangle = a+b+c$ $semi\;Perimeter \;of \;an \;Right \;Triangle = \frac{a+b+c}{2}$ where, b is the Base of Right Triangle. h is the Hypotenuse of Right Triangle. a is the Hight of Right Triangle. $Area \;of \;an \;Equilateral \;Triangle = \frac{\sqrt{3}}{4}a^{2}$ $Perimeter \;of \;an \;Equilateral \;Triangle = 3a$ $Semi \;Perimeter \;of \;an \;Equilateral \;Triangle = \frac{3a}{2}$ $Height \;of \;an \;Equilateral \;Triangle = \frac{\sqrt{3}}{2}a$ Where, a is the side of an equilateral triangle. h is the altitude of an equilateral triangle. $\large perimeter\;of\;a\;rhombus =4\times Side$ $\large Area\;of\;a\;Rhombus = A = \frac{1}{2} \times d_{1} \times d_{2}$ Where, d1 and d2 are the diagonals of the rhombus bisect each other 90-degree angle. $\ Area\;of\;a\;Parallelogram = b\times h$ $\ Perimeter\;of\;Parallelogram = 2\left(b+h\right)$ Where, b: Base h: Height. $\large Perimeter\;of\;a\;Trapezoid\;=a+b+c+d$ Where, a, b, c, d are the lengths of each side. $\large Area\;of\;a\;Trapezoid\; = \frac{1}{2} \times h \times (a + b)$ Where:, h = height a = the short base b = the long base $\large Surface\;area\;of\;Cube=6a^{2}$ $\large Volume\;of\;a\;cube=a^{3}$ Where, a is the side length of the cube. $\large Surface\;area\;of\;Cuboid = 2(lb + bh + hl)$ $\large Volume\;of\;a\;Cuboid = h \times l \times w$ Where, l: Height h: Legth w: Depth $\large Diameter\;of\;a\;sphere=2r$ $\large Circumference\;of\;a\;sphere=2\pi r$ $\large Surface\;area\;of\;a\;sphere=4\pi r^{2}$ $\large Volume\;of\;a\;sphere=\frac{4}{3}\: \pi r^{3}$ Where, r: Radius $\large Curved\;Surface\;area\;of\;a\;Hemisphere =4\pi r^{2}$ $\large Total\;Surface\;area\;of\;a\;Hemisphere =3\pi r^{2}$ $\large Volume\;of\;a\;Hemisphere =\frac{2}{3}\: \pi r^{3}$ Where, r: Radius $\large Curved\;Surface\;area\;of\;a\;Cylinder =2\pi rh$ $\large Total\;Surface\;area\;of\;a\;Cylinder =2\pi r(r+h)$ $\large Volume\;of\;a\;Cylinder = \pi r^{2} h$ Where, r: Radius h: Height $\large Total\;Surface\;Area\;of\;cone=\pi r \left (s+r \right )$ $\large Vomule\;of\;cone=\frac {1}{3}\pi r^{2}h$ $\large Curved\;Surface\;Area\;of\;cone=\pi rs$ Where, r is the radius of cone. h is the height of cone. s is the slant height of the cone.

#### Summary of Mensuration

We have shared very important formulas for Mensuration which is helps to score in the class 8 exams. If you have any questions and doubts related too Mensuration please let me know through comment or mail as well as social media. When you understand the formulas behind each Mensuration topics then it would be easier to solve the most complex problems related to a Mensuration too.

Chapter-wise Maths Formulas for Class 8

### NCERT Solutions Class 8 Maths By Chapters

• Chapter 1 – Rational Numbers
• Chapter 2 – Linear Equations in One Variable
• Chapter 3 – Understanding Quadrilaterals
• Chapter 4 – Practical Geometry
• Chapter 5 – Data Handling
• Chapter 6 – Squares and Square Roots
• Chapter 7 – Cubes and Cube Roots
• Chapter 8 – Comparing Quantities
• Chapter 9 – Algebraic Expressions and Identities
• Chapter 10 – Visualising Solid Shapes
• Chapter 11 – Mensuration
• Chapter 12 – Exponents and Powers
• Chapter 13 – Direct and Inverse Proportions
• Chapter 14 – Factorisation
• Chapter 15 – Introduction to Graphs
• Chapter 16 – Playing with Numbers