6th Class Mathematics Mensuration Mensuration Basics

Mensuration Basics

Category : 6th Class




  •                    The branch of geometry that deals with the measurement of length, area, or volume.



  •                   The length of the sides enclosing the figure is called its perimeter.



  •                   The amount of space in the boundary of plane figure.
  •                    Unit of area is square unit.


Some important formula of Mensuration



  •                     Perimeter of square\[=4\times side=4a\]
  •                    Area of square\[={{(side)}^{2}}={{(a)}^{2}}\]
  •                    Diagonal of square \[=\sqrt{2}\times \] side





  •                    Perimeter of rectangle \[=2(l+b)\]
  •                   Area of rectangle \[=l\times b\] sq. unit
  •                   Diagonal of rectangle \[=\sqrt{{{1}^{2}}+{{b}^{2}}}\]


  •                    Perimeter of triangle = sum of all side
  •                    Area of triangle \[=\frac{1}{2}\times \]base \[\times \]height sq. unit

\[=\frac{1}{2}\times AD\times BC\]sq. unit

  •                    Area of equilateral triangle \[=\frac{\sqrt{3}}{4}\times {{\left( side \right)}^{2}}\]

\[=\frac{\sqrt{3}}{4}\times {{\left( a \right)}^{2}}\]sq. unit



  •                     Circumference of circle \[=2\pi r\]
  •                     Area of circle\[=\pi {{r}^{2}}\]sq. unit

  •                    Area of quadrant \[=\frac{1}{4}\pi {{r}^{2}}\]sq. unit

Note\[\to \pi =\frac{22}{7},\,3.14\]

  •                   Area of semicircle \[=\frac{1}{2}\pi {{r}^{2}}\] sq. unit


  •                    Area of ring \[=\pi \left( {{R}^{2}}-{{r}^{2}} \right)\] sq. unit

Example 1: The area of a rectangle is\[270\,\,c{{m}^{2}}\]. If its length is 30 cm, find its breadth.

Solution: We have

Area of rectangle\[=270\,\,c{{m}^{2}}\]

Length of rectangle \[=30\,\,cm\]

\[\therefore \]Breadth of rectangle\[=\left( \frac{Area}{Length} \right)cm\]

\[=\left( \frac{Area}{Length} \right)cm\]

Example 2: Find the area of shaded region.

Solution: Area square\[={{(side)}^{2}}\]

\[\Rightarrow \]\[{{(14)}^{2}}=196\,\,c{{m}^{2}}\]

Now, diameter \[=14\,\,cm\]


Area\[=\pi {{r}^{2}}=\frac{22}{7}\times 7\times 7=154\,\,c{{m}^{2}}\]

Area of shaded region



Example 3: Find the area of triangle, whose base and height are 13 cm and 14 cm.

Solution: Area of triangle\[=\frac{1}{2}\times \]base\[\times \]height\[=\frac{1}{2}\times 13\times 14\,\,c{{m}^{2}}\]



Notes - Mensuration Basics
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