Mensuration Basics
Category : 6th Class
MENSURATION BASICS
MENSURATION
PERIMETER
AREA
Some important formula of Mensuration
SQUARE:-
RECTANGLE
TRIANGLE
\[=\frac{1}{2}\times AD\times BC\]sq. unit
\[=\frac{\sqrt{3}}{4}\times {{\left( a \right)}^{2}}\]sq. unit
CIRCLE
Note\[\to \pi =\frac{22}{7},\,3.14\]
RING
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 \[=14\,\,cm\]
Radius\[=\frac{14}{2}=7\,\,cm\]
Area\[=\pi {{r}^{2}}=\frac{22}{7}\times 7\times 7=154\,\,c{{m}^{2}}\]
Area of shaded region
\[=196\,\,c{{m}^{2}}-154\,\,c{{m}^{2}}\]
\[=42\,\,c{{m}^{2}}\]
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\[=\frac{1}{2}\times 13\times 14\,\,c{{m}^{2}}\]
\[=91\,\,c{{m}^{2}}\]
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