# 6th Class Mathematics Mensuration Mensuration Basics

Mensuration Basics

Category : 6th Class

MENSURATION BASICS

MENSURATION

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

PERIMETER

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

AREA

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

Some important formula of Mensuration

SQUARE:-

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

RECTANGLE

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

TRIANGLE

•                    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

CIRCLE

•                     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

RING

•                    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 $=14\,\,cm$

Radius$=\frac{14}{2}=7\,\,cm$

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

$=196\,\,c{{m}^{2}}-154\,\,c{{m}^{2}}$

$=42\,\,c{{m}^{2}}$

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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