7th Class Mathematics Mensuration Mensuration

Mensuration

Category : 7th Class

Mensuration

Standard Units of Area

The inter relationship among various units of measurement of area are listed below.

$1{{m}^{2}}~~~~~~~=\text{ }\,\,\left( 100\text{ }\times \text{ }100 \right)\text{ }c{{m}^{2}}=\text{ }{{10}^{4}}c{{m}^{2}}$

$1\text{ }{{m}^{2}}~~~~~~~=\text{ }\left( 10\text{ }\times \text{ }10 \right)\text{ }d{{m}^{2}}=\text{ }100\text{ }d{{m}^{2}}$

$1\text{ }d{{m}^{2}}~~~~~~=\text{ }\left( 10\text{ }\times \text{ }10 \right)\text{ }c{{m}^{2}}=\text{ }100\text{ }c{{m}^{2}}$

$1\text{ }da{{m}^{2}}~~~~~=\text{ }\left( 10\text{ }\times \text{ }10 \right)\text{ }r{{n}^{2}}=\text{ }100\text{ }{{m}^{2}}$

$1\text{ }h{{m}^{2}}~~~~~~=\text{ }\left( 100\text{ }\times \text{ }100 \right)\text{ }{{m}^{2}}=\text{ }{{10}^{4}}{{m}^{2}}$

$1\text{ }k{{m}^{2}}~~~~~~=\text{ }\left( 1000\text{ }\times \text{ }1000 \right)\text{ }{{m}^{2}}=\text{ }{{10}^{6}}{{m}^{2}}$

$1\text{ }hectare~~=\text{ }\,10000{{m}^{2}}$

$1\,\,k{{m}^{2}}~~~~~~=\,\,\,100\text{ }hectare$

Formula Related to Perimeter and Area

•              $Area\text{ }of\text{ }a\text{ }triangle=\frac{1}{2}\times b\times h$
•              $Area\text{ }of\text{ }an\text{ }equilateral\text{ }triangle=\frac{\sqrt{3}}{4}\times {{a}^{2}}$
•       Perimeter of a rectangle $=\text{ }2\text{ }\left( Length\text{ }\times \text{ }breadth \right)$
•             $Area\,of\,a\,rectangle\,\,=\,\,Length\,\,\times \,\,breadth$
•             $Diagonal\,\,of\,\,a\,\,rec\operatorname{tangle}\,\,=\sqrt{length{{)}^{2}}+{{(breadth)}^{2}}}$
•             Perimeter of a square $=\text{ }4\text{ }\times \text{ }side$
•             $Area\text{ }of\text{ }a\text{ }square\text{ }=\text{ }sid{{e}^{2}}=\,\,\frac{1}{2}\times \text{ }{{\left( diagonal \right)}^{2}}$
•             $Side\,\,of\,a\,square=\sqrt{area}$
•             Diagonal of a square $=\text{ }side\text{ }\times \text{ }\sqrt{2}$
•             Perimeter of a parallelogram = 2 $\times$sum of length of adjacent sides.
•            Area of a parallelogram = $base\text{ }\times \text{ }corresponding\text{ }height.$
•            Perimeter of a rhombus = $~4\text{ }\times \text{ }side$
•            Area of a rhombus $=\,\,base\text{ }\times \text{ }vertical\text{ }height$
•           Area of a rhombus $=\left( \frac{1}{2} \right)~\,\times \text{ }product\text{ }of\text{ }diagonals$
•           Circumference of a circle = $2\pi r$
•           Area of a circle = $\pi {{r}^{2}}$
•           The volume of a cuboid $=\text{ }length\text{ }\times \text{ }breadth\text{ }\times \text{ }height$
•           The volume of a cube = (length)3

•              Example:

Find the area of a triangle whose sides are 15 cm, 9 cm and 2 cm.

(a) 48 $c{{m}^{2}}$                                                               (b) 80 $c{{m}^{2}}$

(c) 54 $c{{m}^{2}}$                                                               (d) 78 $c{{m}^{2}}$

(e) None of these

Ans.     (c)

Explanation: Here $a\text{ }=\text{ }15\text{ }cm,\text{ }b\text{ }=\text{ }9\text{ }cm\text{ }and\text{ }c\text{ }=\text{ }12\text{ }cm$

Also, ${{a}^{2}}=\text{ }{{b}^{2}}+\text{ }{{c}^{2}}\Rightarrow$ The given triangle is a right triangle.

$\therefore \text{ }Area\text{ }of\text{ }the\text{ }right\text{ }triangle\text{ }=\text{ }\frac{1}{2}\text{ }\times \text{ }9\text{ }\times \text{ }12\text{ }=\text{ }54\text{ }c{{m}^{2}}$

•               Example:

The dimensions of the floor of a room are 15 m and 20 m« How many square tiles each of length 20 cm are required to furnish the floor?

(a) 2,400                                                           (b) 5,200

(c) 7,500                                                            (d) 5,250

(e) None of these

Ans.     (c)

Explanation:  $Area\text{ }of\text{ }the\text{ }room\text{ }=\text{ }15\text{ }m\text{ }\times \text{ }20\text{ }m$

$=\text{ }1500\text{ }cm\text{ }\times \text{ }2000\text{ }cm\text{ }=\text{ }3\text{ }\times \text{ 1}{{\text{0}}^{6}}c{{m}^{2}}$

$Area\text{ }of\text{ }a\text{ }tile\text{ }=\text{ }20\text{ }cm\text{ }\times \text{ }20\text{ }cm\text{ }=\text{ }400\text{ }c{{m}^{2}}$

Total number of tiles required $=\,\,\frac{3\times {{10}^{6}}}{400}\,\,=\,\,\frac{30000}{4}=\,\,7,500$

Other Topics

Notes - Mensuration

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