Area
Category : 5th Class
All the geometrical shapes occupies some space. The occupied space by a geometrical shape is called area of that geometrical shape.
Shaded part in the above figures represent area.
Area of a Triangle
Area of a triangle \[=\frac{1}{2}\times \] base x height.
Height
In a triangle, the length of the perpendicular which is drawn from vertex to the opposite side is called height of the triangle.
Base
In a triangle, the length of the side of the triangle on which perpendicular is drown is called base.
Area of a triangle \[=\frac{1}{2}\times \] base \[\times \] height
In triangle ABC
Height = CD and base = AB
Area of the triangle \[ABC=\times AB\times CD.\]
Find the area of the following figure:
Explanation
Area of a triangle \[=\frac{1}{2}\times \] base \[\times \]height
In triangle PQR
Area of the triangle \[\text{PQR=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ QR }\!\!\times\!\!\text{ PS=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ 4cm }\!\!\times\!\!\text{ 7cm14c}{{\text{m}}^{\text{2}}}\]
Area of a Rectangle
Area of a rectangle = length \[\times \] breadth.
Length
The longer side of a rectangle is called length of the rectangle.
Breadth
The shorter side of a rectangle is called breadth of the rectangle.
In the rectangle PQRS
Length of rectangle = Longer side PQ = RS = 7 cm
Breadth of the rectangle = Shorter side
QR = SP = 5 cm
Area of the rectangle PQRS = Length \[\times \] Breadth
= PQ \[\times \] QR.
Find the area of the following figure:
In rectangle ABCD
Length = AB = 6 cm
Breadth = BC = 4 cm
Thus area of the rectangle 3 \[\text{ABCD=AB }\!\!\times\!\!\text{ BC=6 cm }\!\!\times\!\!\text{ 4 cm=24 c}{{\text{m}}^{\text{2}}}\text{.}\]
Area of a Square
Area of a square\[~=\text{ }sid{{e}^{2}}.\]
In the square PQRS PQ=QR=RS=SP Area of the square \[PQRS\text{ }=\text{ }PQ\text{ }\times \text{ }PQ\] \[=P{{Q}^{2}}.\]
Find the area of the following figure:
In square ABCD
AB = BC = CD = DA = 5 cm
Area of the square \[\text{ABCD }=\text{ A}{{\text{B}}^{\text{2}}}\]
\[{{(5cm)}^{2}}=25c{{m}^{2}}\]
Area of a Circle
Area of the circle \[=\pi {{r}^{2}}\]
To find the area of a circle, square of the radius is multiplied by the constant \[\pi .\]
Find the area of the circle whose radius is 11.9 cm.
Solution:
Area of a circle \[\pi {{r}^{2}}\]
\[=\frac{22}{7}\times 11.9\text{ }cm\text{ }\times 11.9\text{ }cm\]
\[=445.06\text{ }c{{m}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec