The perimeter of the circle is equal to the circumference of the circle.
Circumference = 2πr
= 2 x 3.14 x 5
= 31.4 cm
Radius = 5cm
Area = $πr^2$ = 3.14 x 5 x 5 = 78.5 sq.cm
The height of the largest triangle inscribed will be equal to the radius of the semi-circle and base will be equal to the diameter of the semi- circle.
Area of triangle = $\dfrac{1}{2}$ x base x height
= $\dfrac{1}{2}$ x 2r x r
= $r^2$
Length of an arc = $\dfrac{θ}{360°} × (2πr)$
Length of an arc AB = $\dfrac{60°}{360°} × 2 × \dfrac{22}{7} × 21$
= $\dfrac{1}{6} × 2 × \dfrac{22}{7} × 21$
Arc AB Length = 22cm
The angle subtended by the arc = 60°
So, area of the sector = $\dfrac{60°}{360°} × πr^2$ $cm^2$
= $\dfrac{441}{6} × \dfrac{22}{7}$ $cm^2$
= 231 $cm^2$
The area of a sector = $\dfrac{θ}{360°} × π r^2$
Given, θ = p
So, area of sector = $\dfrac{p}{360} × π R^2$
Multiplying and dividing by 2 simultaneously,
= $\dfrac{\dfrac{p}{360}}{(π R^2)}×\dfrac{2}{2}$
= $\dfrac{p}{720} × 2πR^2$
Given,
Area of a circle = $154 cm^2$
$πr^2 = 154$
$\dfrac{22}{7} × r^2 = 154$
$r^2 = \dfrac{154 × 7}{22}$
$ r^2 = 7 × 7$
r = 7 cm
Perimeter of circle = $2πr = 2 × \dfrac{22}{7} × 7 = 44$ cm
If θ is the angle (in degrees) of a sector of a circle of radius r, then the area of the sector is $\dfrac{2πrθ}{360}$
If the sum of the circumferences of two circles with radii $R_1$ and $R_2$ is equal to the circumference of a circle of radius R, then $R_1 + R_2$ = R.
Here,
$R_1 = \dfrac{36}{2} = 18$ cm
$R_2 = \dfrac{20}{2} = 10$ cm
$R = R_1 + R_2 = 18 + 10 = 28$ cm
Therefore, the radius of the required circle is 28 cm
According to the given,
Perimeter of circle = Area of circle
2πr = $πr^2$
r = 2
Therefore, radius = 2 units