Tag: maths

Area of a sector $=\dfrac{\theta}{360^0} \times \pi r^2 =\dfrac{30}{360} \times \pi (7)^2 = \dfrac{49 \pi}{12}$

Area of sector formed from angle $\theta$ = $\frac{\theta}{260} \pi r^2$, where r is the radius of the circle
Now, if angle is 120, 150 then the ratio of area of sector will be:
= $\frac{\frac{120}{360} \pi r^2}{\frac{150}{360} \pi r^2}$
= $\frac{120}{150}$ = 4:5

Area of a sector of a circle of radius 'r' and angle  $ \theta = \frac {

\theta  }{ 360 } \pi {r}^{2}$
Given, $ \frac { 60 }{ 360 } \times \frac {22}{7} \times {r}^{2} = \frac {66}{7}  {cm}^{2} $
$ {r}^{2} = 18 $
Now, area of sector with angle {300}^{o} $ = \frac

{ 300 }{ 360 } \times \frac {22}{7} \times {r}^{2} = \frac

{ 300 }{ 360 } \times \frac {22}{7} \times 18 = \frac {330}{7}  {cm}^{2} $

Required area $\displaystyle =2\times \frac{115^{\circ}}{360^{\circ}}\pi \left ( 25 \right )^{2}$
$\displaystyle =2\times \frac{115}{360}\times \frac{22}{7}\times 625$

$1ℓ=1000\quad { cm }^{ 3 }\ 5.28ℓ=5.28\times 1000\quad { cm }^{ 3 }\ \quad \quad \quad \quad =5280\quad { cm }^{ 2 }$

$2\times 5=10$.
Sum of decimal places $= 4$
$\therefore 0.002\times 0.5=0.001$

$3.14\times { 10 }^{ 6 }=3.14\times 1000000=3140000$

$A)\dfrac{0.1}{0.03}=\dfrac{10}{3}=3.333$

LCM of the denominators of given rational numbers i.e. 5, 11, 7, 3, 6, and 8 = 9240