Find the area of a ring whose outer and inner radii are 19 cm and 6 cm respectively.
A) \[330c{m^2}\]
B) \[310c{m^2}\]
C) \[320c{m^2}\]
D) \[350c{m^2}\]

Hint:
Here we need to find the area of the ring, for that, we will find the area of outer circle using the formula of area of circle. Then we will find the area of the inner circle using the formula. Then we will calculate the difference of area of inner and outer circle to get the required area of the ring.

Formula used:
We will use the formula of area of the circle,\[{\rm{Area}} = \pi {r^2}\].

Complete step by step solution:
First we will find the area of the outer circle of given radius of 19 cm.
Now, we will substitute the value of radius of outer circle in the formula \[{\rm{Area}} = \pi {r^2}\]. Therefore, we get
\[ \Rightarrow Area = \pi {\left( {19} \right)^2}\]
Simplifying the expression, we get
\[ \Rightarrow Area = \dfrac{{22}}{7} \times 361 = \dfrac{{7942}}{7}c{m^2}\]
Now, we will find the area of the inner circle of a given radius of 16 cm.
Substituting the value of radius of inner circle in the formula \[{\rm{Area}} = \pi {r^2}\], we get
\[ \Rightarrow Area = \pi {\left( {16} \right)^2}\]
Simplifying the expression, we get
\[ \Rightarrow Area = \dfrac{{22}}{7} \times 256 = \dfrac{{5632}}{7}c{m^2}\]
Now, we will find the area of the ring, which is equal to the difference of the area of the outer circle and area of the inner circle.
Area of ring \[ = \] Area of other circle \[ - \] Area of inner circle
Substituting the values of areas in the above equation, we get
\[ \Rightarrow \] Area of ring \[ = \dfrac{{7942}}{7}c{m^2} - \dfrac{{5632}}{7}c{m^2}\]
On subtracting the terms, we get
\[ \Rightarrow \]Area of ring \[ = \dfrac{{2310}}{7}c{m^2}\]
On further simplification, we get
\[ \Rightarrow \] Area of ring \[ = 330c{m^2}\]

Thus, the correct option is option A.

Note:
Here we have subtracted the fractional terms \[\dfrac{{7942}}{7}\] and \[\dfrac{{5632}}{7}\]. Here we can see that the denominator of these two fractions are the same, so we have directly subtracted the terms of the numerator keeping the denominator the same. But if two fractions have the different denominators, then to subtract them, we will take LCM so as to make the denominator the same.