Question

The radius of a section of a circle is 3.5 cm and the length of its arc is 2.2 cm. Find the area of the arc.

Answer 1

Hint: We know that the area of a sector of a circle is \[A=\dfrac{\theta }{360}\times \pi {{r}^{2}}\] where $\theta$ is angle in degrees and r is radius of circle of which the sector is a part of, but we do not know \[\theta \] so we can use relation \[{{\theta }^{c}}=\dfrac{\pi }{180}\times \theta \], where \[{{\theta }^{c}}\] is angle in radian, we are aware about relation between s and \[{{\theta }^{c}}\] i.e. \[s=r{{\theta }^{c}}\](s is length of arc) Using these three relations we can find area of that sector.

Complete step-by-step solution:

Now, we know that area of a sector is
\[A=\dfrac{\theta }{360}\times \pi {{r}^{2}}\]
Where $\theta$ is the angle in degrees of the sector from the supposed circle of the same area
Now we have to use the relation between the length of arc and angle in radian,
\[s=r{{\theta }^{c}}\]
And the relationship between angle in degrees and radian is –
\[{{\theta }^{c}}=\dfrac{\pi }{180}\times \theta \]
Using above three relations we can find the solutions,
Since \[{{\theta }^{c}}=\dfrac{s}{r}\]
\[\dfrac{s}{r}=\dfrac{\pi }{180}\times \theta \]
\[\Rightarrow \dfrac{2.2}{3.5}=\dfrac{\pi }{180}\times \theta \]
\[\Rightarrow \dfrac{22}{35}=\dfrac{\pi }{180}\times \theta \]
\[\Rightarrow \dfrac{22}{35}=\dfrac{{22}/{7}\;}{180}\times \theta \]
\[\Rightarrow \theta =36{}^\circ \]
Area of arc is where r is given –
\[\Rightarrow A=\dfrac{1}{10}\times \pi {{(3.5)}^{2}}\]
\[\Rightarrow A=1.225\pi \] sq. cm
Hence the answer is 3.85 sq. cm

Note: The student must make sure he knows what is the relationship between arc length, angle in degree, angle in radian, etc. The common mistakes committed by students include doing wrong substitution in the equation, mistaking radian angle for degree angle. The student must remember all subsections from chapter trigonometry.