Question

The perimeter of sector OAB of a circle with centre O and radius 5.6 cm, is 15.6 cm. Find length of the arc AB. Also find the value of \(\theta\).

Hint:

Express decimal values as fractions to make cancellation straightforward:
\[ 4.4 = \frac{44}{10} \quad \text{and} \quad 5.6 = \frac{56}{10} \] Substituting these into the equation:
\[ \frac{44}{10} = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times \frac{56}{10} \] Allows the fractions and decimals to cancel out perfectly with minimal effort!

Answer 1

Step 1: Find the arc length from the perimeter.
Perimeter of sector $= 2r + l$, so $15.6 = 2(5.6) + l$, giving $l = 15.6 - 11.2 = 4.4$ cm.
Step 2: Switch to the radian form of the arc length formula.
In radians, $l = r\theta$, so $\theta_{\text{rad}} = \frac{l}{r} = \frac{4.4}{5.6} = \frac{11}{14}$.
Step 3: Convert this angle to degrees.
\[ \theta_{\text{deg}} = \frac{11}{14} \times \frac{180 \times 7}{22} = \frac{11 \times 180 \times 7}{14 \times 22} = \frac{13860}{308} = 45^\circ \]
\[ \boxed{l = 4.4\ \text{cm}, \ \theta = 45^\circ} \]