A chord subtends an angle of 72º at the centre of a circle of radius 24.5m. Calculate the perimeter of the minor segment. [Take π = 227

Question 1 Report

A chord subtends an angle of 72º at the centre of a circle of radius 24.5m. Calculate the perimeter of the minor segment. [Take π = $\frac{22}{7}$

Answer

From ΔOKB

sin 36° = $\frac{| K B |}{24.5}$

|KB| = 0.5878 x 24.5

=14.4011m

Or

Length of arc = $\frac{72}{360} * 2 * \frac{22}{7} * 24.7$

=30.8m

Perimeter = 2(14.4011) + 30.8

= 59.6022

= 59.6m

Answer Details

From ΔOKB

sin 36° = $\frac{| K B |}{24.5}$

|KB| = 0.5878 x 24.5

=14.4011m

Or

Length of arc = $\frac{72}{360} * 2 * \frac{22}{7} * 24.7$

=30.8m

Perimeter = 2(14.4011) + 30.8

= 59.6022

= 59.6m

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