Area of a segment
A segment is the section between a chord and an arc. It is essentially a sector with the triangle cut out, so we need to use our knowledge of triangles here as well.
To calculate the area of a segment, we will need to do three things:
- find the area of the whole sector
- find the area of the triangle within the sector
- subtract the area of the triangle from the area of the sector
Example
1. \(\text{Area of sector =}~\frac{40}{360} \times \pi \times {8}^{2}\)
\(\text{= 22.340...}\)
\(\text{= 22.34 cm}^{2}~\text{(to two decimal places)}\)
2. This is a non right-angled triangle, so we will need to use the formula:
\(\text{Area of triangle =}~\frac{1}{2}~\text{ab}~\text{sin C}\)
In this formula, \(\text{a}\) and \(\text{b}\) are the two sides which form the angle \(\text{C}\). So \(\text{a}\) and \(\text{b}\) are both 8cm, and \(\text{C}\) is 40⁰.
\(\text{Area of triangle =}~\frac{1}{2} \times {8} \times {8} \times \text{sin 40}\)
\(\text{= 20.569...}\)
\(\text{= 20.57 cm}^{2}~\text{(to two decimal places)}\)
3. To find the area of the shaded segment, we need to subtract the area of the triangle from the area of the sector.
\(\text{Area of segment = Area of sector - Area of triangle}\)
\(\text{= 22.34 - 20.57}\)
\(\text{= 1.77 cm}^{2}\)
Question
Find the area of the shaded segment:
1. \(\text{Area of sector =}~\frac{35}{360} \times \pi \times {80}^{2}\)
\(\text{= 1954.768...}\)
\(\text{= 1954.77 m}^{2}~\text{(to two decimal places)}\)
2. \(\text{Area of triangle =}~\frac{1}{2} \times {80} \times {80} \times \text{sin 35}\)
\(\text{= 1835.444...}\)
\(\text{= 1835.44 m}^{2}~\text{(to two decimal places)}\)
3. \(\text{Area of segment = Area of sector - Area of triangle}\)
\(\text{= 1954.77 - 1835.44}\)
\(\text{= 119.33 m}^{2}\)
