To find the area of a sector given the radius [tex]\( r \)[/tex] and the angle [tex]\( \theta \)[/tex] in radians, we use the formula:
[tex]\[
\text{Area of a sector} = \frac{1}{2} r^2 \theta
\][/tex]
Given:
- [tex]\( r = \frac{9}{2} \)[/tex]
- [tex]\( \theta = \frac{5 \pi}{6} \)[/tex]
First, let's calculate the area by substituting the given values into the formula:
[tex]\[
\text{Area} = \frac{1}{2} \left(\frac{9}{2}\right)^2 \cdot \frac{5 \pi}{6}
\][/tex]
We know from the calculations:
[tex]\[
\left(\frac{9}{2}\right)^2 = \left(\frac{9}{2}\right) \times \left(\frac{9}{2}\right) = \frac{81}{4}
\][/tex]
Thus:
[tex]\[
\text{Area} = \frac{1}{2} \cdot \frac{81}{4} \cdot \frac{5 \pi}{6}
\][/tex]
Simplify step-by-step:
[tex]\[
\text{Area} = \frac{1}{2} \cdot \frac{81}{4} \cdot \frac{5 \pi}{6} = \frac{81}{8} \cdot \frac{5 \pi}{6} = \frac{81 \times 5 \pi}{8 \times 6} = \frac{405 \pi}{48}
\][/tex]
Simplify the fraction [tex]\(\frac{405}{48}\)[/tex]:
[tex]\[
\frac{405}{48} = \frac{405 \div 3}{48 \div 3} = \frac{135}{16}
\][/tex]
Therefore, the area of the sector in terms of [tex]\(\pi\)[/tex] is:
[tex]\[
\text{Area} = \frac{135 \pi}{16} \text{ sq units}
\][/tex]
Thus, the area of the sector when [tex]\( r = \frac{9}{2} \)[/tex] and [tex]\( \theta = \frac{5 \pi}{6} \)[/tex] radians is:
[tex]\[
\frac{135 \pi}{16} \text{ sq units}
\][/tex]
