Question

I need help on 11A, I got the answer as 5.046cm but the book says it's 8.37 cm

Answer 1

Answer:

$\textsf{Area of a triangle}=\dfrac{1}{2} ab \sin C$

(where $a$ and $b$ are the sides, and $C$ is the included angle)

$\textsf{Area of a sector}= \dfrac{\theta}{360} \pi r^2$

(where $\theta$ is the angle in degrees and $r$ is the radius)

Therefore, using the above formulas:

$\begin{aligned}\textsf{Area of segment } \rm (A) & = \textsf{area of sector}- \textsf{area of triangle}\\& = \dfrac{\theta}{360} \pi r^2 - \dfrac{1}{2}r^2 \sin \theta\\\end{aligned}$

Given:

• $\theta$ = 90°
• A = 20 cm²

Substitute the given values into the formula and solve for r:

$\begin{aligned}\implies 20 & = \dfrac{90}{360} \pi r^2 - \dfrac{1}{2}r^2 \sin 90\\\\20 & = \dfrac{1}{4} \pi r^2 - \dfrac{1}{2}r^2 (1)\\\\20 & = \left(\dfrac{1}{4} \pi -\dfrac{1}{2}\right)r^2\\\\20 & = \left(\dfrac{\pi -2}{4}\right)r^2\\\\r^2 & = 20\left(\dfrac{4}{\pi -2}\right)\\\\r^2 & =\dfrac{80}{\pi-2}\\\\r & =\sqrt{\dfrac{80}{\pi-2}}\\\\r & = 8.37\: \sf (2\:dp)\end{aligned}$