Question

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

Answer 1

`\textsf{Area of a triangle}=(1)/(2) ab \sin C`

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

`\textsf{Area of a sector}= (\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}\\& = (\theta)/(360) \pi r^2 - (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 & = (90)/(360) \pi r^2 - (1)/(2)r^2 \sin 90\\\\20 & = (1)/(4) \pi r^2 - (1)/(2)r^2 (1)\\\\20 & = \left((1)/(4) \pi -(1)/(2)\right)r^2\\\\20 & = \left((\pi -2)/(4)\right)r^2\\\\r^2 & = 20\left((4)/(\pi -2)\right)\\\\r^2 & =(80)/(\pi-2)\\\\r & =\sqrt{(80)/(\pi-2)}\\\\r & = 8.37\: \sf (2\:dp)\end{aligned}`