Trigs : Areas

Trigs : Areas

Area Formulas:

* Area of segment = Area of Sector - Area of Triangle 
*Area of Triangle =  $\frac{1}{2}ab\sin C$  (C is in Degrees)
*Area of Sector =  $\frac{1}{2}r^2\theta$  (theta in radians)

Examples

1. Find the area of the following sector.
 $180=\pi$  

 $1^{\degree}=\frac{\pi}{180}$  
 $\therefore45\degree=\frac{\pi}{180}\times45$  
 $=\frac{1}{4}\pi$  

 $A=\frac{1}{2}r^2\theta$  
 $=\frac{1}{2}\left(15.5\right)^2\left(\frac{\pi}{4}\right)$  
 $=94.3cm^2\ or\ \frac{961}{32}\pi cm^2$  

2. Find the area of the sector.

  $A=\frac{1}{2}r^2\theta$  

 $=\frac{1}{2}\left(6.8\right)^2\left(\frac{\pi}{3}\right)$  
 $=25.2cm^2or\ \frac{578}{75}\pi\ cm^2$  

3. Find the area of the major sector.

 $A=\frac{1}{2}r^2\theta$  
 $=\frac{1}{2}\left(10.2\right)^2\left(\frac{4}{3}\pi\right)$  
 $=\frac{1734}{25}\pi\ cm^2\ or\ 218cm^2$  

Question

4.The diagram shows a sector OABC of a circle with center O. OA = OC = 10.4cm, angle AOC = 120 degrees.

1. Calculate the length of the arc ABC correct to 3s.f.

2. Calculate the area of the shaded segment ABC.

Solution 

1.                                                  2.

  $180^{\degree}=\pi$                           Area of sector =  $\frac{1}{2}r^2\theta$  

 $1\degree=\frac{\pi}{180}$                            $=\frac{1}{2}\left(10.4\right)^2\left(\frac{2}{3}\pi\right)$  
 $\therefore120\degree=\frac{\pi}{180}\times120$         $=\frac{2704}{45}\pi\ cm^2or\ 113.26cm^2$  
 $=\frac{2}{3}\pi$                               Area of Triangle = $\frac{1}{2}ab\sin C$  

 $s=r\theta$                              $=\frac{1}{2}\left(10.4\right)\left(10.4\right)\sin\left(120\right)$  
 $=10.4\times\frac{2}{3}\pi$                   $=46.83cm^2$  
 $21.8cm\ \left(3s.f.\right)$                Area of segment =  $113.26-46.8$  = $66.4cm^2$  

Question

5. The diagram shows an equilateral triangle ABC with sides of length 6cm.

P is the midpoint of AB. Q is the mid point of AC. APQ is a sector, calculate the are of the shaded region.

Solution

 $180\degree=\pi$                               Area of Triangle = $=\frac{1}{2}ab\sin C$  
 $1\degree=\frac{\pi}{180}$                              $=\frac{1}{2}\left(6\right)\left(6\right)\sin60$  
 $\therefore60\degree=\frac{\pi}{180}\times60$              $=15.6cm^2$  
 $=\frac{1}{3}\pi$                                     Area of shaded region = $15.6-4.7$  
                                             $=10.9cm^2$  
Area of sector  $=\frac{1}{2}r^2\theta$  
 $=\frac{1}{2}\left(3\right)^2\left(\frac{1}{3}\pi\right)$  
 $=4.7cm^2$  

Question

6.The diagram shows a triangle PQR in which PQ = 10cm, QR = 14cm. and QPR = 0.7radians

a. Find the size of angle PRQ in radians to 2d.p.

b. Find the area of the shaded region.

Note: The point S lies on PR such that PS =10cm

Solution

a. 

 $1^c=\frac{180}{\pi}$                                $1\degree=\frac{\pi}{180}$                
 $0.7^c=\frac{180}{\pi}\times0.7$            $27.39=\frac{\pi}{180}\times27.39$  
 $=40.11\degree$                                      $=0.48^c$  

 $\frac{\sin A}{a}=\frac{\sin B}{b}$  
 $\frac{\sin\ A}{10}=\frac{\sin\left(40.11\right)}{14}$  
 $\sin\ A=\ \frac{\sin\left(40.11\right)}{14}\times10$  
 $A=\ \sin^{-1}\left(\frac{\sin\left(40.11\right)}{14}\times10\right)$  
=27.39 degrees 

Soltuion p2

b.
Area of sector =  $\frac{1}{2}r^2\theta$                   Area of shaded region =  $45.1-35$  
 $=\frac{1}{2}\left(10\right)^2\left(0.7\right)$                          $=10.1cm^2$  
 $=35cm^2$  

Area of triangle = $\frac{1}{2}ab\sin C$  
 $=\frac{1}{2}\left(10\right)\left(14\right)\sin\left(40.11\right)$  
 $=45.1cm^2$