# 5.2.6 Circles, PT3 Focus Practice

 
Question 14:
In the Diagram, OKLM is a sector of a circle and OAB is a quadrant of a circle with common centre O.

It is given that OA = 8 cm, ∠KOM = 90o and OK : OB = 3 : 2.
Using      π=     22   7        , calculate
(a) the area, in cm2, of the shaded region,
(b) the perimeter, in cm, of the shaded region.

Solution:
 (a)
           OB=OA=8 cm and           OK:OB=3:2                  OK       OB      =    3    2                       OK    8      =    3    2                OK=    3    2      ×8           OK=12                      Area of the shaded region           =(                      270     o                        360     o               ×       22    7       ×       12    2         )(                      90     o                        360     o               ×       22    7       ×    8    2         )           =339    3    7      50    2    7                =289    1    7           cm    2                 

(b)
         Length of arc KLM           =             270    o                   360    o            ×2×       22    7      ×12           =56    4    7       cm                      Length of arc AB           =             90    o                   360    o            ×2×       22    7      ×8           =12    4    7       cm                      Perimeter of shaded region           =Length of arc KLM+Length of            arc AB+MO+OK+OA+OB           =56    4    7      +12    4    7      +12+12+8+8           =109    1    7       cm            

 

Question 15:
Diagram shows sectors OABC and ODE with the common centre respectively.

Calculate
(a) the perimeter, in cm, of the whole diagram,
(b) the area, in cm2, of the shaded region.

Solution:
(a)

(b)