10.2 Circles I, PT3 Focus Practice


10.2 Circles I, PT3 Focus Practice
Question 1:
Diagram below shows a circle with centre O.
 
The radius of the circle is 35 cm.
Calculate the length, in cm, of the major arc AB.
( Use π = 22 7 )
 
Solution:
Angle of the major arc AB = 360o – 144o= 216o
Length of major arc A B = 216 o 360 o × 2 π r = 216 o 360 o × 2 × 22 7 × 35 = 132 cm



Question 2:
In diagram below, O is the centre of the circle. SPQ and POQ are straight lines.
 
The length of POis 8 cm and the length of POQ is 18 cm.
Calculate the length, in cm, of SPT.

Solution
:
Radius = 18 – 8 = 10 cm
PT2 = 102 – 82
  = 100 – 64
   = 36
PT = 6 cm
Length of SPT= 6 + 6
= 12 cm


Question 3:
Diagram below shows two circles. The bigger circle has a radius of 14 cm with its centre at O.
The smaller circle passes through O and touches the bigger circle.
 
Calculate the area if the shaded region.
( Use π = 22 7 )
 
Solution:
Area of bigger circle = π R 2 = 22 7 × 14 2 Radius, r of smaller circle = 1 2 × 14 = 7 cm Area of smaller circle = π r 2 = 22 7 × 7 2 Area of shaded region = ( 22 7 × 14 2 ) ( 22 7 × 7 2 ) = 616 154 = 462 cm 2
 


Question 4:
Diagram below shows two sectors. ABCD is a quadrant and BED is an arc of a circle with centre C.
Calculate the area of the shaded region, in cm2.
( Use π = 22 7 )

Solution
:
The area of sector C B E D = 60 o 360 o × π r 2 = 60 o 360 o × 22 7 × 14 2 = 102 2 3 cm 2

The area of quadrant A B C D = 1 4 × π r 2 = 1 4 × 22 7 × 14 2 = 154 cm 2


Area of the shaded region = 154 102 2 3 = 51 1 3 cm 2


Question 5:
Diagram below shows a square KLMN. KPN is a semicircle with centre O.
Calculate the perimeter, in cm, of the shaded region.
( Use π = 22 7 )

Solution
:
K O = O N = O P = 7 cm P N = 7 2 + 7 2 = 98 = 9.90 cm Arc length K P = 1 4 × 2 × 22 7 × 7 = 11 cm
Perimeter of the shaded region
= KL + LM +MN + NP + Arc length PK
= 14 + 14 +14 + 9.90 + 11
= 62.90 cm

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