10.2.1 Circles I, PT3 Focus Practice


10.2.1 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 PO is 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 of 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 ) =616154 =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 CBED = 60 o 360 o ×π r 2 = 60 o 360 o × 22 7 × 14 2 =102 2 3  cm 2

The area of quadrant ABCD = 1 4 ×π r 2 = 1 4 × 22 7 × 14 2 =154  cm 2


Area of the shaded region =154102 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
:
KO=ON=OP=7 cm PN= 7 2 + 7 2 = 98 =9.90 cm Arc length KP = 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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