10.2.3 Circles I, PT3 Focus Practice


Question 11:
The Town Council plans to build an equilateral triangle platform in the middle of a roundabout. The diameter of circle RST is 24 m and the perpendicular distance from R to the line ST is 18 m. as shown in Diagram below.
Find the perimeter of the platform.

Solution:

Given diameter = 24 m
hence radius = 12 m
O is the centre of the circle.
Using Pythagoras’ theorem:
x 2 = 12 2 6 2 x= 14436   =10.39 m TS=RS=RT  =10.39 m ×2  =20.78 m Perimeter of the platform TS+RS+RT =20.78×3 =63.34 m

Question 12:
Amy will place a ball on top of a pillar in Diagram below. Table below shows the diameters of three balls X, and Z.


Which ball X, or Z, can fit perfectly on the top of the pillar? Show the calculation to support Amy’s choice.

Solution:

Let the radius of the top of the pillar=r cm. O is the centre of the circle. In Δ OQR, r 2 = ( r4 ) 2 + 8 2  ( using Pythagoras' theorem ) r 2 = r 2 8r+16+64 r 2 = r 2 8r+80 r 2 r 2 +8r=80 8r=80 r= 80 8 r=10 cm Therefore, diameter =2×10 =20 cm Ball Y with diameter 20 cm can fit perfectly  on top of the pillar.

Question 13:
Diagram below shows a rim of a bicycle wheel with a diameter of 26 cm. Kenny intends to build a holder for the rim.

Which of the rim holder, X, or Z, can fit the bicycle rim perfectly? Show the calculation to support your answer.

Solution:

Let the radius of the rim holder=r cm. O is the centre of the circle. In Δ OQR, r 2 = ( r8 ) 2 + 12 2  ( using Pythagoras' theorem ) r 2 = r 2 16r+64+144 r 2 = r 2 16r+208 r 2 r 2 +16r=208 16r=208 r= 208 16 r=13 cm Therefore, diameter =2×13 =26 cm Rim holder Z with diameter 26 cm can fit the bicycle perfectly.


10.2.2 Circles I, PT3 Focus Practice


Question 6:
In the diagram below, CD is an arc of a circle with centre O.


Determine the area of the shaded region.
( Use π= 22 7 )

Solution:

Area of sector=Area of circle× 72 o 360 o                       = 22 7 × ( 10 ) 2 × 72 o 360 o                       = 440 7  cm 2 Area of ΔOBD= 1 2 ×6×8                        =24  cm 2 Area of shaded region= 440 7 24                                   =38 6 7  cm 2


Question 7:
In diagram below, ABC is a semicircle with centre O.

Calculate the area, in cm2 , of the shaded region.
( Use π= 22 7 )

Solution:
ACB= 90 o AB= 6 2 + 8 2     = 100     =10 cm Radius=10÷2            =5 cm The shaded region =( 1 2 × 22 7 ×5×5 )( 1 2 ×6×8 ) =39 2 7 24 =15 2 7  cm 2

Question 8:
In diagram below, ABC is an arc of a circle centre O

The radius of the circle is 14 cm and AD = 2 DE.
Calculate the perimeter, in cm, of the whole diagram.
( Use π= 22 7 )

Solution:
Length of arc ABC = 3 4 ×2πr = 3 4 ×2× 22 7 ×14 =66 cm Perimeter of the whole diagram =16+8+8+66 =98 cm

[adinserter block="3"]
Question 9:
In diagram below, KLMN is a square and KLON is a quadrant of a circle with centre K.



Calculate the area, in cm2, of the coloured region.
( Use π= 22 7 )

Solution:
Area of the coloured region = 45 o 360 o ×π r 2 = 45 o 360 o × 22 7 × 14 2 =77  cm 2


[adinserter block="3"]
Question 10:
Diagram below shows two quadrants, AOC and EOD with centre O.



Sector AOB and sector BOC have the same area.
Calculate the area, in cm2, of the coloured region.
( Use π= 22 7 )

Solution:
Area of the sector AOB=Area of the sector BOC Therefore, AOB=BOC                             = 90 o ÷2                             = 45 o Area of the coloured region = 45 o 360 o × 22 7 × 16 2 =100 4 7  cm 2

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

10.1 Circles I


10.1 Circles I
 
10.1.1 Parts of a Circle
1. A circle is set of points in a plane equidistant from a fixed point.

2. 
Parts of a circle:
(a)    The centre, O, of a circle is a fixed point which is equidistant from all points on the circle.




(b)
   A sector is the region enclosed by two radii and an arc.




(c)
    An arc is a part of the circumference of a circle.


(d)
   A segment is an area enclosed by an arc and a chord.
 

10.1.2 Circumference of a Circle
   circumference=πd,      where d=diameter                            =2πr,     where r=radius                                    π(pi)= 22 7      or   3.142
Example:
Calculate the circumference of a circle with a diameter of 14 cm. ( π = 22 7 )

Solution
:
Circumference = π × Diameter = 22 7 × 14 = 44 cm



10.1.3 Arc of a Circle
The length of an arc of a circle is proportional to the angle at the centre.
      Length of arc Circumference = Angle at centre 360 o        
Example:
 
Calculate the length of the minor arc AB of the circle above. ( π = 22 7 )

Solution
:
Length of arc Circumference = Angle at centre 360 o Length of arc A B = 120 o 360 o × 2 × 22 7 × 7 = 14 2 3 cm


10.1.4 Area of a Circle

   Area of a circle = π× ( radius ) 2                                =π r 2
 
Example:
Calculate the area of each of the following circles that has
(a) a radius of 7 cm,
(b)   a diameter of 10 cm.
( π = 22 7 )  

Solution
:
(a)
Area of a circle = π r 2 = 22 7 × 7 × 7 = 154 cm 2

(b)
Diameter of circle = 10 cm Radius of circle = 5 cm Area of circle = π r 2 = 22 7 × 5 × 5 = 78.57 cm 2


10.1.5 Area of a Sector
The area of a sector of a circle is proportional to the angle at the centre.
      Area of sector Area of circle = Angle at centre 360 o      

Example
:







Area of sector A B C = 72 o 360 o × 22 7 × 7 × 7 = 30 4 5 cm 2