5.2.2 Circles, PT3 Focus Practice


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


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

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