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1.5.3 Circular Measure Long Questions (Question 5 & 6)


Question 5:
Diagram below shows a circle PQRT, centre and radius 5 cm. AQB is a tangent to the circle at Q. The straight lines, AO and BO, intersect the circle at P and R respectively. 
OPQR is a rhombus. ACB is an arc of a circle at centre O.
Calculate
(a) the angle x , in terms of π ,
(b) the length , in cm , of the arc ACB ,   
(c) the area, in cm2,of the shaded region.



Solution:
(a)
Rhombus has 4 equal sides, therefore OP = PQ = QR = OR = 5 cm
OR is radius to the circle, therefore OR = OQ = 5 cm

Triangles OQR and OQP are equilateral triangle,
Therefore,  ∠ QOR=  ∠QOP = 60o
 ∠ POR = 120o
x = 120o × π/180o
x = 2π/ 3 rad

(b) 
cos  ∠ AOQ= OQ / OA
cos 60o = 5 / OA
OA = 10 cm

Length of arc, ACB,
s = r θ
Arc ACB = (10) (2π / 3)
Arc ACB = 20.94 cm

(c)
Area of shaded region = 1 2 r 2 ( θsinθ ) ( change calculator to Rad mode ) = 1 2 ( 10 ) 2 ( 2π 3 sin 2π 3 ) =50( 2.0940.866 ) =61.40  cm 2

Question 6:

In the diagram above, AXB is an arc of a circle centre O and radius 10 cm with  ∠AOB = 0.82 radian. AYB is an arc of a circle centre P and radius 5 cm with  ∠APB = θ.
Calculate:
(a) the length of the chord AB,
(b) the value of θ in radians,
(c) the difference in length between the arcs AYB and AXB.



Solution:
(a)
1 2 AB=sin0.41×10( Change the calculator to Rad mode ) 1 2 AB=3.99 The length of chord AB=3.99×2=7.98 cm.

(b)
Let  1 2 θ=α, θ=2α sinα= 3.99 5 α=0.924 rad θ=0.924×2=1.848 radian.

(c)
Using s =
Arcs AXB = 10 × 0.82 = 8.2 cm
Arcs AYB = 5 × 1.848 = 9.24 cm

Difference in length between the arcs AYB and AXB
= 9.24 – 8.2
= 1.04 cm

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