1.5.3 Circular Measure Long Questions (Question 5 & 6)

Question 5:

Diagram below shows a circle PQRT, centre O 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 = 60^o

∠ POR = 120^o

x = 120^o × π/180^o

x = 2π/ 3 rad

(b)

cos ∠ AOQ= OQ / OA

cos 60^o = 5 / OA

OA = 10 cm

Length of arc, ACB,

s = r θ

Arc ACB = (10) (2π / 3)

Arc ACB = 20.94 cm

(c)

\begin{array}{l} Area of shaded region \\ = \frac{1}{2} r^{2} (θ - \sin θ) (change calculator to Rad mode) \\ = \frac{1}{2} {(10)}^{2} (\frac{2 π}{3} - \sin \frac{2 π}{3}) \\ = 50 (2.094 - 0.866) \\ = 61.40 {cm}^{2} \end{array}

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)

\begin{array}{l} \frac{1}{2} A B = \sin 0.41 \times 10 \to (Change the calculator to Rad mode) \\ \frac{1}{2} A B = 3.99 \\ ∴ The length of chord A B = 3.99 \times 2 = 7.98 c m . \end{array}

(b)

\begin{array}{l} Let \frac{1}{2} θ = α, θ = 2 α \\ \sin α = \frac{3.99}{5} \\ α = 0.924 rad \\ ∴ θ = 0.924 \times 2 = 1.848 radian . \end{array}

(c)

Using s = rθ

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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