**Question 5**:

Diagram below shows a circle

*PQRT*, centre*O*and radius 5 cm.*AQB*is a tangent to thecircle 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 cm

^{2},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)**

**Question 6**:

In the diagram above,

Calculate:

*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)**

$\begin{array}{l}\frac{1}{2}AB=\mathrm{sin}0.41\times 10\to \left(\text{ChangethecalculatortoRadmode}\right)\\ \frac{1}{2}AB=3.99\\ \therefore \text{Thelengthofchord}AB=3.99\times 2=7.98\text{}cm.\end{array}$

**(b)**

$\begin{array}{l}\text{Let}\frac{1}{2}\theta =\alpha ,\text{}\theta =2\alpha \\ \mathrm{sin}\alpha =\frac{3.99}{5}\\ \alpha =0.924\text{rad}\\ \therefore \theta =0.924\times 2=1.848\text{radian}\text{.}\end{array}$

**(c)**

Using

*s*=*rθ*Arcs

*AXB*= 10 × 0.82 = 8.2 cmArcs

*AYB*= 5 × 1.848 = 9.24 cmDifference in length between the arcs

*AYB*and*AXB*= 9.24 – 8.2

=

**1.04 cm**