**10.2.1 Circles I, PT3 Focus Practice**

Question 1:

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*.
$\left(\text{Use}\pi =\frac{22}{7}\right)$

*Solution***:**

Angle of the major arc

$\begin{array}{l}\text{Length of major arc}AB\\ =\frac{{216}^{o}}{{360}^{o}}\times 2\pi r\\ =\frac{{216}^{o}}{{360}^{o}}\times 2\times \frac{22}{7}\times 35\\ =132\text{cm}\end{array}$
*AB*= 360^{o}– 144^{o}= 216^{o}**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*.

SolutionSolution

**:**

Radius = 18 – 8 = 10 cm

*PT*

^{2 }= 10

^{2}– 8

^{2}

= 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.

$\left(\text{Use}\pi =\frac{22}{7}\right)$

*Solution***:**

$\begin{array}{l}\text{Areaofbiggercircle}=\pi {R}^{2}=\frac{22}{7}\times {14}^{2}\\ \text{Radius,}r\text{ofsmallercircle}=\frac{1}{2}\times 14=7\text{cm}\\ \text{Areaofsmallercircle}=\pi {r}^{2}=\frac{22}{7}\times {7}^{2}\\ \therefore \text{Areaofshadedregion}\\ \text{=}\left(\frac{22}{7}\times {14}^{2}\right)-\left(\frac{22}{7}\times {7}^{2}\right)\\ =616-154\\ =462{\text{cm}}^{2}\end{array}$

**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 cm

^{2}.
$\left(\text{Use}\pi =\frac{22}{7}\right)$

SolutionSolution

**:**

$\begin{array}{l}\text{Theareaofquadrant}ABCD\\ =\frac{1}{4}\times \pi {r}^{2}\\ =\frac{1}{4}\times \frac{22}{7}\times {14}^{2}\\ =154{\text{cm}}^{\text{2}}\end{array}$

$\begin{array}{l}\text{Areaoftheshadedregion}\\ =154-102\frac{2}{3}\\ =51\frac{1}{3}{\text{cm}}^{2}\end{array}$

**Question 5:**

Diagram below shows a square

*KLMN*.*KPN*is a semicircle with centre*O*.
Calculate the perimeter, in cm, of the shaded region.

$\left(\text{Use}\pi =\frac{22}{7}\right)$

SolutionSolution

**:**

$\begin{array}{l}KO=ON=OP=7\text{cm}\\ PN=\sqrt{{7}^{2}+{7}^{2}}\\ \text{}=\sqrt{98}\\ \text{}=9.90\text{cm}\\ \\ \text{Arclength}KP\\ =\frac{1}{4}\times 2\times \frac{22}{7}\times 7\\ =11\text{cm}\end{array}$

Perimeter of the shaded region

=

*KL + LM + MN + NP +*Arc length*PK*= 14 + 14 +14 + 9.90 + 11

= **62.90 cm**