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# Solved Examples on the Areas of Sectors

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For SAT Students
Any question labeled SAT-C is a question that allows a calculator.
Any question labeled SAT-NC is a question that does not allow a calculator.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
State the formulas and reasons for applicable steps
Negative angles are not allowed
Show all work
Except stated otherwise, use:

$d = diameter \\[3ex] r = radius \\[3ex] A = area\;\;of\;\;sector \\[3ex] L = length\;\;of\;\;arc \\[3ex] \theta = central\:\:angle \\[3ex] \pi = \dfrac{22}{7} \\[5ex] RAD = radians \\[3ex] ^\circ = DEG = degrees \\[3ex] A = \dfrac{\pi r^2\theta}{360} ...\theta \:\:in\:\:DEG \\[5ex] \theta = \dfrac{360A}{\pi r^2} ...\theta \:\:in\:\:DEG \\[5ex] r = \dfrac{360A}{\pi\theta} ...\theta \:\:in\:\:DEG \\[5ex] A = \dfrac{r^2\theta}{2} ...\theta \:\:in\:\:RAD \\[5ex] \theta = \dfrac{2A}{r^2} ...\theta \:\:in\:\:RAD \\[5ex] r = \sqrt{\dfrac{2A}{\theta}} ...\theta \:\:in\:\:RAD \\[5ex] Circumference\:\:of\:\:a\:\:circle = 2\pi r = \pi d \\[3ex] A = \dfrac{Lr}{2} \\[5ex] r = \dfrac{2A}{L} \\[5ex] L = \dfrac{2A}{r}$

(1.)

This implies that:

$r = 3\dfrac{1}{5} = \dfrac{16}{5} \\[5ex] L = 4 \\[3ex] \theta = \dfrac{L}{r} \\[5ex] \theta = 4 \div \dfrac{16}{5} \\[5ex] \theta = 4 * \dfrac{5}{16} \\[5ex] \theta = \dfrac{4 * 5}{16} \\[5ex] \theta = \dfrac{20}{16} = \dfrac{5}{4} RAD$
(2.) JAMB A sector of a circle has an area of $55\;cm^2$
If the radius of the circle is $10\;cm$, calculate the angle of the sector.

$\left[\pi = \dfrac{22}{7}\right] \\[5ex] A.\;\; 45^\circ \\[3ex] B.\;\; 63^\circ \\[3ex] C.\;\; 75^\circ \\[3ex] D.\;\; 90^\circ \\[3ex]$

$A = \dfrac{\pi r^2\theta}{360} \\[5ex] 360A = \pi r^2 \theta \\[3ex] \pi r^2 \theta = 360A \\[3ex] \theta = 360A \div \pi r^2 \\[3ex] \theta = (360 * 55) \div \left(\dfrac{22}{7} * 10 * 10\right) \\[5ex] \theta = 360 * 55 * \dfrac{7}{22} * \dfrac{1}{10} * \dfrac{1}{10} \\[5ex] \theta = \dfrac{360 * 55 * 7}{22 * 10 * 10} \\[5ex] \theta = \dfrac{36 * 5 * 7}{2 * 1 * 10} \\[5ex] \theta = \dfrac{18 * 7}{2} \\[5ex] \theta = 9 * 7 \\[3ex] \theta = 63^\circ$
(3.)

$L = \dfrac{2\pi r\theta}{360} ...\theta \:\:in\:\:DEG \\[5ex] \pi = \dfrac{22}{7} \\[5ex] r = 6400 km \\[3ex] \theta = 12^\circ \\[3ex] L = \left(2 * \dfrac{22}{7} * 6400 * 12\right) \div 360 \\[5ex] L = 2 * \dfrac{22}{7} * 6400 * 12 * \dfrac{1}{360} \\[5ex] L = \dfrac{2 * 22 * 6400 * 12}{7 * 360} \\[5ex] L = \dfrac{3379200}{2520} \\[5ex] L = 1340.952381 \\[3ex] L \approx 1341 \:km$
(4.) JAMB An arc subtends an angle $50^\circ$ at the centre of circle of radius $6\;cm$
Calculate the area of the sector formed.

$\left[\pi = \dfrac{22}{7}\right] \\[5ex] A.\;\; \dfrac{100}{7}\;\;cm^2 \\[5ex] B.\;\; \dfrac{110}{7}\;\; cm^2 \\[5ex] C.\;\; \dfrac{80}{7}\;\;cm^2 \\[5ex] D.\;\; \dfrac{90}{7}\;\;cm^2 \\[5ex]$

$A = \dfrac{\pi r^2\theta}{360} ...\theta \:\:in\:\:DEG \\[5ex] A = (\pi * r^2 * \theta) \div 360 \\[5ex] A = \dfrac{22}{7} * \dfrac{6}{1} * \dfrac{6}{1} * \dfrac{50}{1} * \dfrac{1}{360} \\[5ex] A = \dfrac{22 * 5}{7} \\[5ex] A = \dfrac{110}{7}\;\; cm^2$
(5.)

(6.) JAMB A circular arc subtends angle $150^\circ$ at the centre of a circle of radius $12\;cm$
Calculate the area of the sector of the arc.

$A.\;\; 30\pi\;cm^2 \\[3ex] B.\;\; 60\pi\;cm^2 \\[3ex] C.\;\; 120\pi\;cm^2 \\[3ex] D.\;\; 150\pi\;cm^2 \\[3ex]$

$A = \dfrac{\pi r^2\theta}{360} ...\theta \:\:in\:\:DEG \\[5ex] A = (\pi * r^2 * \theta) \div 360 \\[5ex] A = \dfrac{\pi}{1} * \dfrac{12}{1} * \dfrac{12}{1} * \dfrac{150}{1} * \dfrac{1}{360} \\[5ex] A = \dfrac{\pi * 12 * 15}{3} \\[5ex] A = \pi * 12 * 5 \\[3ex] A = 60\pi\;\; cm^2$
(7.)

(8.) JAMB A chord of a circle subtends an angle of $120^\circ$ at the centre of a circle of diameter $4\sqrt{3}\;cm$
Calculate the area of the major sector.

$A.\;\; 8\pi\;cm^2 \\[3ex] B.\;\; 32\pi\;cm^2 \\[3ex] C.\;\; 4\pi\;cm^2 \\[3ex] D.\;\; 16\pi\;cm^2 \\[3ex]$

$2r = d \\[3ex] r = \dfrac{d}{2} \\[5ex] d = 4\sqrt{3} \\[3ex] r = \dfrac{4\sqrt{3}}{2} = 2\sqrt{3} \\[5ex]$ Let us draw the diagram

$\underline{Major\;\;sector\;\;ADB} \\[3ex] Reflex\;\; \angle \phi + 120 = 360 ...\angle s \;\;at\;\;a\;\;point \\[3ex] \phi = 360 - 120 = 240 \\[3ex] \phi = 240^\circ = \dfrac{240}{180}\pi = \dfrac{4}{3}\pi = \dfrac{4\pi}{3} \\[5ex] A = \dfrac{r^2\phi}{2} ...\phi \:\:in\:\:RAD \\[5ex] A = \dfrac{1}{2} * r^2 * \phi \\[3ex] A = \dfrac{1}{2} * \left(2\sqrt{3}\right)^2 * \dfrac{4\pi}{3} \\[5ex] A = \dfrac{1}{2} * 4 * 3 * \dfrac{4\pi}{3} \\[5ex] A = 2 * 4\pi \\[3ex] A = 8\pi cm^2$