Make the most of NESA Maths Reference Sheet and test your maths skills with these sample questions and explanations for measurement and financial maths!
The NESA Maths Reference Sheet is a great resource… if you know how to use it! Navigate measurement and financial mathematics with our Ultimate NESA Maths Reference Sheet Guide.
While memorisation has its place in learning, Matrix recommends that students learn to derive their responses and learn how to apply these formulae correctly. As a bonus, we’ve included a nifty HSC Maths Cheatsheet for you to download and print out!
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Click on the following formulas to see what they mean and apply them to a practice question!
Measurement  
Length
\(l=\frac{\theta}{360}\times2\pi r\)

Surface area

Financial Mathematics 
\(A= P(1+r)^n\)
\(S_n=\frac{n}{2} [2a+(n1)d] = \frac{n}{2}(a+l)\) 
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Next section: Area
Example 1:
Find the length of the minor arc AB, such that the length of radius OA is 30 cm. Leave your answer as an exact value.
Solution 1:
Substitute \(\theta = \text{115}\) and \(r = \text{30}\) into \(l=\frac{\theta}{360}\times2\pi r\):
\begin{align*} 
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Next section: Surface Area
Example 2:
Given that the following circle has a radius of 5 m, find the area of the major sector AOB. Leave your answer as an exact value.
Solution 2:
Substitute \(\theta=330\) and \(r=5\) into \(A= \frac{\theta}{360} \times \pi r^2\):
\begin{align*} 
Example 3:
What is the area of the shape below? Note that all the measurements were made in millimetres.
Solution 3:
Substitute \(h=12\), \(a=16\) and \(b=20\) into \(A=\frac{h}{2} (a+b) \):
\begin{align*} 
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Next section: Volume
Example 4:
Gemma is designing a can for her new spaghetti recipe. In order to figure out how much pink paint she needs to completely cover the can, she needs to know what the total surface area of the cylindrical can is.
Knowing that the dimensions below are given in centimetres, what is the surface area of Gemma’s can? Leave your answer as an exact value.
Solution 4:
Substitute \(h=9\) and \(r=4\) into \(A=2 \pi r^2 + 2 \pi r h \):
\begin{align*} 
Example 5:
Biological cells with larger surface area to volume ratios are able to more efficiently diffuse oxygen into the cell and waste material out.
If we approximate a cell to be the shape of a sphere, what is the surface area of a cell with a diameter of 0.07 mm? Leave your answer as an exact value.
Solution 5:
Substitute \(r= \frac{0.07}{2}\) (since the radius is half of the diameter) into \(A = 4 \pi r^2\): \begin{align*} 
Use  Formula  Variables 
Volume of a pyramid/cone  \(V= \frac{1}{3}Ah\)  \begin{align*} V &= \text{volume of pyramid/cone}\\ A &= \text{base area}\\ h &= \text{height}\\ \end{align*} 
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Next section: Financial Mathematics
Example 6:
Ricky is planning her birthday party. She wants to prank her friends by putting water balloons inside all of the party hats.
Assuming that the party hats are a perfect cone with a radius of 5 cm and a height of 12 cm, how much water does she need to fill one party hat? Leave your answer as an exact value.
Solution 6:
To find the area of the circular base, substitute \(r=5\) into \(A= \pi r^2\): \begin{align*}
Substitute \(A= 25 \pi\) and \(h=12\) into \(V= \frac{1}{3}Ah\): \begin{align*} 
Example 7:
Gordon is making meatballs. Assuming that 1 cm^{3} = 1 gram, how much mince does he need to make a perfectly spherical meatball with a diameter of 3 cm? Leave your answer as an exact value.
Solution 7:
Substitute \(r= \frac{3}{2}\) (since the radius is half of the diameter) into \(V= \frac{4}{3} \pi r^3\):
\begin{align*} 
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Next section: Sequences and series
Example 8:
Jacky’s bank is offering her a fantastic interest rate of 3% per annum, compounded every three months. If she deposits $12 360 into her savings account today and resolves to not take away from or add to these funds, what will be her savings account balance in 12 months time? Provide your answer to the nearest two decimal places.
Solution 8:
\begin{align*} A & = \text{Jacky’s savings account balance in 12 months time: this is what we are trying to find!} \\ P & = 12 \ 360 \\ \end{align*}The interest rate of 3% per year is compounded every three months. So, we need to divide the 3% interest rate by the number of three month periods in 12 months to find the compound interest rate per three month period:\begin{align*} r & = \frac{3 \text{%}}{ \frac{12}{3} } \\ & = \frac{0.03}{4} \\ ∴r & = 0.0075 \\ \end{align*}Since the compounding period is three months and Jacky plans to leave her savings account balance as it is for 12 months:\begin{align*} n & = \frac{12}{3} \\ ∴n & = 4 \\ \end{align*}Hence, substitute \(P=12 \ 360\), \(r=0.0075\) and \(n=4\) into \(A= P(1+r)^n\):\begin{align*} A &= 12 \ 360 \times (1+0.0075)^4 \\ &= 12 \ 360 \times 1.0075^4 \\ &= 12 \ 734.99 \text{ (to the nearest two decimal places)} \\ ∴ \text{Jacky’s savings account balance in 12 months time} &= $12 \ 734.99\\ \end{align*} 
Need to brush up on Financial Mathematics? Learn more and download your free worksheet here.
Note: A series is the sum of the elements in a sequence.
In an arithmetic progression, each term is determined by adding a constant to the preceding term.
Example 9:
Consider the following arithmetic sequence:
\( 3, 15, 27, 39… \)
What is the 23^{rd} term of this sequence?
Solution 9:
Substitute \(n=23\), \(a=3\) and \(d=12\) into \(T_n = a+(n1)d\)
\begin{align*} 
Example 10:
The sum of the first 16 terms of an arithmetic series is 632. Given the first term of the series is 2, write down the first 4 terms of the series.
Solution 10:
Substitute \(S_n=632\), \(n=16\) and \(a=2\) into \(S_n=\frac{n}{2} [2a+(n1)d] = \frac{n}{2}(a+l)\)
\begin{align*} \begin{align*} 
In a geometric progression, each term is found by multiplying the previous term by a constant.
Example 11:
A basketball player is bouncing a ball. On each bounce, the ball reaches a height that is 85% of its previous height. The total distance that the ball has travelled just before they make contact with the ball for the 8^{th} time is 6 metres.
a) At what height did the basketball player begin bouncing the ball? Give your answer in metres to the nearest three decimal places.
b) Hence or otherwise, find the total distance that the ball travels if the basketball player continues bouncing the ball until it eventually does not lift off the ground. Give your answer in metres to the nearest two decimal places.
Solution 11 a):
Solution 11 b):
In part a), we found that the total distance d travelled by the ball is \(d = 2 \times S_n – T_1 – T_n\). Need a refresher? Click here to scroll up to the explanation in part a).
In part a), we were interested in the distance that the ball travels before the 8th bounce. So, we used \(n = 8\). In part b) though, we are trying to find how far the ball can travel if it was allowed to bounce forever. So here, we let \(n = ∞\). Substitute \(n = ∞\) into \(d = 2 \times S_n – T_1 – T_n\): \begin{align*} Since: \begin{align*} \begin{align*} 
Review ‘Series and Sequences’ and download your free worksheet here.
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