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2020 Ordinary Level - Paper One
Section A
Question 1
(a) John works as part of a sales team. He earns a basic rate of €\(12.60\) per hour.
In addition to his hourly pay, he earns a commission of \(22\%\) on any sales he makes above €\(200\) each week.
(i) During a particular week, John worked \(45\) hours at the basic rate and made sales amounting to €\(350\). Find John’s gross pay for this week.
(ii) During the following week John worked \(51\) hours. This included \(3\) hours on Sunday.
If John works on a Sunday he receives \(1.5\) times the basic rate for those hours.
His gross pay for that week was €\(713.20\).
Find the amount of sales John made in that week.
(i) \(600\mbox{ euro}\)
(ii) \(435\mbox{ euro}\)
(i)
\begin{align}45\times12.60+(350-200)\times0.22=600\mbox{ euro}\end{align}
(ii)
\begin{align}\frac{713.20-(51-3)\times12.60-3\times(12.60\times1.5)}{0.22}+200=435\mbox{ euro}\end{align}
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(b) John pays tax at the Standard Rate of \(20\%\) and at the Higher Rate of \(40\%\).
He has a weekly Tax Credit of €\(26\).
The weekly Standard Rate Cut-off Point is €\(678\).
Find John’s net income for the week where his salary was €\(713.20\).
\(589.52\mbox{ euro}\)
\begin{align}713.20-678\times0.2-(713.20-678)\times0.4+26=589.52\mbox{ euro}\end{align}
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Question 2
(a) Solve the equation:
\begin{align}\frac{9x-6}{2}=\frac{3x-14}{3}+\frac{9x}{4}\end{align}
\(x=-\dfrac{4}{3}\)
\begin{align}\frac{9x-6}{2}=\frac{3x-14}{3}+\frac{9x}{4}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{6(9x-6)}{12}=\frac{4(3x-14)+3(9x)}{12}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6(9x-6)=4(3x-14)+3(9x)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}54x-36=12x-56+27x\end{align}
\begin{align}\downarrow\end{align}
\begin{align}54x-12x-27x=36-56\end{align}
\begin{align}\downarrow\end{align}
\begin{align}15x=-20\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=-\frac{20}{15}\\&=-\frac{4}{3}\end{align}
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(b) Solve the simultaneous equations:
\begin{align}3x-y&=4\\4x^2-3xy&=4\end{align}
\(x=\dfrac{2}{5}\) and \(y=-\dfrac{14}{5}\) or \(x=2\) and \(y=2\)
\begin{align}3x-y&=4\\4x^2-3xy&=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=3x-4\end{align}
\begin{align}4x^2-3xy=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4x^2-3x(3x-4)=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4x^2-9x^2+12x=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5x^2-12x+4=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(5x-2)(x-2)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=\dfrac{2}{5}\) or \(x=2\)
\begin{align}\downarrow\end{align}
\begin{align}y&=3x-4\\&=3\left(\frac{2}{5}\right)-4\\&=-\frac{14}{5}\end{align}
or
\begin{align}y&=3x-4\\&=3(2)-4\\&=2\end{align}
\[\,\]
First Solution
\(x=\dfrac{2}{5}\) and \(y=-\dfrac{14}{5}\)
\[\,\]
Second Solution
\(x=2\) and \(y=2\)
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Question 3
\(z_1=3-4i\), \(z_2=-2+i\) and \(z_3=2iz_2\), where \(i^2=-1\).
(a)
(i) Write \(z_3\) in the form \(a+bi\), where \(a,b\in\mathbb{Z}\).
(ii) Plot \(z_1\), \(z_2\) and \(z_3\) on the given Argand Diagram.
Label each point clearly.
(iii) Find \(|z_1|\).
(i) \(-2-4i\)
(ii)
(iii) \(5\)
(i)
\begin{align}z_3&=2iz_2\\&=2i(-2+i)\\&=-4i+2i^2\\&=-2-4i\end{align}
(ii)
(iii)
\begin{align}|z_1|&=\sqrt{3^2+(-4)^2}\\&=\sqrt{9+16}\\&=\sqrt{25}\\&=5\end{align}
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(b) If \(z_1\times z_4=29+3i\), write \(z_4\) in the form \(a+bi\), where \(a,b\in\mathbb{R}\).
\(3+5i\)
\begin{align}z_4&=\frac{29+3i}{z_1}\\&=\frac{29+3i}{3-4i}\\&=\frac{29+3i}{3-4i}\times\frac{3+4i}{3+4i}\\&=\frac{87+116i+9i-12}{9-12i+12i+16}\\&=\frac{75+125i}{25}\\&=3+5i\end{align}
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Question 4
(a) The Golden Gate Bridge in San Francisco is constantly being repainted. It is estimated that the surface area of exposed steel
that needs to be painted is approximately \(10\) million square feet.
(i) Given that \(1\) metre is equal to \(3.28\) feet, convert the surface area of the bridge into square metres, giving your answer in the form \(a\times 10^n\), where \(1\leq a<10\) and \(n\in\mathbb{N}\).
Give the value of \(a\) correct to two significant figures.
(ii) A litre of the paint used on the Golden Gate Bridge will cover approximately \(5\) square
metres. This paint comes in \(25\) litre tins. Find the minimum number of tins of paint that will be needed to paint the entire bridge.
(i) \(9.3\times10^5\mbox{ square metres}\)
(ii) \(7{,}440\mbox{ tins}\)
(i)
\begin{align}10{,}000{,}000\mbox{ square feet}&=\frac{10{,}000{,}000}{3.28^2}\mbox{ square metres}\\&\approx929{,}506\mbox{ square metres}\\&\approx9.3\times10^5\mbox{ square metres}\end{align}
(ii)
\begin{align}\frac{\frac{930{,}000}{5}}{25}=7{,}440\mbox{ tins}\end{align}
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(b) Solve the equation \(2^{9x-1}=8^{2x}\).
\(x=\dfrac{1}{3}\)
\begin{align}2^{9x-1}=8^{2x}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2^{9x-1}=2^{3(2x)}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}9x-1=2(3x)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}9x-1=6x\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=\frac{1}{3}\end{align}
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Question 5
(a) Solve the equation \(x^2-3x-4=0\).
\(x=-1\) or \(x=4\)
\begin{align}x^2-3x-4=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x+1)(x-4)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=-1\) or \(x=4\)
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(b)
(i) Complete the table below to show the value of the function \(f(x)=-x^2-x+6\) for each of the given values of \(x\).
| \(x\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|---|---|---|
|
\(f(x)\) |
|
|
\(4\) |
|
|
|
|
(ii) Hence draw the graph of \(f\) in the domain \(-4\leq x\leq2\), where \(x\in\mathbb{R}\), on the diagram below.
(iii) On the same diagram as the function \(f\), draw the graph of the function \(g(x)=f(x)-2\), in the domain \(-2\leq x\leq 4\), where \(x\in\mathbb{R}\).
Label the graphs of \(f(x)\) and \(g(x)\) clearly.
(i)
| \(x\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|---|---|---|
|
\(f(x)\) |
\(-6\) |
\(0\) |
\(4\) |
\(6\) |
\(6\) |
\(4\) |
\(0\) |
(ii)
(iii)
(i)
| \(x\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|---|---|---|
|
\(f(x)\) |
\(-6\) |
\(0\) |
\(4\) |
\(6\) |
\(6\) |
\(4\) |
\(0\) |
(ii)
(iii)
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Question 6
(a)
(i) Differentiate the function \(f(x)=4x^3-3x^2+x-7\), where \(x\in\mathbb{R}\), with respect to \(x\).
(ii) Find the slope of the tangent to the graph of \(f(x)=4x^3-3x^2+x-7\)
at the point \((1,-5)\).
(iii) Hence find the equation of the tangent to the graph at this point.
(i) \(12x^2-6x+1\)
(ii) \(7\)
(iii) \(y=7x-12\)
(i)
\begin{align}f'(x)=12x^2-6x+1\end{align}
(ii)
\begin{align}f'(1)&=12(1^2)-6(1)+1\\&=7\end{align}
(iii)
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-(-5)=7(x-1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y+5=7x-7\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=7x-12\end{align}
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(b) The function \(g(x)=2x^2+px+q\), where \(p,q\in\mathbb{Z}\), and \(x\in\mathbb{R}\).
Given that \(g(2)=6\) and \(g'(3)=9\), find the value of \(p\) and the value of \(q\).
Note: \(g'(3)\) is the value of the derivative of \(g(x)\) at \(x=3\).
\(p=-3\) and \(q=4\)
\begin{align}g(2)=6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2(2^2)+p(2)+q=6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2p+q=-2\end{align}
and
\begin{align}g'(3)=9\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4(3)+p=9\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p&=9-12\\&=-3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}q&=-2-2p\\&=-2-2(-3)\\&=4\end{align}
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Section B
Question 7
(a) Pat buys a new car for €\(32{,}000\).
He trades in his old car and is given an allowance of €\(20{,}000\) by the garage.
He borrows the balance of the money from the credit union.
His fixed monthly repayment over three years is €\(44.366\) per month.
(i) How much money does Pat pay in total to the credit union for the loan?
(ii) Show the amount that Pat repays as a percentage of the amount that he borrows from the credit union is \(133.1\%\), correct to one decimal place.
(i) \(15{,}971.76\mbox{ euro}\)
(ii) The answer is already in the question!
(i)
\begin{align}443.66\times36=15{,}971.76\mbox{ euro}\end{align}
(ii)
\begin{align}\frac{15{,}971.6}{12{,}000}\times100\approx133.1\%\end{align}
as required.
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(b) A sum of money is invested at \(r\%\) per annum compound interest for \(3\) years.
At the end of the \(3\) years the value of the investment has increased by \(33.1\%\).
Find the value of \(r\).
\(10%\)
\begin{align}(1+r)^3=1.331\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1+r&=\sqrt[3]{1.331}\\&=1.1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r&=0.1\\&=10\%\end{align}
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(c)
(i) It is estimated that the value of cars depreciates at a compound rate of \(20\%\) per year.
Use this percentage to find the value of Pat’s car after three years (original price €\(32{,}000\)).
(ii) Pat’s friend Caitlín bought a new car three years ago.
Its value also depreciated by \(20\%\) per year.
It is now worth €\(17{,}920\).
Find the original value of the car.
(i) \(16{,}384\mbox{ euro}\)
(ii) \(35{,}000\mbox{ euro}\)
(i)
\begin{align}32{,}000\times0.8^3=16{,}384\mbox{ euro}\end{align}
(ii)
\begin{align}\frac{17{,}920}{0.8^3}=35{,}000\mbox{ euro}\end{align}
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Question 8
A swimmer is on a starting block at the beginning of a race. When she dives off the block until she resurfaces, the level of the swimmer relative to the level of the water is given by the function:
\begin{align}h(x)=\frac{1}{60}x^2-\frac{1}{4}x+\frac{3}{5}\end{align}
In the function, \(x\) is the horizontal distance in metres of the swimmer from the block, \(0\leq x\leq 12\), where \(x\in\mathbb{R}\) and \(h(x)\) is measured in metres.
(a) Find the height of the block above the water.
\(\dfrac{3}{5}\mbox{ m}\)
\begin{align}h(0)&=\frac{1}{60}(0^2)-\frac{1}{4}(0)+\frac{3}{5}\\&=\frac{3}{5}\mbox{ m}\end{align}
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(b)
(i) Show that the swimmer is on the surface of the water (i.e. \(h(x)=0\)) when she is \(12\) metres from the starting block.
(ii) Find the horizontal distance, in metres, from the starting block to the point where the swimmer enters the water.
(i) The answer is already in the question!
(ii) \(x=3\mbox{ m}\)
(i)
\begin{align}h(12)&=\frac{1}{60}(12^2)-\frac{1}{4}(12)+\frac{3}{5}\\&=\frac{3}{5}\mbox{ m}\\&=0\end{align}
as required.
(ii)
\begin{align}h(x)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{60}x^2-\frac{1}{4}x+\frac{3}{5}=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2-15x+36=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-3)(x-12)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=3\) or \(x=12\)
\begin{align}\downarrow\end{align}
\begin{align}x=3\mbox{ m}\end{align}
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(c)
(i) Find \(h'(x)\), the derivative of \(h(x)=\dfrac{1}{60}x^2-\dfrac{!}{4}x+\dfrac{3}{5}\).
(ii) Use your answer to Part (c)(i) to find the horizontal distance (\(x\)), in metres, from the starting block to the point at which the swimmer reaches her greatest depth.
(iii) Hence find this greatest depth.
(i) \(h'(x)=\frac{1}{30}x-\frac{1}{4}\)
(ii) \(7.5\mbox{ m}\)
(iii) \(\dfrac{27}{80}\mbox{ m}\)
(i)
\begin{align}h'(x)=\frac{1}{30}x-\frac{1}{4}\end{align}
(ii)
\begin{align}\frac{1}{30}x-\frac{1}{4}=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{30}{4}\\&=7.5\mbox{ m}\end{align}
(iii)
\begin{align}h(7.5)&=\frac{1}{60}(7.5)^2-\frac{1}{4}(7.5)+\frac{3}{5}\\&=\frac{27}{80}\mbox{ m}\end{align}
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(d) In the \(2016\) Summer Olympics, Michael Phelps, won the \(200\mbox{ m}\) Butterfly final in a time of \(1\) minute and \(53.36\) seconds ( \(1:53.36\)). The time it took to swim each of the four \(50\mbox{ m}\) sections of the race (split time) is given in the table below. The percentage increase in the time taken to swim one section is also given. Thus, it took the swimmer \(14.7\%\) longer to swim the second \(50\) metres compared to the first \(50\) metres.
| Race Section | Split time (secs) | Percentage Increase |
|---|---|---|
|
\(0\) - \(50\mbox{ m}\) |
\(24.85\) |
\ |
|
\(50\) - \(100\mbox{ m}\) |
\(28.5\) |
\(14.7\%\) |
|
\(100\) - \(150\mbox{ m}\) |
\(29.33\) |
|
|
\(150\) - \(200\mbox{ m}\) |
\(30.68\) |
|
|
Total |
\(1\colon53.36\) |
\ |
(i) Complete the table by finding the percentage increase in split time by comparing each
split time with the previous one.
Give your answers correct to one decimal place.
(ii) Another swimmer in the race completed the first \(50\mbox{ m}\) in a time of \(25.01\) seconds.
His subsequent \(50\mbox{ m}\) split times increased at the same rates as each of Michael Phelps’ times. Using the table below, or otherwise, find the difference between his finishing time and that of Michael Phelps.
Give your answers in seconds correct to \(2\) decimal places.
| Race Section | Split time (secs) | Percentage Increase |
|---|---|---|
|
\(0\) - \(50\mbox{ m}\) |
\(25.01\) |
\ |
|
\(50\) - \(100\mbox{ m}\) |
|
\(14.7\%\) |
|
\(100\) - \(150\mbox{ m}\) |
|
|
|
\(150\) - \(200\mbox{ m}\) |
|
|
|
Total |
|
\ |
(i)
| Race Section | Split time (secs) | Percentage Increase |
|---|---|---|
|
\(0\) - \(50\mbox{ m}\) |
\(24.85\) |
\ |
|
\(50\) - \(100\mbox{ m}\) |
\(28.5\) |
\(14.7\%\) |
|
\(100\) - \(150\mbox{ m}\) |
\(29.33\) |
\(2.9\%\) |
|
\(150\) - \(200\mbox{ m}\) |
\(30.68\) |
\(4.6\%\) |
|
Total |
\(1\colon53.36\) |
\ |
(ii)
| Race Section | Split time (secs) | Percentage Increase |
|---|---|---|
|
\(0\) - \(50\mbox{ m}\) |
\(25.01\) |
\ |
|
\(50\) - \(100\mbox{ m}\) |
\(28.686\) |
\(14.7\%\) |
|
\(100\) - \(150\mbox{ m}\) |
\(29.518\) |
\(14.7\%\) |
|
\(150\) - \(200\mbox{ m}\) |
\(30.876\) |
\(4.6\%\) |
|
Total |
\(1\colon54\colon09\) |
\ |
\begin{align}0.73\mbox{ s}\end{align}
(i)
| Race Section | Split time (secs) | Percentage Increase |
|---|---|---|
|
\(0\) - \(50\mbox{ m}\) |
\(24.85\) |
\ |
|
\(50\) - \(100\mbox{ m}\) |
\(28.5\) |
\(14.7\%\) |
|
\(100\) - \(150\mbox{ m}\) |
\(29.33\) |
\(2.9\%\) |
|
\(150\) - \(200\mbox{ m}\) |
\(30.68\) |
\(4.6\%\) |
|
Total |
\(1\colon53.36\) |
\ |
(ii)
| Race Section | Split time (secs) | Percentage Increase |
|---|---|---|
|
\(0\) - \(50\mbox{ m}\) |
\(25.01\) |
\ |
|
\(50\) - \(100\mbox{ m}\) |
\(28.686\) |
\(14.7\%\) |
|
\(100\) - \(150\mbox{ m}\) |
\(29.518\) |
\(14.7\%\) |
|
\(150\) - \(200\mbox{ m}\) |
\(30.876\) |
\(4.6\%\) |
|
Total |
\(1\colon54\colon09\) |
\ |
\begin{align}1\colon54\colon09-1\colon53\colon36&\approx0.73\mbox{ s}\end{align}
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Question 9
The following sequence of patterns is created using matchsticks to form equilateral triangles.
(a) Complete the table below to show the number of matchsticks required to make each of the first six patterns of the above sequence.
| Pattern Number | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
|
Number of matchsticks |
\(3\) |
\(7\) |
|
|
|
|
| Pattern Number | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
|
Number of matchsticks |
\(3\) |
\(7\) |
\(11\) |
\(15\) |
\(19\) |
\(23\) |
| Pattern Number | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
|
Number of matchsticks |
\(3\) |
\(7\) |
\(11\) |
\(15\) |
\(19\) |
\(23\) |
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(b)
(i) How many matchsticks are required to make pattern \(10\) of the sequence?
(ii) Find a formula for \(T_n\), the number of matchsticks required to make pattern \(n\) of the sequence.
(iii) Pattern \(k\) has \(147\) matchsticks, where \(k\in\mathbb{N}\). Find the value of \(k\).
(i) \(39\)
(ii) \(T_n=4n-1\)
(iii) \(37\)
(i)
\begin{align}3+(10-1)4=39\end{align}
(ii)
\begin{align}T_n&=3+(n-1)4\\&=3+4n-4\\&=4n-1\end{align}
(iii)
\begin{align}T_k=147\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4k-1=147\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k&=\frac{147+1}{4}\\&=37\end{align}
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(c)
(i) Find a formula for \(S_n\), the total number of matchsticks required to make the first \(n\) patterns.
(ii) Find the total number of complete patterns in the sequence that can be made using \(820\) matchsticks.
(i) \(S_n=n+2n^2\)
(ii) \(20\)
(i)
\begin{align}S_n&=\frac{n}{2}[a+(n-1)d]\\&=\frac{n}{2}[6+(n-1)(4)]\\&=\frac{n}{2}(6+4n-4)\\&=\frac{n}{2}(2+4n)\\&=n+2n^2\end{align}
(ii)
\begin{align}S_n=820\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n+2n^2=820\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2n^2+n-820=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(2n+41)(n-20)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n=20\end{align}
(\(n\) is positive)
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(d)
(i) The table below shows the number of triangles formed by each pattern for the first
two patterns. Complete the table to show the number of triangles formed for patterns
three to six.
| Pattern Number | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
|
Number of Triangles |
\(1\) |
\(3\) |
|
|
|
|
(ii) The area of each triangle is \(4\sqrt{3}\mbox{ cm}^2\).
Find, correct to the nearest \(\mbox{cm}^2\), the combined total area covered by the first \(15\) patterns in the sequence.
(i)
| Pattern Number | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
|
Number of Triangles |
\(1\) |
\(3\) |
\(5\) |
\(7\) |
\(9\) |
\(11\) |
(ii) \(1{,}559\mbox{ cm}^2\)
(i)
| Pattern Number | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
|
Number of Triangles |
\(1\) |
\(3\) |
\(5\) |
\(7\) |
\(9\) |
\(11\) |
(ii)
\begin{align}(4\sqrt{3})(S_{15})&=(4\sqrt{3})\left[\frac{15}{2}(2(1)+(15-1)(2))\right]\\&\approx1{,}559\mbox{ cm}^2\end{align}
