Question 1 (5 Marks)
Question 1(a) : 3 marks
Identify the three expressions that simplify to $6 x-5$. Drag and drop the three expressions to the allocated space below.
▶️Answer/Explanation
Ans:
Question 1(b) : 2 marks
Identify the two expressions that simplify to $9 x^2-4$. Drag and drop the two expressions to the allocated space below.
▶️Answer/Explanation
Ans:
Expression 1: $(3x+2)(3x-2)$
This is a difference of squares expression. When you multiply these two binomials using the formula for the difference of squares $(a^2 – b^2) = (a + b)(a – b)$, you get:
$(3x+2)(3x-2) = (3x)^2 – (2)^2 = 9x^2 – 4.$
Expression 2: $(x-1)(9x+4)+5x$
Let’s first expand the product $(x-1)(9x+4)$ using the distributive property:
$(x-1)(9x+4) = 9x^2 + 4x – 9x – 4 = 9x^2 – 4-5x.$
Now, add the term $5x$ to the expression:
$$(x-1)(9x+4)+5x = (9x^2 – 4) + 5x-5x = 9x^2 – 4.$$
So, the expression $(x-1)(9x+4)+5x$ simplifies to $9x^2 – 4$.
To summarize, both of the provided expressions indeed simplify to $9x^2 – 4$:
1. $(3x+2)(3x-2)$ simplifies to $9x^2 – 4$ due to the difference of squares.
2. $(x-1)(9x+4)+5x$ simplifies to $9x^2 – 4$ after expanding and combining terms.
Question 2 (6 Marks)
Question 2(a) : 3 marks
Given that $B A C$ and $A C B$ are angles in radians. Show that the length of side $B C$ is $16.33 \mathrm{~cm}$ to the nearest two decimal places.
▶️Answer/Explanation
Ans:
The given equation involves trigonometric ratios and a proportion. Let’s solve for the value of \(BC\).
The trigonometric ratios are:
\(\sin 45^\circ = \frac{1}{\sqrt{2}}\) (since \(\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}\))
\(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
The equation is:
\(\frac{\sin 45^\circ}{BC} = \frac{\sin 60^\circ}{20 \, \text{cm}}\)
Substitute the trigonometric values:
\(\frac{\frac{1}{\sqrt{2}}}{BC} = \frac{\frac{\sqrt{3}}{2}}{20 \, \text{cm}}\)
Now, cross-multiply to solve for \(BC\):
\(BC \cdot \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2} \cdot 20 \, \text{cm}\)
Simplify:
\(BC = \frac{\sqrt{3}}{\sqrt{2}} \cdot 20 \, \text{cm} = \sqrt{\frac{3}{2}} \cdot 20 \, \text{cm} = 10 \sqrt{6} \, \text{cm}\)
So, the value of \(BC\) is \(10 \sqrt{6}\) cm.
Question 2(b) : 1 marks
Triangle $A B C$ is reflected in the horizontal line $B C$ as shown in Diagram 1. The shape in Diagram 1 is reflected in the vertical line passing through point $C$.
Select the figure which shows the final shape after this reflection.
▶️Answer/Explanation
Ans:
$\text{option B}$
Question 2(c) : 2 marks
Given that $\mathrm{AC}=22.31 \mathrm{~cm}$, determine the perimeter of the final shape formed after the reflections.
▶️Answer/Explanation
Ans:
$B C$ is $16.33 \mathrm{~cm}$
$4 \times (\rm{AB+BC+AC})$
$4 \times (\rm{20+16.33+22.31})$
$234.56\mathrm{~cm})$
Question 3 (8 Marks)
Question 3(a) : 2 marks
In a group of 60 students:
34 study Extended mathematics or Physics
In the Venn diagram below
Set A represents the number of students who study Extended mathematics Set $B$ represents the number of students who study Physics.
Determine the missing values and complete the Venn diagram below.
▶️Answer/Explanation
Ans:
$\text{34=9+A}$
$\text{A=25}$
For study neither $60-(25+9+12)=14$
Question 3(b) : 1 marks
Describe the region $A \cap B^{\prime}$ in context.
▶️Answer/Explanation
Ans:
In the context of the given Venn diagram and the information provided:
- Set A represents the number of students who study Extended Mathematics.
- Set B represents the number of students who study Physics.
The notation \(A \cap B^{\prime}\) represents the intersection of Set A and the complement of Set B, which includes all the elements that are in Set A but not in Set B. In simpler terms, it refers to the students who study Extended Mathematics but do not study Physics.
So, the region \(A \cap B^{\prime}\) in the Venn diagram corresponds to the group of students who are studying Extended Mathematics but are not studying Physics, that is equal to $25$. These students have chosen to study Extended Mathematics exclusively without Physics.
Question 3(c) : 2 marks
One student is selected at random from the group. Given that the student studies Physics, determine the probability that the student can participate in the competition.
▶️Answer/Explanation
Ans:
To determine the probability that a randomly selected student can participate in the competition, we need to find the ratio of the number of students who can participate to the total number of students in the group.
From the given information:
Total number of students = 60
Number of students studying both Extended mathematics and Physics (participants) = $A \cap B = 9$
The probability (P) that a randomly selected student can participate in the competition is given by:
P = Number of participants / Total number of students
$\rm{P} =\frac{ 9 }{ 60}$
Simplifying the fraction, we get:
$P = \frac{3 }{ 20}$
Therefore, the probability that a randomly selected student can participate in the competition is 3/20.
Question 3(d) : 3 marks
Three students are selected at random from the group. Find the probability that they can participate in the competition.
▶️Answer/Explanation
Ans:
To find the probability that three randomly selected students can participate in the competition, we need to calculate the probability of each student being able to participate and then multiply those probabilities together.
Given:
Number of students studying both Extended mathematics and Physics (participants) = 9
Total number of students = 60
The probability that a single student can participate is given by:
\[ P(\text{single student}) = \frac{\text{Number of participants}}{\text{Total number of students}} \]
\[ P(\text{single student}) = \frac{9}{60} \]
To find the probability that all three randomly selected students can participate, we multiply their individual probabilities together:
\[ P(\text{three students}) = P(\text{single student}) \times P(\text{single student}) \times P(\text{single student}) \]
\[ P(\text{three students}) = \left(\frac{9}{60}\right) \times \left(\frac{9}{60}\right) \times \left(\frac{9}{60}\right) \]
Simplifying the expression, we get:
\[ P(\text{three students}) = \frac{9^3}{60^3} \]
Therefore, the probability that three randomly selected students can participate in the competition is $\frac{9^3}{60^3}$, which can be further simplified if desired.
Question 3(e) : 2 marks
▶️Answer/Explanation
Ans:
Selecting students for the competition randomly may not be the most practical approach, considering the specific criteria required for participation. In this case, the competition requires participants who are studying both Extended mathematics and Physics.
If students are chosen randomly, there is no guarantee that the selected students will meet the participation criteria. The probability of randomly selecting three students who can participate, as calculated earlier, is quite low. This means that the likelihood of selecting three students who fulfill the requirements by chance alone is relatively small.
To ensure that the competition participants meet the necessary criteria, it would be more practical to implement a selection process based on the students’ subject enrollment records or their expressed interest in participating. This way, the school can ensure that only eligible students are considered for the competition, increasing the chances of finding suitable participants.
Question 4 (6 Marks)
The students have conducted an experiment in their maths class to test the claims of the advertisements. In the experiment, students have tested nine batteries from GeneriCell and measured the lifetime of the batteries. The results are shown in Table 1 to the nearest minute. A higher number indicates that the battery has a longer lifetime.
Question 4(a) : 1 marks
Using the box-and-whisker plot, write down the percentage of batteries with a lifetime between 488 and 506 minutes from GeneriCell.
▶️Answer/Explanation
Ans:
$50 \%$
Question 4(b) : 3 marks
The experiment is repeated for the Maximizer brand. The times are recorded in Table 2 to the nearest minute.
On the canvas provided, draw a box-and-whisker plot to summarize the data given in Table 2 for Maximizer. The draggable lines can be resized as required.
▶️Answer/Explanation
Ans:
The five values required of the blue box and whisker plot: Min 478, LQ 494, Median 502, UQ 503, Max 509
Question 4(c) : 2 marks
Using your box-and-whisker plots:
Identify one reason that supports GeneriCell in their advertisement claim and one reason that supports Maximizer in their advertisement claim.
GeneriCell………………
Maximizer………………
▶️Answer/Explanation
Ans:
Higher upper quartile or $25 \%$ are above 506 for GeneriCell while $25 \%$ above their503 for Maximizer WTTEACCEPT better upper quartile or higher Q3
ACCEPT most Genericell batteries last 506 or less while most Maximizer last their 503 or less
ACCEPT more Generic last over their503
Question 5 (7 marks)
Question 5(a) : 3 marks
The equation, $x+\frac{1}{2} x=5^2-x$ can be described in words as
“The sum of a number and its half is the same as the difference between the square of five and the number.”
Calculate the value of the number.
▶️Answer/Explanation
Ans:
\(x + \frac{1}{2} x = 5^2 – x\).
Now, let’s solve for \(x\):
Combine the terms on the left side: \(x + \frac{1}{2} x = \frac{3}{2} x\).
Substitute \(5^2 = 25\) into the equation: \(\frac{3}{2} x = 25 – x\).
Add \(x\) to both sides: \(\frac{5}{2} x = 25\).
Now, solve for \(x\):
\[x = \frac{25}{\frac{5}{2}} = 10\]
So, the value of the number is \(x = 10\).
Question 5(b) : 4 marks
In another case,
“The sum of a number and its square is the same as 14”.
Find all possible values of the number.
▶️Answer/Explanation
Ans:
“The sum of a number and its square is the same as 14” can be written as:
\[x + x^2 = 14\]
This equation represents the relationship described in the sentence. Now, let’s solve for \(x\):
\[x^2 + x – 14 = 0\]
The quadratic formula is given by: \(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\), where the quadratic equation is in the form \(ax^2 + bx + c = 0\).
In this case, \(a = 1\), \(b = 1\), and \(c = -14\). Plugging these values into the formula:
\[x = \frac{-1 \pm \sqrt{1^2 – 4 \cdot 1 \cdot (-14)}}{2 \cdot 1}\]
Simplifying the expression inside the square root:
\[x = \frac{-1 \pm \sqrt{1 + 56}}{2}\]
\[x = \frac{-1 \pm \sqrt{57}}{2}\]
So, the two possible values of \(x\) are:
\[x = \frac{-1 + \sqrt{57}}{2}\]
\[x = \frac{-1 – \sqrt{57}}{2}\]
These are the correct values of \(x\) that satisfy the equation \(x^2 + x – 14 = 0\).
Question 6 (14 marks)
In this question, you will make calculations for changes that individuals can make to save water on a daily basis.
Question 6(a) : 1 marks
Write down the missing percentage for drinking and cooking on the pie chart.
▶️Answer/Explanation
Ans:
Sum of all percentage is equal to $100$
$(100-40-9-12)\%$=remaining
$=39 \%$
Question 6(b) : 4 marks
It is estimated that the daily water usage per person is 120 litres.
A regular shower has a flow rate of 8 Litres (L) per minute.
Using your answer from part (b), determine the duration of a regular shower.
▶️Answer/Explanation
Ans:
To determine the amount of water used for showering, we need to calculate $40\%$ of the daily water usage per person, which is $\rm{120~ litres}$.
To find $40\%$ of a quantity, we can multiply that quantity by $0.40$ (or $40\%$).
Water used for showering $= 120 \text{ liters} \times 0.40 = 48 \text{ liters}$.
Therefore, the amount of water used for showering is 48 liters.
Now, to determine the duration of a regular shower, we can divide the amount of water used by the flow rate:
\[\text{Duration of shower} = \frac{\text{Amount of water used}}{\text{Flow rate}}\]
Plugging in the values:
\[\text{Duration of shower} = \frac{48 \text{ liters}}{8 \text{ L/min}}\]
Simplifying the expression:
\[\text{Duration of shower} = 6 \text{ minutes}\]
Therefore, the duration of a regular shower is 6 minutes.
Question 6(c) : 2 marks
In a water-saving condition the flow rate of a shower can be reduced to be $5 \mathrm{~L}$ per minute. Given that the duration of the shower does not change:
Determine the amount of water used for showering in a water-saving condition.
▶️Answer/Explanation
Ans:
To determine the amount of water used for showering in a water-saving condition, we can use the new flow rate of 5 L per minute and the duration of the shower.
Let’s denote the amount of water used for showering in the water-saving condition as $x$ liters.
Using the equation:
\[\text{Amount of water used} = \text{Flow rate} \times \text{Duration}\]
In the original condition, we found that the duration of the shower is 6 minutes. So, in the water-saving condition, the duration of the shower remains the same.
Substituting the values into the equation:
\[x = 5 \text{ L/min} \times 6 \text{ min}\]
Simplifying the expression:
\[x = 30 \text{ liters}\]
Therefore, in a water-saving condition, the amount of water used for showering is 30 liters.
Question 6(d) : 3 marks
The amount of water used by the washing machine is $14.4 \mathrm{~L}$. The eco-setting for washing machines reduces water by $5 \%$.
Calculate the amount of water used by the washing machine in the eco-setting.
▶️Answer/Explanation
Ans:
To calculate the amount of water used by the washing machine in the eco-setting, we need to reduce the original amount by 5%.
Let’s denote the original amount of water used by the washing machine as $x$ liters.
According to the given information, the eco-setting reduces water usage by $5\%$, which means it uses $95\%$ of the original amount.
To calculate the amount of water used in the eco-setting, we can multiply $95\%$ (or $0.95$) by the original amount of water used:
Amount of water used in the eco-setting $= 0.95 \times 14.4 \text{ liters}$
Calculating the value:
Amount of water used in the eco-setting $= 13.68 \text{ liters}$
Therefore, the amount of water used by the washing machine in the eco-setting is 13.68 liters (or 13.7 liters rounded to one decimal place).
Question 6(f) : 1 marks
Suggest the order of activities in which it is most important to save water. Drag and drop the activities in the appropriate order.
▶️Answer/Explanation
Ans:
To determine the order of activities to prioritize water savings, we need to consider the percentage of water used by each activity and the potential for water savings.
Order of activities, from highest to lowest priority for water savings:
Showering (40%): The highest percentage of water is used for showering, so reducing the flow rate to 5 litres per minute can result in significant water savings.
Dishwasher (9%): The dishwasher uses a relatively lower percentage of water compared to other activities. However, using the eco setting and saving 5% of water can still contribute to water conservation.
Washing machine (12%): The washing machine also accounts for a significant portion of water usage. By using the eco setting and saving 5% of water, water conservation can be achieved.
Drinking and cooking (39%): Although drinking and cooking account for a significant percentage of water usage, there is no specified water-saving condition for this activity. However, adopting water-efficient practices such as using only the necessary amount of water during cooking and minimizing water wastage can still contribute to conservation efforts.
So, the suggested order of activities to prioritize water savings is:
- Showering
- Dishwasher
- Washing machine
- Drinking and cooking
Question 6(g) : 3 marks
Justify your chosen order in part (e). You should refer to your answers from previous parts.
▶️Answer/Explanation
Ans:
1. Showering (40%): Showering was prioritized first because it accounts for the highest percentage of water usage (40%). By reducing the flow rate to 5 liters per minute, significant water savings can be achieved without compromising personal hygiene.
2. Dishwasher (9%): The dishwasher was chosen as the second priority because it uses a relatively lower percentage of water compared to other activities. However, by using the eco setting and saving 5% of water, it still presents an opportunity for water conservation. The water-saving condition for the dishwasher allows for efficient water usage during this activity.
3. Washing machine (12%): The washing machine was placed as the third priority due to its higher water usage percentage (12%). By utilizing the eco setting and saving 5% of water, water conservation can be practiced during laundry cycles. This can result in substantial water savings considering the frequency at which washing machines are used.
4. Drinking and cooking (39%): Drinking and cooking were placed last in the order since there was no specified water-saving condition mentioned for this activity. However, it is still important to adopt water-efficient practices such as using only the necessary amount of water during cooking and minimizing water wastage in daily kitchen activities.
In summary, the order was chosen based on the percentage of water usage for each activity and the potential for water savings with the given water-saving conditions. Prioritizing showering, dishwasher, washing machine, and then drinking and cooking allows for effective water conservation while considering the impact on daily activities.
Question 7 (19 marks)
The following question explains the factors that are important to consider in car park design.
Question 7(a) : 4 marks
$\mathbf{Angled parking scenario}$
The car park is designed to fit cars with maximum dimensions as shown in the table. $\mathrm{DE}$ and $\mathrm{DF}$ are equal. $\mathrm{FG}$ is perpendicular to $\mathrm{DE}$.
The angled parking scenario is modelled in $\mathbf{Diagram 1}$ below.
Given that $\mathrm{D E}$ is $5.3$.
Calculate the value of the tuming area shaded in green.
▶️Answer/Explanation
Ans:
Given:
Central angle \(\theta = 75^\circ\)
Outer radius \(R = 5.3\) m
Inner radius \(r = 3.5\) m
The formula for the area of a sector is \(A = \frac{\theta}{360^\circ} \pi r^2\).
First, calculate the area of the sector with the outer radius:
\[A_{\text{outer}} = \frac{75^\circ}{360^\circ} \pi (5.3 \, \text{m})^2\]
Calculate the area of the sector with the inner radius:
\[A_{\text{inner}} = \frac{75^\circ}{360^\circ} \pi (3.5 \, \text{m})^2\]
Now, calculate the difference in the areas:
\[\text{Difference} = A_{\text{outer}} – A_{\text{inner}}\]
\[
\begin{align*}
A_{\text{outer}} &= \frac{75}{360} \pi \times (5.3)^2 \\
&= \frac{5}{24} \pi \times 28.09 \\
&\approx 18.35 \, \text{m}^2 \\
\\
A_{\text{inner}} &= \frac{75}{360} \pi \times (3.5)^2 \\
&= \frac{5}{24} \pi \times 12.25 \\
&\approx 6.39 \, \text{m}^2 \\
\\
\text{Difference} &= A_{\text{outer}} – A_{\text{inner}} \\
&\approx 18.35 – 6.39 \\
&\approx 10.36 \, \text{m}^2 \\
\end{align*}
\]
So, the difference in the area of the sector is approximately \(10.36 \, \text{m}^2\).
Question 7(b) : 5 marks
Find the value of the minimum lane width EG to the nearest one decimal place.
▶️Answer/Explanation
Ans:
To solve the equation using the sine formula, we have:
\[\frac{\text{DG}}{\sin 15} = \frac{5.3}{\sin 90}\]
First, let’s evaluate the value of \(\sin 15\) and \(\sin 90\):
\(\sin 15 \approx 0.2588\) (rounded to four decimal places)
\(\sin 90 = 1\)
Now we can substitute these values into the equation:
\[\frac{\text{DG}}{0.2588} = \frac{5.3}{1}\]
Simplifying the equation:
\[\text{DG} = 0.2588 \times 5.3\]
Calculating the value:
\[\text{DG} \approx 1.36964\]
Therefore, the value of DG is approximately 1.36964.
\( \begin{gathered}\rm{D E=D G+E G} \\ 5 \cdot 3=1.4+\rm{E G} \\ \rm{E G}=3.9 \mathrm{~m}\end{gathered} \)
Question 7(c) : 10 marks
You are the designer in a planning department, below is your brief for a car park design.
Design a layout for the car park with the dimensions provided in the diagram.
In your answer, you should:
$\bullet$ identify the relevant factors you considered in your design
$\bullet$ justify with calculations that your design is making the best use of the available width of the car park
$\bullet$ justify the degree of accuracy of your design
$\bullet$ illustrate the design on one of the diagrams below.
 |  |
▶️Answer/Explanation
Ans:
To design a car park that maximizes the number of parking spaces and makes the best use of the available width, several factors were considered.
Car size: The dimensions of cars were taken into account to determine the space required for each parking spot. This ensures that the parking spaces are large enough to accommodate vehicles comfortably.
Distance between cars: Sufficient space was allocated between parking spaces to allow for the opening of car doors and easy movement between parked vehicles.
Lane width: The minimum lane width requirement of 5.5 m for perpendicular parking and 4.0 m for angled parking was followed to ensure proper circulation and maneuvering of vehicles within the car park.
Angle of parking: The decision to use either perpendicular parking or angled parking was made based on the available width and the ability to maximize the number of parking spaces.
Calculations were performed to determine the maximum number of cars that can be accommodated in each type of parking layout while adhering to the minimum lane width requirement.
In terms of accuracy, the design aims to maximize the number of parking spaces while maintaining appropriate lane widths. However, it’s important to note that the actual layout and number of parking spaces may vary based on additional factors such as the shape of the available space and any specific requirements or constraints not mentioned in the given brief.
To illustrate the design, I have chosen to provide a layout for perpendicular parking with the maximum number of cars that can fit within the given width. Please refer to the diagram below:
 |  |
Question 8 (33 marks)
The line $\rm{y=2 x+1}$ is shown on the coordinate axes below. Vertical line segments can be added between the line and the $\rm{x}$ axis, one unit apart horizontally. In this question, you will investigate areas of different trapeziums formed.
Question 8(a) : 2 marks
For stage 3 , show that the area of the trapezium is 12 units squared.
▶️Answer/Explanation
Ans:
To show that the area of the trapezium is 12 square units, we can use the formula for the area of a trapezium:
\[ \text{Area} = \frac{1}{2} (a + b) \times h \]
where \( a \) and \( b \) are the lengths of the parallel sides, and \( h \) is the distance between the parallel sides.
In this case, we are given that the lengths of the parallel sides are $5$ and $7$, and the distance between the parallel sides is $1$.
Substituting the given values into the formula, we have:
\[ \text{Area} = \frac{1}{2} (1 + 7) \times 3 \]
Simplifying the expression:
\[ \text{Area} = \frac{1}{2} \times 8 \times 3 \]
\[ \text{Area} = 12 \]
Therefore, the area of the trapezium is indeed 6 square units.
Question 8(b) : 1 marks
Write down the missing values in the table up to stage 6 .
▶️Answer/Explanation
Ans:
It seems like the pattern is that the area of the trapezium is increasing by consecutive odd numbers. Each time, the next odd number is added to the previous area to get the area at the next stage.
Calculating the missing values:
For stage 5: \(20 + 2 \times 5 = 20 + 10 = 30\)
For stage 6: \(30 + 2 \times 6 = 30 + 12 = 42\)
So, the completed table is:
Question 8(c) : 2 marks
Describe in words two patterns you see in the table for $\mathrm{A}$.
▶️Answer/Explanation
Ans:
1.Arithmetic Progression with Common Difference 2:
The differences between consecutive terms in the “Area of trapezium” column form an arithmetic progression (AP) with a common difference of 2. This means that each term is obtained by adding 2 to the previous term. For example, the differences between consecutive terms are 4, 6, 8, and so on, each time increasing by 2. This arithmetic progression pattern leads to the sequence of area values in the table.
2. Triangular Number Sequence:
The values in the “Area of trapezium” column correspond to the triangular number sequence. Triangular numbers are the sums of consecutive natural numbers. In this case, the area values represent the cumulative sum of the first \(n\) odd numbers, where \(n\) is the stage number. This is why the areas follow the sequence \(2, 6, 12, 20, 30, 42, \ldots\), which are the first few triangular numbers: \(1, 3, 6, 10, 15, 21, \ldots\).
In summary, the patterns observed in the table are the increasing of areas by consecutive even numbers and the representation of triangular numbers in the sequence of area values.
Question 8(d) : 2 marks
Write down a general rule for $\rm{A}$ in terms of $n$.
▶️Answer/Explanation
Ans:
**Method 1: Using Arithmetic Progression**
The sum of the first \(n\) positive integers can be represented as an arithmetic progression:
\[1 + 2 + 3 + \ldots + n = \frac{n \cdot (n+1)}{2}\]
Now, let’s consider the area \(A\) in terms of \(n\):
\[A = 2 + 4 + 6 + \ldots + (2n)\]
Each term in this sequence is an even number, and we can factor out 2 from each term:
\[A = 2 \cdot (1 + 2 + 3 + \ldots + n)\]
Substitute the formula for the sum of the first \(n\) positive integers:
\[A = 2 \cdot \frac{n \cdot (n+1)}{2} = n \cdot (n+1)\]
**Method 2: Using Induction**
You can also prove the formula using mathematical induction. The base case is when \(n = 1\), and you’ve already established that \(A = 2\) for this case.
Assume that the formula \(A = n \cdot (n+1)\) holds for \(n = k\). Now, consider the case for \(n = k+1\):
\[A = 2 + 4 + 6 + \ldots + (2k) + (2k+2)\]
Using the assumption, replace the sum up to \(2k\) with \(k \cdot (k+1)\):
\[A = k \cdot (k+1) + (2k+2)\]
Simplify:
\[A = k^2 + k + 2k + 2 = (k+1) \cdot (k+2)\]
So, by induction, if the formula holds for \(n = k\), it also holds for \(n = k+1\), which completes the proof.
Both methods demonstrate that the area \(A\) of the trapezium in terms of \(n\) is indeed \(n \cdot (n+1)\).
Question 8(e) : 3 marks
Verify your general rule for $\rm{A}$.
▶️Answer/Explanation
Ans:
Let’s compare these values with the formula \(A = n \cdot (n+1)\):
For stage 1: \(n = 1\), \(n \cdot (n+1) = 1 \cdot (1+1) = 2\)
For stage 2: \(n = 2\), \(n \cdot (n+1) = 2 \cdot (2+1) = 6\)
For stage 3: \(n = 3\), \(n \cdot (n+1) = 3 \cdot (3+1) = 12\)
For stage 4: \(n = 4\), \(n \cdot (n+1) = 4 \cdot (4+1) = 20\)
For stage 5: \(n = 5\), \(n \cdot (n+1) = 5 \cdot (5+1) = 30\)
For stage 6: \(n = 6\), \(n \cdot (n+1) = 6 \cdot (6+1) = 42\)
As we can see, the values of \(A\) obtained from the formula \(n \cdot (n+1)\) perfectly match the values given in the table for each stage.
This verification confirms that the general rule \(A = n \cdot (n+1)\) is correct and accurately describes the area of the trapezium at each stage.