QUESTION 12
iii
Easy
Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage. Find a rule to determine the number of matchsticks required for the \(n^{th}\) stage.
SOLUTION
1
Assume the stage number and number of matchsticks
Let \(x\) be the stage number and \(y\) be the number of matchsticks.
2
Write down the linear equation
From the pattern, we notice:
When \(x\) increases by 1, \(y\) increases by 5.
So the polynomial is linear and has the form \(y = mx + c\),
where \(m\) is the slope(rate of increase) and \(c\) is the constant term.
The increase in each stage is \(11 − 6 = 5\), so \(m = 5\)
Taking \(x = 1\) and \(y = 6\) and substituting in the linear equation, we get
\(6 = 5(1) + c\)
⇒ \(c = 1\)
So the equation becomes
\(y = 5x + 1\)
When \(x\) increases by 1, \(y\) increases by 5.
So the polynomial is linear and has the form \(y = mx + c\),
where \(m\) is the slope(rate of increase) and \(c\) is the constant term.
The increase in each stage is \(11 − 6 = 5\), so \(m = 5\)
Taking \(x = 1\) and \(y = 6\) and substituting in the linear equation, we get
\(6 = 5(1) + c\)
⇒ \(c = 1\)
So the equation becomes
\(y = 5x + 1\)
3
Write this in stage notation
To find the number of matchsticks required for the \(n^{th}\) stage, we can use the rule
\(T_{n} = 5n + 1\)
\(T_{n} = 5n + 1\)
🏆
Final Answer : \(T_{n} = 5n + 1\)
Concept Note
The number of matchsticks at any stage \(n\) can be written as a linear equation:
\(y = mx + c\), where
\(m = \)rate of increase(slope)
\(x = \)stage number
\(c = \)constant term