The following sequence of patterns is created using matchsticks to form equilateral triangles - Leaving Cert Mathematics - Question 9 - 2020

To find the number of matchsticks for pattern 10, we can observe the pattern: Each subsequent pattern increases by 4 matchsticks. Therefore:

egin{align*} T_n &= 3 + (n - 1) imes 4 \ T_{10} &= 3 + (10 - 1) imes 4 \ T_{10} &= 3 + 36 \ T_{10} &= 39. ext{Thus, 39 matchsticks are required.} \end{align*}

Based on the observed pattern, the formula for $T_n$ is:

$T_n = 3 + (n - 1) imes 4$

This captures the relationship between the pattern number and the number of matchsticks required.

Using the formula found in the previous step, we set:

$147 = 3 + (k - 1) imes 4$

Solving gives:

egin{align*} 147 - 3 &= (k - 1) imes 4 \ 144 &= (k - 1) imes 4 \ 36 &= k - 1 \ k &= 37. ext{Thus, } k = 37.

To find the total number of matchsticks required for the first $n$ patterns:

$S_n = \sum_{i=1}^n T_i = \sum_{i=1}^n [3 + (i - 1)4]$

This simplifies to:

$S_n = n[6 + (n - 1) \cdot 4] / 2 = 2n^2 + 2n.$

Setting up the equation:

$2n^2 + 2n = 820$

This can be rearranged to:

$2n^2 + n - 820 = 0.$

Using the quadratic formula to solve for $n$:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 4 \cdot 2 \cdot 820}}{4} = 20.$

Thus, a total of 20 complete patterns can be made.

The number of triangles for the first two patterns is given as:

• Pattern 1: 1
• Pattern 2: 3

The pattern shows that each additional pattern adds another triangle:

• Pattern 3: 5
• Pattern 4: 7
• Pattern 5: 9
• Pattern 6: 11

Thus, the complete table is:

Pattern Number	1	2	3	4	5	6
Number of Triangles	1	3	5	7	9	11

The area of each triangle is given as $4\sqrt{3} \text{ cm}^2$.

The total number of triangles formed in the first 15 patterns can be calculated by noticing the sequence formed:

Total triangles from patterns 1 to 15 is: $1 + 3 + 5 + ... + 29 = \frac{15}{2}[1 + 29] = 15 \cdot 15 = 225$.

Thus, the combined area is: $A = 225\cdot 4\sqrt{3} = 900\sqrt{3} \text{ cm}^2$.

Calculating numerically: $\approx 1558.84572 \text{ cm}^2 ext{ rounded to } 1560 ext{ cm}^2.$