Expressing Patterns Algebraically Worksheet | Fun and Engaging ... - Free Printable

Let's solve the problem step by step for Question 2 and Question 3, based on the pattern of matchsticks.

We are asked to:
1. Count the number of matchsticks used in each pattern.
2. Fill in the table with the number of matchsticks (M) for Pattern Numbers 1 to 4.
3. Find a formula that expresses the number of matchsticks algebraically as a function of the pattern number P.
4. Determine how many matchsticks are added each time (M = ?) and how the pattern number increases (P = ?).
5. Draw the next pattern (Pattern 4).

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🔷 Problem 2: House-Shaped Patterns

Each "house" is made of:
- A square base (4 sticks)
- A triangle roof (2 sticks, sharing one side with the square)

But notice: the triangle shares one side with the square, so it only adds 2 new sticks.

So, one house uses:
- Square: 4 sticks
- Triangle: 2 additional sticks (top two sides)
- Total: 6 sticks

Now let’s count for each pattern:

#### ✔ Pattern 1: One house
- 6 matchsticks

#### ✔ Pattern 2: Two houses joined
- First house: 6 sticks
- Second house: shares one vertical side with first house
- So second house needs only 5 new sticks (because one side is shared)
- Total: 6 + 5 = 11 sticks

Wait — let’s double-check carefully.

Actually, when two houses are joined side-by-side:
- The vertical side between them is shared.
- Each house has:
- 4 sides for the square
- 2 for the roof (but the top of the roof is separate)

Let’s break down one house:
- Bottom horizontal: 1
- Left vertical: 1
- Right vertical: 1
- Top horizontal: 1 → total 4 for square
- Roof: two diagonal sticks → 2 more
- But the roof sits on the top of the square, so no shared sides with the roof.

But when you join two houses side by side, they share the right wall of the first and left wall of the second.

So:
- First house: 6 sticks (4 square + 2 roof)
- Second house: shares one vertical stick (the left wall), so it needs:
- Bottom: 1
- Right wall: 1
- Top: 1
- Roof: 2
- Left wall is shared → not needed
→ That’s 4 new sticks? Wait, let’s draw it mentally.

Better: Count all visible sticks.

Let’s do this systematically.

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#### 🔹 Counting Matchsticks for Problem 2

Pattern 1 (One house):
- Square: 4 sticks
- Roof (triangle): 2 sticks
- Total: 6

Pattern 2 (Two houses side-by-side):
- They share one vertical side.
- So:
- Left house: full 6
- Right house: missing one vertical side (shared), so adds 5?
- But wait: the shared side is internal, so we don’t count it twice.

Total sticks:
- Bottom: 2 horizontal sticks
- Verticals: 3 (left, middle, right)
- Top: 2 horizontal (for squares)
- Roofs: 2 triangles × 2 sticks = 4

Wait — better to list:

| Stick type | Count |
|-----------|-------|
| Bottom horizontal | 2 (one per house) |
| Top horizontal (square) | 2 |
| Verticals: left, middle, right | 3 |
| Roof diagonals: 2 per house | 4 |

Total: 2 + 2 + 3 + 4 = 11

Yes.

Pattern 3 (Three houses):
- Add another house to the right.
- Now, the third house shares its left wall with the second house.
- So again, we add 5 new sticks? Let's count:

Bottom: 3 horizontal
Top (square): 3 horizontal
Verticals: 4 (left, mid1, mid2, right)
Roof diagonals: 3 × 2 = 6

Total: 3 + 3 + 4 + 6 = 16

Alternatively, from previous:
- Pattern 2: 11
- Pattern 3: add 5 more? 11 + 5 = 16 → yes.

Wait: difference:
- P1: 6
- P2: 11 → +5
- P3: 16 → +5

So it's increasing by +5 each time.

So:
- M₁ = 6
- M₂ = 11
- M₃ = 16
- M₄ = 21

So the sequence is: 6, 11, 16, 21 → arithmetic sequence with common difference +5

So:
- M = 5P + 1 ?

Check:
- P = 1: 5(1) + 1 = 6 → ✔️
- P = 2: 5(2) + 1 = 11 → ✔️
- P = 3: 5(3) + 1 = 16 → ✔️
- P = 4: 5(4) + 1 = 21 → ✔️

✔ So the formula is:
M = 5P + 1

And the change in M is +5, and P increases by +1.

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✔ Answer for Problem 2:

| Pattern Number, P | 1 | 2 | 3 | 4 |
|--------------------|---|---|---|---|
| Number of matchsticks, M | 6 | 11 | 16 | 21 |

M = 5
P = +1

> (Note: The blank under M likely means "increase by", so M = 5 means each pattern adds 5 matchsticks.)

And the next pattern (P=4) should be 4 houses in a row.

Draw: four connected houses, each sharing a vertical side with the next.

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🔷 Problem 3: Diamond-Shaped Patterns

Each shape looks like a diamond made of two triangles stacked vertically.

Let’s analyze:

Pattern 1: One diamond (two triangles)
- Upward triangle: 3 sticks
- Downward triangle: 3 sticks
- But they share a horizontal base, so total sticks = 3 + 3 - 1 = 5

Wait — actually, let’s count:

- Top point: 1 stick
- Left side: 1
- Right side: 1
- Bottom point: 1
- But if it's two triangles sharing a side, then:
- Up triangle: 3 sticks
- Down triangle: 3 sticks
- Shared side: 1 (middle horizontal)
- So total: 3 + 3 - 1 = 5

Yes.

So Pattern 1: 5 sticks

Pattern 2: Two diamonds side-by-side

They share a vertical side?

Look at the image: two diamonds joined at their vertical center line.

So:
- First diamond: 5 sticks
- Second diamond: shares one vertical edge with the first
- So second diamond adds only 4 new sticks?

Let’s count total:

- Top points: 2
- Left sides: 2
- Right sides: 2
- Bottom points: 2
- Middle vertical: 1 (shared)
- Horizontal connections: 2 (top and bottom of the shared line?)

Wait — better to visualize:

Each diamond is a rhombus made of two equilateral triangles.

When joined side-by-side, they share one vertical edge.

So:
- First diamond: 5 sticks
- Second diamond: shares one side → adds 4 new sticks
- Total: 5 + 4 = 9

Pattern 3: Three diamonds

Add another diamond to the right, sharing a vertical edge with the second.
- Adds 4 more sticks
- Total: 9 + 4 = 13

So sequence:
- P1: 5
- P2: 9 → +4
- P3: 13 → +4

So difference is +4

Then:
- P4: 13 + 4 = 17

So:
- M₁ = 5
- M₂ = 9
- M₃ = 13
- M₄ = 17

This is an arithmetic sequence: M = 4P + 1

Check:
- P=1: 4(1)+1 = 5 → ✔️
- P=2: 8+1=9 → ✔️
- P=3: 12+1=13 → ✔️
- P=4: 16+1=17 → ✔️

✔ So formula: M = 4P + 1

Change in M: +4, P increases by +1

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✔ Answer for Problem 3:

| Pattern Number, P | 1 | 2 | 3 | 4 |
|--------------------|---|---|---|---|
| Number of matchsticks, M | 5 | 9 | 13 | 17 |

M = 4
P = +1

Next pattern (P=4): draw 4 diamonds in a row, each sharing a vertical side with the next.

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✔ Final Answers:

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#### Problem 2:
- Table:
- P: 1 → M: 6
- P: 2 → M: 11
- P: 3 → M: 16
- P: 4 → M: 21
- M = 5
- P = +1

#### Problem 3:
- Table:
- P: 1 → M: 5
- P: 2 → M: 9
- P: 3 → M: 13
- P: 4 → M: 17
- M = 4
- P = +1

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📌 Summary:

| Problem | Formula | M increase | P increase |
|--------|---------|------------|------------|
| 1 (Example) | M = 3P + 1 | +3 | +1 |
| 2 | M = 5P + 1 | +5 | +1 |
| 3 | M = 4P + 1 | +4 | +1 |

The key idea: look for the common difference in the number of matchsticks, and use that to find the algebraic rule.

Let me know if you'd like the drawings sketched out!