Question

find the perimeter when 80 triangles are put together in the pattern shown below. assume that all triangle sides are 1 cm long. the perimeter is □ cm.

Explanation:

Step1: Analyze the pattern

From the figure, when we have \( n \) triangles (assuming the pattern is a sequence of triangles forming a larger shape, likely a parallelogram - like structure with two triangles per "layer" or a pattern where the number of triangles is even, let's assume the number of triangles \( n \) is even. Let \( n = 2k \), where \( k \) is the number of pairs of triangles. For example, when \( n = 2 \) (2 triangles), the perimeter is \( 4 \) cm? Wait, no, let's look at the given figure. The figure shows 4 triangles (maybe? Wait, the problem says 80 triangles. Let's find the pattern.

Wait, let's consider the number of triangles: when we have \( n \) triangles (equilateral, side 1 cm) arranged in a pattern where they form a sort of parallelogram with two triangles per "column" or row. Wait, let's take small cases.

Case 1: 2 triangles (arranged to form a rhombus). Perimeter: 4 cm? Wait, no, two equilateral triangles with side 1 cm, joined along a side, form a rhombus with side 1 cm, perimeter \( 4\times1=4 \)? Wait, no, two equilateral triangles joined along a side: the shape has 4 sides? Wait, no, each triangle has 3 sides. When joined, the common side is internal. So for 2 triangles: total outer sides. Each triangle contributes 2 sides (since one side is shared). So \( 2\times2 + 2\times1 \)? Wait, maybe better to look at the number of triangles as \( n \), and find the perimeter formula.

Wait, the problem says 80 triangles. Let's assume the pattern is such that the number of triangles \( n \) is even, \( n = 2k \). Let's take \( k = 1 \) (2 triangles): perimeter? Let's draw it. Two equilateral triangles, base to base, forming a rhombus. Each side is 1 cm. The perimeter: top, bottom, left, right. Wait, no, the rhombus has 4 sides, each 1 cm, so perimeter 4. For \( k = 2 \) (4 triangles): arranged in two rows of two? Wait, the figure shows 4 triangles (maybe 4 triangles forming a larger parallelogram). Let's count the perimeter. For 4 triangles: let's see, the horizontal sides: top and bottom, each has 2 segments? Wait, no, maybe the formula is: when there are \( n \) triangles (even, \( n = 2k \)), the perimeter is \( 2 + 2k \)? Wait, no, let's check with \( n = 2 \): \( 2 + 2\times1 = 4 \), which matches. For \( n = 4 \): \( 2 + 2\times2 = 6 \)? Wait, no, maybe not. Wait, another approach: each triangle has side 1 cm. When we arrange them in a pattern where they form a shape with height (number of triangles vertically) and width. Wait, the figure shows a vertical stack? Wait, the figure has 4 triangles, arranged in two columns? Wait, maybe the number of triangles \( n \), and the perimeter is \( (n/2)+2 + (n/2)+2 \)? No, that would be \( n + 4 \), which doesn't make sense. Wait, let's look at the figure again. The figure is a parallelogram made of 4 equilateral triangles (two on top, two on bottom, or four in a 2x2 diamond shape). Wait, for 4 triangles, the perimeter: let's count the outer sides. Each side of the large parallelogram: the slant sides: 2 cm each? No, each small triangle has side 1 cm. Wait, maybe the pattern is that for \( n \) triangles (where \( n \) is even, \( n = 2k \)), the perimeter is \( 2 + 2k \) horizontal sides? No, maybe the formula is \( P = 2 + n/2 + 2 + n/2 \)? Wait, no, let's take \( n = 2 \): perimeter 4. \( n = 4 \): perimeter 6? Wait, no, if you have 4 triangles (arranged as two rows of two, forming a larger rhombus), the perimeter would be 6? Wait, no, each side of the larger rhombus would be 2 cm? No, the small triangles have side 1 cm. Wait, maybe I'm overcomplicating. Let's assume that the number of triangles is \( n \), and the perimeter formula is \( P = 2 + n/2 + 2 + n/2 = n + 4 \)? No, for \( n = 2 \), \( 2 + 4 = 6 \), which is wrong. Wait, maybe the correct pattern is: when you have \( n \) triangles (even, \( n = 2k \)), the perimeter is \( 2 + 2k \) for the top and bottom, and 2 for the sides? Wait, no. Let's look at the problem again: "80 triangles are put together in the pattern shown below. Assume that all triangle sides are 1 cm long."

Wait, maybe the pattern is that each pair of triangles (2 triangles) contributes 2 to the perimeter, plus 2 for the ends. So for \( k \) pairs (total \( n = 2k \) triangles), the perimeter is \( 2k + 2 \)? Wait, no, for \( k = 1 \) (2 triangles), perimeter 4: \( 2(1) + 2 = 4 \), yes. For \( k = 2 \) (4 triangles), perimeter 6: \( 2(2) + 2 = 6 \), yes. For \( k = 3 \) (6 triangles), perimeter 8: \( 2(3) + 2 = 8 \). So the pattern is \( P = 2k + 2 \), where \( k = n/2 \) (since \( n = 2k \)). So for \( n = 80 \), \( k = 40 \). Then \( P = 2(40) + 2 = 82 \)? Wait, no, wait: when \( k = 1 \) (2 triangles), perimeter 4: \( 2(1) + 2 = 4 \). When \( k = 2 \) (4 triangles), perimeter 6: \( 2(2) + 2 = 6 \). So the formula is \( P = 2k + 2 \), where \( k = n/2 \). So \( n = 80 \), \( k = 40 \), so \( P = 2(40) + 2 = 82 \)? Wait, but let's check with the figure. The figure shows 4 triangles (k=2), so perimeter should be 6. Let's count the outer sides: each triangle has side 1 cm. The figure: 4 triangles arranged in a vertical stack (two on top, two on bottom, forming a parallelogram). The top and bottom sides: each has 2 segments (1 cm each), so 2 cm each. The left and right sides: each has 2 segments? No, wait, the left side: how many segments? If it's a parallelogram made of 4 equilateral triangles, the left side would have 2 sides of 1 cm each? No, the height (vertical) would be 2 triangles, so the left side is 2 cm? Wait, no, the perimeter: top: 2 cm, bottom: 2 cm, left: 2 cm, right: 2 cm? No, that would be 8 cm, which is wrong. Wait, I must have the pattern wrong.

Alternative approach: Let's look at the number of triangles and the perimeter.

- 2 triangles: perimeter 4 (as two triangles joined along a side, forming a rhombus with 4 sides of 1 cm each? No, two equilateral triangles joined along a side: the shape has 4 sides? Wait, no, each triangle has 3 sides. When joined, the common side is internal, so total outer sides: 3 + 3 - 2 = 4. So perimeter 4.

- 4 triangles: arranged as two pairs, joined along the base. So the shape is a larger rhombus? Wait, 4 equilateral triangles, each with side 1 cm, arranged in a 2x2 diamond shape. The perimeter: let's count the outer sides. The top side: 2 cm (two segments of 1 cm), bottom side: 2 cm, left side: 2 cm, right side: 2 cm? No, that's 8 cm, but that can't be. Wait, maybe the pattern is different. Maybe the triangles are arranged in a linear pattern, like a chain. But the figure shows a vertical stack.

Wait, the problem says "the pattern shown below" – the figure is a parallelogram made of 4 equilateral triangles (two in the top row, two in the bottom row, forming a larger parallelogram with base 2 cm and height 2 cm? No, each small triangle has side 1 cm. So the base of the large parallelogram is 2 cm (two small triangle bases), and the slant side is 2 cm (two small triangle sides). Then the perimeter would be \( 2\times(2 + 2) = 8 \) cm for 4 triangles. But that contradicts the earlier thought.

Wait, maybe the key is that each triangle (except the ones on the ends) contributes 1 to the perimeter, and the ends contribute 2 each. Wait, for 2 triangles: perimeter 4. For 4 triangles: perimeter 6. For 6 triangles: perimeter 8. So the pattern is \( P = n + 2 \), where \( n \) is the number of triangles? No, 2 + 2 = 4, 4 + 2 = 6, 6 + 2 = 8. Yes! That works. So for \( n \) triangles, perimeter \( P = n + 2 \)? Wait, no, 2 triangles: 2 + 2 = 4, correct. 4 triangles: 4 + 2 = 6, correct. 6 triangles: 6 + 2 = 8, correct. So the formula is \( P = n + 2 \)? Wait, but let's check with the figure. If the figure has 4 triangles, perimeter 6, which matches \( 4 + 2 = 6 \). Then for 80 triangles, perimeter \( 80 + 2 = 82 \)? Wait, but that seems too simple. Wait, maybe the pattern is that when you have \( n \) triangles arranged in a column (stacked vertically), each triangle after the first adds 1 to the perimeter. Wait, first triangle: perimeter 3. Second triangle: joined to the first, so perimeter 3 + 1 = 4. Third triangle: joined to the second, perimeter 4 + 1 = 5. Fourth triangle: joined to the third, perimeter 5 + 1 = 6. Ah! So the pattern is \( P = n + 2 \) when \( n \) is the number of triangles? Wait, no, first triangle (n=1): 3, which is 1 + 2 = 3. Second triangle (n=2): 4 = 2 + 2. Third (n=3): 5 = 3 + 2. Fourth (n=4): 6 = 4 + 2. Yes! So the formula is \( P = n + 2 \). Wait, but the figure shows 4 triangles, perimeter 6, which is 4 + 2 = 6. So that works. Then for 80 triangles, perimeter is \( 80 + 2 = 82 \).

Wait, but let's confirm with the first few terms:

- n=1: perimeter 3 (triangle, 3 sides)
- n=2: joined along a side, so perimeter 3 + 3 - 2 = 4 (which is 2 + 2)
- n=3: joined to the second, perimeter 4 + 1 = 5 (3 + 2)
- n=4: joined to the third, perimeter 5 + 1 = 6 (4 + 2)

Yes, so the pattern is \( P = n + 2 \), where \( n \) is the number of triangles. Therefore, for \( n = 80 \), \( P = 80 + 2 = 82 \).

Step2: Apply the formula

Given \( n = 80 \) triangles, using the pattern \( P = n + 2 \), we substitute \( n = 80 \):

\( P = 80 + 2 = 82 \)

Answer:

82