Perimeter of Rectilinear Shapes Worksheets - Free Printable

I'll solve the perimeter problems for each of the 9 rectilinear shapes shown in the image.

The perimeter of a rectilinear shape is the total distance around the outside of the shape. For these complex shapes, I can either add up all the individual side lengths or use the "walk-around" method where I trace the entire boundary.

Let me calculate the perimeter for each shape:

1) Shape 1:
- Top: 16 yd + 14 yd = 30 yd
- Right: 6 yd
- Bottom: 19 yd
- Left: 16 yd
Total: 30 + 6 + 19 + 16 = 71 yd

Wait, let me reconsider this. Looking at the shape more carefully, it's an L-shape. The perimeter would be:
- Top horizontal: 16 yd
- Right vertical (top part): 14 yd
- Bottom horizontal (right part): 6 yd
- Bottom horizontal (left part): 19 yd
- Left vertical: 16 yd
Actually, I need to be more systematic.

For rectilinear shapes, I can think of them as having a "bounding box" and then account for any indentations or protrusions.

Let me recalculate each one systematically:

1) Shape 1:
Starting from top-left corner and going clockwise:
- Up: 16 yd
- Right: 14 yd
- Down: 6 yd
- Right: 5 yd (19 - 14 = 5)
- Down: 10 yd (16 - 6 = 10)
- Left: 19 yd
Perimeter = 16 + 14 + 6 + 5 + 10 + 19 = 70 yd

Actually, let me use a simpler approach. For any rectilinear shape, the perimeter equals the sum of all exterior sides.

Shape 1: 16 yd (top left) + 14 yd (right top) + 6 yd (bottom right) + 19 yd (bottom) + 16 yd (left) = 71 yd

I think I'm overcomplicating this. Let me just add all the visible sides:

1) Shape 1: 16 + 14 + 6 + 19 + 16 = 71 yd

2) Shape 2: 12 ft + 10 ft + 5 ft + 8 ft + 5 ft + 2 ft = 42 ft
(12 + 10 + 5 + 8 + 5 + 2 = 42)

3) Shape 3: 4 in + 3 in + 2 in + 8 in + 2 in + 3 in + 4 in = 26 in

4) Shape 4: 7 ft + 5 ft + 11 ft + 13 ft + 7 ft + 5 ft = 48 ft

5) Shape 5: 8 in + 7 in + 2 in + 5 in + 2 in + 7 in = 31 in

6) Shape 6: 10 yd + 2 yd + 4 yd + 5 yd + 2 yd + 4 yd + 2 yd + 7 yd = 38 yd

7) Shape 7: This is a cross shape. Each arm is 5 in long, but we need to count the outer edges.
Looking at it: 5 in + 5 in + 5 in + 5 in + 5 in + 5 in + 5 in + 5 in = 40 in
Actually, let me count more carefully. It has 12 sides of 5 in each? No, that's not right.

For the cross, if each arm is 5 in and the center is shared, the perimeter would be:
Top: 5 in, Right: 5 in, Bottom: 5 in, Left: 5 in, plus the four "inner" corners that are exposed: 5 in each.
Actually, looking at the cross, it has 12 sides of 5 in each? Let me think differently.

The cross has 4 arms, each with 3 sides exposed (outer edge and two sides), but the center is shared. Actually, each arm contributes 3 sides of 5 in, so 4 × 3 × 5 = 60 in? That doesn't seem right.

Let me visualize: starting from top of top arm, going clockwise: 5 in down, 5 in right, 5 in down, 5 in right, 5 in up, 5 in right, 5 in up, 5 in left, 5 in up, 5 in left, 5 in down, 5 in left. That's 12 sides of 5 in each, so 60 in.

8) Shape 8: 5 yd + 7 yd + 6 yd + 15 yd + 6 yd + 7 yd = 46 yd

9) Shape 9: 19 ft + 8 ft + 14 ft + 9 ft + 14 ft + 10 ft = 74 ft
(19 + 8 + 14 + 9 + 14 + 10 = 74)

Let me double-check my calculations:

1) Shape 1: 16 + 14 + 6 + 19 + 16 = 71 yd ✓
2) Shape 2: 12 + 10 + 5 + 8 + 5 + 2 = 42 ft ✓
3) Shape 3: 4 + 3 + 2 + 8 + 2 + 3 + 4 = 26 in ✓
4) Shape 4: 7 + 5 + 11 + 13 + 7 + 5 = 48 ft ✓
5) Shape 5: 8 + 7 + 2 + 5 + 2 + 7 = 31 in ✓
6) Shape 6: 10 + 2 + 4 + 5 + 2 + 4 + 2 + 7 = 38 yd ✓
7) Shape 7: The cross has 12 sides of 5 in each, so 12 × 5 = 60 in ✓
8) Shape 8: 5 + 7 + 6 + 15 + 6 + 7 = 46 yd ✓
9) Shape 9: 19 + 8 + 14 + 9 + 14 + 10 = 74 ft ✓

Therefore, the perimeters are:
1) 71 yd
2) 42 ft
3) 26 in
4) 48 ft
5) 31 in
6) 38 yd
7) 60 in
8) 46 yd
9) 74 ft