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Area and Perimeter of Composite Shapes Concept CW worksheet | Live ... - Free Printable

Area and Perimeter of Composite Shapes Concept CW worksheet | Live ...

Educational worksheet: Area and Perimeter of Composite Shapes Concept CW worksheet | Live .... Download and print for classroom or home learning activities.

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Show Answer Key & Explanations Step-by-step solution for: Area and Perimeter of Composite Shapes Concept CW worksheet | Live ...
Let’s solve each problem one by one. We’ll break down each shape into simpler parts (like rectangles, triangles, semicircles), calculate their areas, and add or subtract as needed. Remember: use π = 3.14 and round to two decimal places.

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Problem 1)
Shape: Rectangle + Semicircle on the right side.

- Rectangle: length = 12 yd, width = 4 yd → Area = 12 × 4 = 48 sq yd
- Semicircle: diameter = 8 yd → radius = 4 yd
Area of full circle = πr² = 3.14 × 4² = 3.14 × 16 = 50.24
Half of that = 50.24 ÷ 2 = 25.12 sq yd

Total area = 48 + 25.12 = 73.12 sq yd

Check: Makes sense — rectangle is big, semicircle adds a bit more.

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Problem 2)
Shape: Triangle + Semicircle on the right.

- Triangle: base = 10 in, height = 11 in → Area = (base × height) ÷ 2 = (10 × 11) ÷ 2 = 110 ÷ 2 = 55 sq in
- Semicircle: diameter = 10 in → radius = 5 in
Area of full circle = 3.14 × 5² = 3.14 × 25 = 78.5
Half = 78.5 ÷ 2 = 39.25 sq in

Total area = 55 + 39.25 = 94.25 sq in

Check: Triangle and semicircle both contribute significantly — answer looks reasonable.

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Problem 3)
Shape: Arrow = Triangle (left) + Rectangle (right)

First, find dimensions:

- Total length = 12 ft
- Rectangle part: width = 6 ft (given), so triangle base must be 12 - 6 = 6 ft? Wait — look again.

Actually, the arrow has:
- A rectangular body: height = 6 ft, width = ? Let’s see total horizontal length is 12 ft, and the “point” sticks out. The flat part on top/bottom near the point is labeled 2 ft — that means the triangle’s base is 6 ft + 2 ft + 2 ft? No.

Wait — better way:

The entire shape can be seen as:
- A rectangle in the middle: 6 ft tall, and how wide? From the diagram, the vertical sides are 6 ft, and the horizontal segment at the bottom says “6 ft” for the rectangle part? Actually, let's read labels:

Label says: from left tip to start of rectangle = 12 ft? No — it says “←12 ft→” under the whole shape. And above the rectangle part, there’s a label “2 ft” on the top edge sticking out? Actually, looking carefully:

It’s an arrow pointing left. The rectangle part is 6 ft high and 6 ft wide (since bottom label says “6 ft” under the rectangle). Then the triangle on the left has base = 6 ft (same as rectangle height) and height = 12 ft - 6 ft = 6 ft? But wait — the 12 ft is the total length from tip to end of rectangle.

So:

- Rectangle: 6 ft (height) × 6 ft (width) = 36 sq ft
- Triangle: base = 6 ft (matches rectangle height), height = 12 ft - 6 ft = 6 ft → Area = (6 × 6)/2 = 36/2 = 18 sq ft

But wait — the top has a little notch? Label says “2 ft” on the top edge of the triangle part? Actually, re-examining: the shape is symmetric? The “2 ft” might indicate that the triangle extends 2 ft beyond the rectangle on top and bottom? That would mean the triangle’s base is 6 ft + 2 ft + 2 ft = 10 ft? But then height?

This is confusing. Let me reinterpret based on standard problems.

Alternative interpretation: The arrow is made of a rectangle 6 ft by 6 ft, and a triangle attached to its left side with base = 6 ft (same as rectangle height) and height = 12 ft - 6 ft = 6 ft. But then why is there a “2 ft” label? Looking at the image description: “2 ft” is written on the top slanted part? Actually, perhaps the 2 ft is the overhang — meaning the triangle’s base is wider than the rectangle.

Standard approach in such diagrams: if the rectangle is 6 ft tall, and the triangle’s base spans the full height plus 2 ft on top and 2 ft on bottom? That doesn’t make sense.

Wait — perhaps the “2 ft” is the length of the small horizontal segment at the top of the triangle? Meaning the triangle is not full height? I think I need to assume the simplest case.

Looking at common textbook problems: often, the arrow is composed of a rectangle and a triangle where the triangle’s base equals the rectangle’s height, and the total length is given.

Given: total length = 12 ft, rectangle width = 6 ft (from bottom label), so triangle height = 12 - 6 = 6 ft. Rectangle height = 6 ft, so triangle base = 6 ft.

Then area = rectangle (6×6=36) + triangle (½×6×6=18) = 54 sq ft.

But the “2 ft” label — perhaps it’s indicating that the triangle’s apex is offset? Or maybe it’s a typo? Another possibility: the 2 ft is the amount the triangle extends beyond the rectangle on each side vertically? So triangle base = 6 + 2 + 2 = 10 ft, and height = 12 - 6 = 6 ft? Then triangle area = ½ × 10 × 6 = 30, rectangle = 6×6=36, total 66.

But that seems too big. Let me check online or standard problems — since this is a common type.

Actually, upon second thought: in many such worksheets, the "2 ft" is the length of the small horizontal line at the top of the triangular part, which suggests that the triangle is not right-angled but has a flat top? No, it's an arrowhead.

Perhaps the shape is: a rectangle 6 ft by 6 ft, and on the left, a triangle with base 6 ft and height 6 ft, but the "2 ft" is irrelevant? That can't be.

Wait — looking back at user's image description: "3) [arrow shape] with labels: 2 ft on the top edge of the triangle part, 6 ft for the rectangle height, 6 ft for the rectangle width (bottom), and 12 ft total length."

I think the correct interpretation is:

- The rectangle is 6 ft high and 6 ft wide.
- The triangle on the left has a base that is the same as the rectangle's height (6 ft), but because of the "2 ft" label, it might mean that the triangle's height is measured differently.

Another idea: the 12 ft is the distance from the tip of the arrow to the right end, and the rectangle starts 2 ft from the top and bottom? So the triangle's base is 6 ft + 2 ft + 2 ft = 10 ft, and the height of the triangle is 12 ft - 6 ft = 6 ft? But then the rectangle is only 6 ft wide, so total length would be 6 (rectangle) + 6 (triangle height) = 12 ft, yes.

So:

- Rectangle: 6 ft (height) × 6 ft (width) = 36 sq ft
- Triangle: base = 10 ft (because it extends 2 ft above and below the rectangle), height = 6 ft → Area = ½ × 10 × 6 = 30 sq ft

Total = 36 + 30 = 66 sq ft

But is the base really 10 ft? The rectangle is 6 ft tall, and if the triangle covers from 2 ft above to 2 ft below, then yes, base = 6 + 2 + 2 = 10 ft.

Yes, that makes sense with the "2 ft" label.

So area = 36 + 30 = 66.00 sq ft

Check: Seems consistent with labels.

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Problem 4)
Shape: L-shaped figure. Can be split into two rectangles.

Option 1: Left rectangle and bottom rectangle.

- Left rectangle: width = 3 yd, height = 10 yd → Area = 3 × 10 = 30 sq yd
- Bottom rectangle: but wait, the right part is indented. Better to do:

Split horizontally or vertically.

Vertical split:
- Left part: 3 yd wide × 10 yd high = 30 sq yd
- Right part: it's a rectangle that is 4 yd wide and 6 yd high? But the total height is 10 yd, and the right part starts after 3 yd, and goes down 6 yd? The label says "4 yd" for the width of the right protrusion, and "6 yd" for its height.

Actually, the shape is like a big rectangle minus a smaller rectangle, or add two rectangles.

Better:
- Full outer rectangle would be 3+4=7 yd wide and 10 yd high, but there's a cutout.

From the diagram: the left column is 3 yd wide and 10 yd tall. Attached to its right, at the bottom, is a rectangle that is 4 yd wide and 6 yd tall. So total area = area of left rect + area of bottom-right rect.

Left rect: 3 × 10 = 30 sq yd
Bottom-right rect: 4 × 6 = 24 sq yd
Total = 30 + 24 = 54 sq yd

Is that correct? The bottom-right rect is attached to the bottom of the left rect, so yes, no overlap.

Alternatively, you could see it as a large rectangle 7 yd by 10 yd minus a rectangle on top-right that is 4 yd by (10-6)=4 yd, so 70 - 16 = 54 sq yd. Same answer.

Good.

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Problem 5)
L-shape again. Dimensions: overall 10 in wide, 12 in tall, with a notch.

Can split into two rectangles.

Option:
- Top rectangle: 10 in wide × 5 in high = 50 sq in
- Bottom rectangle: 5 in wide × (12 - 5) = 5 × 7 = 35 sq in? But wait, the bottom part is only 5 in wide, and height from bottom to the notch is 12 - 5 = 7 in? Yes.

Total = 50 + 35 = 85 sq in

Alternatively, full rectangle 10×12=120, minus the missing part which is 5 in wide and 7 in high? Missing part is on the top-right? No.

Actually, the shape is like a backward L: left part is 5 in wide and 12 in tall, and bottom part is 10 in wide but only 5 in tall? Standard way:

- Vertical part: 5 in (width) × 12 in (height) = 60 sq in
- Horizontal part: but they overlap. Better: the horizontal arm is 10 in long and 5 in high, but the vertical arm is already included.

Correct split:
- Rectangle A: left side, 5 in wide × 12 in high = 60 sq in
- Rectangle B: bottom extension, but since the left 5 in is already counted, the additional part is (10 - 5) = 5 in wide × 5 in high? No, because the bottom is 5 in high, and the vertical part is 12 in, so the horizontal part should be from x=5 to x=10, y=0 to y=5, so size 5 in × 5 in = 25 sq in.

Total = 60 + 25 = 85 sq in. Same as before.

Full rectangle 10×12=120, minus the missing top-right rectangle which is 5 in wide and 7 in high (since 12-5=7), so 5×7=35, 120-35=85. Yes.

Answer: 85.00 sq in

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Problem 6)
Shape: House-like — rectangle on top, triangle on bottom.

Dimensions: total height 14 ft, rectangle height 9 ft, so triangle height = 14 - 9 = 5 ft. Width = 6 ft (same for both).

- Rectangle: 6 ft × 9 ft = 54 sq ft
- Triangle: base = 6 ft, height = 5 ft → Area = (6 × 5)/2 = 30/2 = 15 sq ft

Total = 54 + 15 = 69.00 sq ft

Simple and clear.

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Now, compiling all answers:

1) 73.12
2) 94.25
3) 66.00
4) 54.00
5) 85.00
6) 69.00

All rounded to two decimal places as required.

Final Answer:
1) 73.12
2) 94.25
3) 66.00
4) 54.00
5) 85.00
6) 69.00
Parent Tip: Review the logic above to help your child master the concept of perimeter and area of composite figures worksheet.
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