Free. Exclusive. Just for you.
Four unique services that make learning easier, faster, and smarter - only on our website.

Grade 6 Area Worksheets | Find the Area of Compound Shapes - Free Printable

Grade 6 Area Worksheets | Find the Area of Compound Shapes

Educational worksheet: Grade 6 Area Worksheets | Find the Area of Compound Shapes. Download and print for classroom or home learning activities.

PNG 546×784 28 KB Free · Personal Use
Quality Assured by Worksheets Library Team
Reviewed for educational accuracy and age-appropriateness
ID: #980839
Show Answer Key & Explanations Step-by-step solution for: Grade 6 Area Worksheets | Find the Area of Compound Shapes
Let’s solve each problem step by step. We’ll break each shape into rectangles, find the area of each rectangle, then add them together.

---

Problem 1:

The shape is like an “L”. We can split it into two rectangles:

- Left rectangle: height = 8m, width = ?
The total bottom is 12m, and the right part sticks out 8m → so left part width = 12 - 8 = 4m
Area = 8 × 4 = 32 m²

- Right rectangle: height = 3m, width = 8m
Area = 3 × 8 = 24 m²

Total area = 32 + 24 = 56 m²

Check: Another way — imagine full rectangle 12m wide × 8m tall = 96 m², but we’re missing a top-right rectangle that’s 8m wide × (8-3)=5m tall → 8×5=40 m² → 96 - 40 = 56 m². Same answer!

---

Problem 2:

Again, L-shaped. Split into two rectangles:

Option A:
- Left big rectangle: 20cm high × (18 - 5) = 13cm wide? Wait — let’s look carefully.

Actually, better to split horizontally or vertically.

Let’s do vertical split:

- Left rectangle: full height 20cm, width = 18 - 5 = 13cm → area = 20 × 13 = 260 cm²
- Right small rectangle: height = 10cm, width = 5cm → area = 10 × 5 = 50 cm²
Wait — no! That doesn’t match the drawing.

Looking again: The shape has a “notch” on the top right. So actually:

Better split:

- Bottom rectangle: full width 18cm, height = 20 - 10 = 10cm → area = 18 × 10 = 180 cm²
- Top-left rectangle: width = 18 - 5 = 13cm, height = 10cm → area = 13 × 10 = 130 cm²
Total = 180 + 130 = 310 cm²

Wait — let me double-check with another method.

Imagine full rectangle: 18cm × 20cm = 360 cm²
Missing piece: top-right corner, which is 5cm wide × 10cm tall → 50 cm²
So 360 - 50 = 310 cm²

Yes, correct.

---

Problem 3:

This one looks like a U-shape or has a notch in the middle.

We can think of it as a big rectangle minus a smaller rectangle cut out from the top middle.

Big rectangle: width = 14m, height = 10m → area = 14 × 10 = 140 m²

But wait — the sides are not both 10m. Left side is 10m, right side is 8m. Hmm.

Actually, let’s split into three parts:

Left rectangle: width = 5m, height = 10m → area = 50 m²
Right rectangle: width = ? Total width 14m, left 5m, middle notch 5m → so right width = 14 - 5 - 5 = 4m? But the diagram shows the right part goes up to 8m.

Wait — let’s label:

From left to right:

- Left block: 5m wide, 10m tall → area = 50 m²
- Middle gap: 5m wide, but how deep? The drop is from 10m to... well, the right side is 8m tall, and there's a 4m segment shown inside the notch.

Actually, looking at the diagram:

It says:

Top: 5m (left), then a dip, then a rise, then right side 8m tall.

Inside the dip: horizontal 5m, vertical 4m down.

So perhaps:

Split into:

1. Left rectangle: 5m × 10m = 50 m²
2. Right rectangle: width = 14 - 5 - 5 = 4m? Wait, total width 14m. Left 5m, middle 5m, so right must be 4m. Height = 8m → area = 4 × 8 = 32 m²
3. Middle bottom rectangle: connects left and right under the notch. Width = 5m, height = ? Since left is 10m, and the notch drops 4m, so the base of the notch is at 10 - 4 = 6m from bottom? No.

Wait — better approach:

Think of the whole shape as:

- Full outer rectangle: 14m wide × 10m tall = 140 m²
- Minus the missing rectangle in the top middle.

What’s the size of the missing part?

The notch is 5m wide (given), and how tall? From the top (10m level) down to where? The right side only goes up to 8m, but the notch depth is labeled as 4m — meaning from the top of the left/right shoulders down 4m.

Actually, the diagram shows:

On the right side, after the notch, it rises 4m to reach 8m total height. So the bottom of the notch is at 8 - 4 = 4m above ground? That doesn't make sense.

Let me reinterpret:

Perhaps the shape is made of:

- Left tower: 5m wide × 10m tall
- Right tower: ? wide × 8m tall
- Connecting base between them: 5m wide × ? tall

Total width: 5 (left) + 5 (middle) + x (right) = 14 → x = 4m

Now, the connecting base: since the left is 10m tall and right is 8m tall, and the notch drops 4m from the left top, that means the base under the notch is at height 10 - 4 = 6m? But then the right tower starts at 6m and goes up to 8m? That would mean the right tower is only 2m tall above the base — but it's labeled as 8m total.

I think I'm overcomplicating.

Alternative plan: divide into 3 rectangles:

1. Left vertical: 5m × 10m = 50 m²
2. Right vertical: 4m × 8m = 32 m² (since 14 - 5 - 5 = 4)
3. Horizontal connector at the bottom: spans the middle 5m, and its height is the minimum of the two sides? Or rather, since the notch is 4m deep from the left top, and the right side is 8m, the connector height should be 8m? No.

Wait — here’s a better way:

Draw imaginary lines.

The entire figure can be seen as:

- A large rectangle 14m wide × 8m tall (covering the bottom part up to 8m) → area = 14 × 8 = 112 m²
- Plus a small rectangle on the top left: 5m wide × (10 - 8) = 2m tall → area = 5 × 2 = 10 m²
- Minus the notch? No, because the notch is already accounted for if we do this.

Actually, if we take 14×8 = 112, that includes everything up to 8m height across the whole width. Then above that, only the left 5m has extra 2m (to make 10m). The middle 5m and right 4m don’t go above 8m. But in the diagram, the middle has a dip — wait, no, the dip is within the top part.

I think the key is: the "4m" label is the depth of the notch from the top of the left shoulder.

So:

- Left part: 5m × 10m = 50
- Right part: let’s say width W, height 8m. Total width 14, left 5, middle 5, so W=4 → 4×8=32
- Now the middle part: it’s a rectangle that sits below the notch. Its width is 5m, and its height is... since the left goes down to 10m, and the notch drops 4m, the floor of the notch is at 6m from bottom? But the right side is only 8m tall, so if the connector is at 6m, then the right tower would start at 6m and go to 8m — that’s 2m, but it’s drawn as going all the way down.

I found a simpler way:

Use the subtraction method.

Imagine the bounding box: 14m wide × 10m tall = 140 m²

Now, what is missing? There is a rectangular hole in the top middle.

Width of hole = 5m (given)

Height of hole = ? From the diagram, the left side is 10m, the right side is 8m, and there's a 4m drop indicated in the notch. Actually, the 4m is likely the height of the notch itself — meaning from the top of the adjacent sections down 4m.

Since the left section is 10m, and the notch drops 4m, the bottom of the notch is at 6m. The right section is 8m tall, so if the notch bottom is at 6m, then the right section extends from 6m to 8m — that’s fine.

But the hole is only in the top part. Specifically, the hole is 5m wide and 4m tall, located at the top center.

Is that accurate? Let's see: if we remove a 5m × 4m rectangle from the top middle of the 14×10 rectangle, then:

- Left remaining: 5m wide × 10m tall — good
- Right remaining: 4m wide × 10m tall? But we need it to be only 8m tall. Oh no — that doesn't work.

Unless the right part is also reduced.

Perhaps the shape is symmetric? No.

Another idea: calculate area by adding visible rectangles.

From the diagram:

- Rectangle A: left, 5m x 10m = 50
- Rectangle B: right, 4m x 8m = 32 (width 14-5-5=4)
- Rectangle C: the bottom connector between left and right, spanning the middle 5m, and its height is the same as the right side's base, which is 8m? But then it would overlap.

I recall that in such problems, the "4m" is the depth of the indentation, and the height of the right arm is 8m, so the base of the indentation is at 10 - 4 = 6m from bottom, and the right arm starts at 6m and goes to 8m, so its height above the base is 2m, but total height is 8m, so the base must be at 0? I'm confused.

Let me try numerical values based on standard interpretation.

Upon second thought, in many textbooks, for such a shape:

You can split it as:

1. Left rectangle: 5m * 10m = 50
2. Right rectangle: (14 - 5 - 5) = 4m * 8m = 32
3. Middle rectangle: 5m * (8m) ? Why 8m? Because the right side is 8m, and the left side has a protrusion, but the middle part is filled up to the lower of the two heights? No.

Actually, the middle part is not full height. The notch means that in the middle 5m, the height is less.

Specifically, from the bottom up to some height, it's solid, then there's a gap.

The diagram shows that from the bottom, the shape goes up, then in the middle, it dips down by 4m from the top of the left side.

So, the minimum height in the middle is 10 - 4 = 6m.

And the right side is 8m, which is higher than 6m, so the right side is taller.

To avoid confusion, let's use the following reliable method:

Divide the shape into three parts:

- Part 1: the left column: 5m wide × 10m high = 50 m²
- Part 2: the right column: 4m wide × 8m high = 32 m² (since 14 - 5 - 5 = 4)
- Part 3: the bottom row connecting them: this is a rectangle that is 5m wide (the middle section) and has height equal to the height of the shorter column, which is 8m? But then it would include area that is already in the columns.

No, better: the bottom row should be from y=0 to y=h, where h is the height before the notch.

Since the notch is 4m deep from the top of the left side, and the left side is 10m, the notch starts at y=6m (from bottom).

So, from y=0 to y=6m, the entire width 14m is solid.

Then from y=6m to y=10m, only the left 5m and the right 4m are present; the middle 5m is empty.

So:

- Bottom rectangle: 14m × 6m = 84 m²
- Top-left rectangle: 5m × (10-6) = 5×4 = 20 m²
- Top-right rectangle: 4m × (8-6) = 4×2 = 8 m² [because the right side only goes to 8m, so from y=6 to y=8]

Total area = 84 + 20 + 8 = 112 m²

Let me verify with another method.

Full rectangle 14×10 = 140

Minus the missing parts:

- In the top middle: from y=6 to y=10, x=5 to x=10 (5m wide), but only if the right side were 10m, but it's not.

Actually, the missing part is only in the region where there is no material.

From y=6 to y=10, the middle 5m is missing, but the right part from y=8 to y=10 is also missing because the right side only goes to 8m.

So missing areas:

1. Rectangle A: middle top: 5m wide × 4m high (y=6 to 10) = 20 m²
2. Rectangle B: right top: 4m wide × 2m high (y=8 to 10) = 8 m²

Total missing = 20 + 8 = 28 m²

Full area 140 - 28 = 112 m²

Perfect.

So Problem 3 area = 112 m²

---

Final Answers:

1. 56 m²
2. 310 cm²
3. 112 m²

──────────────────────────────────────

Final Answer:
1. 56
2. 310
3. 112
Parent Tip: Review the logic above to help your child master the concept of composite figures worksheet.
Print Download

How to use

Click Print to open a print-ready version directly in your browser, or use Download to save the file to your device. The ⭐ Answer button generates an AI answer key instantly - useful for teachers who need a quick reference. Need a different version? Our AI Worksheet Generator lets you create a custom worksheet on any topic in seconds.

(view all composite figures worksheet)

Third Grade Area of Composite Shapes Challenge Cards
Finding Area Of Composite Figures Worksheet
Area of Composite Shapes (Compound Figures) Worksheets
Area and Perimeter of Compound Shapes (A)
Area and perimeter of composite figures worksheet | Live Worksheets
Compound Shapes (B) | Cazoom Maths Worksheets
Boost Math Skills With These Engaging Area Of Composite Figures ...
Area of Composite Shapes — Printable Math Worksheet
Geometry Worksheets | Area Worksheets
Compound Area Differentiated Activity for 3rd-5th Grade