Area of Compound Shapes Textbook Exercise - Corbettmaths - Free Printable

Let's solve Question 1 from the image:
We are asked to find the area of each compound shape (a) through (f). Each shape is made up of rectangles, so we can break them down into simpler rectangles, calculate their areas, and then add or subtract as needed.

---

🔷 Strategy:

- Break each shape into rectangles.
- Find the area of each rectangle using:
$$
\text{Area} = \text{length} \times \text{width}
$$
- Add areas for shapes made of multiple rectangles.
- Subtract areas if a part is missing (like in shape (e) and (f)).

---

## ✔ (a)

```
9cm
┌─────────┐
│ │
│ │ 8cm
│ │
├───────┬─┤
│ │ │
│ 5cm │ │
│ │ │
└───────┴─┘
7cm
```

We can split this into two rectangles:

1. Top rectangle: width = 9 cm, height = 8 cm → but wait! The bottom is not aligned. Let's see:

Actually, notice:
- The total height is 8 cm.
- The bottom rectangle has height = 5 cm.
- So the top rectangle has height = 8 − 5 = 3 cm.
- The top rectangle has width = 9 cm.
- The bottom rectangle has width = 7 cm, height = 5 cm.

So:

- Top rectangle: $ 9 \times 3 = 27 \text{ cm}^2 $
- Bottom rectangle: $ 7 \times 5 = 35 \text{ cm}^2 $

✔ Total area = $ 27 + 35 = \boxed{62} \text{ cm}^2 $

---

## ✔ (b)

```
25cm
┌─────────────┐
│ │
│ │ 12cm
│ │
├─────────┬───┤
│ │ │
│ 10cm │ │
│ │ │
└─────────┴───┘
6cm
```

This is a "T" shape.

Break it into:
1. Top rectangle: width = 25 cm, height = 12 − 10 = 2 cm → $ 25 \times 2 = 50 \text{ cm}^2 $
2. Bottom rectangle: width = 6 cm, height = 10 cm → $ 6 \times 10 = 60 \text{ cm}^2 $

Wait — actually, the bottom rectangle is centered, so its width is 6 cm, but the top rectangle spans 25 cm.

But we must ensure alignment. The top rectangle is 25 cm wide and 2 cm tall (since total height is 12 cm, and the stem is 10 cm).

Then the stem is 6 cm wide and 10 cm tall.

So:
- Top: $ 25 \times 2 = 50 $
- Stem: $ 6 \times 10 = 60 $

✔ Total area = $ 50 + 60 = \boxed{110} \text{ cm}^2 $

---

## ✔ (c)

```
4cm
┌──────┐
│ │
│ │
│ │ 10cm
│ │
├──────┼───┐
│ │ │
│ 3cm │ │
│ │ │
└──────┴───┘
13cm
```

Split into:
1. Right rectangle: width = 4 cm, height = 10 cm → $ 4 \times 10 = 40 $
2. Left rectangle: width = 13 − 4 = 9 cm, height = 3 cm → $ 9 \times 3 = 27 $

✔ Total area = $ 40 + 27 = \boxed{67} \text{ cm}^2 $

---

## ✔ (d)

```
2cm
┌──────┐
│ │
│ │ 5cm
│ │
├──────┼───┐
│ │ │
│ 4cm │ │
│ │ │
└──────┴───┘
9cm
↓
3cm
```

This is like an upside-down L-shape.

Break it into:
1. Top rectangle: width = 2 cm, height = 5 cm → $ 2 \times 5 = 10 $
2. Bottom rectangle: width = 9 cm, height = 3 cm → $ 9 \times 3 = 27 $

But wait — the bottom rectangle extends under the top one? Let’s check:

- The top rectangle is 2 cm wide and 5 cm high.
- The bottom rectangle is 9 cm wide and 3 cm high.
- But the total height is 5 + 3 = 8 cm? No — look at the diagram.

Wait: The height of the top part is 5 cm, and the bottom part is 3 cm. The horizontal bar on the left is 4 cm wide.

Let’s re-analyze:

The shape has:
- A left vertical rectangle: width = 4 cm, height = 5 + 3 = 8 cm → $ 4 \times 8 = 32 $
- A right horizontal rectangle: width = 9 − 4 = 5 cm, height = 3 cm → $ 5 \times 3 = 15 $

Wait — no. Actually, the top part is only 2 cm wide, and the bottom is 9 cm wide.

Better approach:

Split into:
1. Top rectangle: width = 2 cm, height = 5 cm → $ 2 \times 5 = 10 $
2. Bottom rectangle: width = 9 cm, height = 3 cm → $ 9 \times 3 = 27 $

But now, the bottom rectangle goes under the top one?

No — actually, the top rectangle is on the right, and the bottom rectangle is below it, but the left side has a 4 cm extension.

Wait — let’s interpret the diagram carefully.

From the labels:
- The top small rectangle is 2 cm wide and 5 cm high.
- Below it, there is a longer base that is 9 cm wide and 3 cm high.
- The left side of the base has a 4 cm section extending upward.

So the shape consists of:
- Top rectangle: 2 cm × 5 cm → $ 10 $
- Left vertical rectangle: 4 cm wide, 3 cm high (but does it go up?) — no, only 3 cm high?
Wait — the total height is 5 + 3 = 8 cm?

Actually, the left side is 4 cm wide and extends from bottom to top?

Look: the left side is labeled 4 cm, and the top rectangle is 2 cm wide, so the bottom rectangle is 9 cm wide, and the top rectangle sits on top of the right side.

So better:

Break into:
1. Bottom rectangle: 9 cm × 3 cm = 27 cm²
2. Top rectangle: 2 cm × 5 cm = 10 cm²

But are they connected? Yes — the top rectangle sits on the right end of the bottom one.

But is there a gap? No — the bottom rectangle is 9 cm wide, and the top rectangle is 2 cm wide, sitting on the far right.

So total area = $ 27 + 10 = \boxed{37} \text{ cm}^2 $

Wait — but what about the left side being 4 cm? That might be a clue.

Wait — perhaps the left vertical part is 4 cm wide and 8 cm tall (5 + 3), and the top right is extra.

Let me redraw mentally:

- From the bottom, a rectangle 9 cm wide and 3 cm high.
- On top of that, only on the right, a rectangle 2 cm wide and 5 cm high.
- But the left side of the figure is 4 cm wide — so maybe the left part is 4 cm wide and 8 cm tall?

Yes! That makes sense.

So:
- Left vertical rectangle: 4 cm × 8 cm = 32 cm²
- Right horizontal rectangle: width = 9 − 4 = 5 cm, but only 3 cm high? Wait — no.

Wait — the top rectangle is 2 cm wide and 5 cm high — so it must be on the right, and the bottom rectangle is 9 cm wide and 3 cm high.

But the left side is 4 cm wide — so from bottom to top, the left column is 4 cm wide and 8 cm tall.

Then the top rectangle is 2 cm wide and 5 cm high — but where?

Wait — the top rectangle is 2 cm wide, and it sits on the right, but above the 3 cm base.

So the total height is 5 + 3 = 8 cm.

So:
- Left rectangle: 4 cm × 8 cm = 32 cm²
- Right rectangle: width = 9 − 4 = 5 cm, height = 3 cm (bottom) + 2 cm (top)? No — the top is only 2 cm wide.

Wait — the top rectangle is only 2 cm wide and 5 cm high — so it's on the right, but how wide is the base?

Let’s try another way.

From the diagram:

- Bottom: 9 cm wide, 3 cm high.
- On top of that, a rectangle 2 cm wide and 5 cm high — but it must be centered or on the right?

But the left side is labeled 4 cm — so likely, the left vertical bar is 4 cm wide and 8 cm tall.

Then the right part is 9 − 4 = 5 cm wide, but the top is only 2 cm wide — so the top rectangle is 2 cm wide and 5 cm high, sitting on the right edge of the bottom rectangle.

But the bottom rectangle is 9 cm wide — so the left part of the bottom rectangle is 4 cm wide (matching the left bar), and the right part is 5 cm wide.

But the top rectangle is only 2 cm wide — so it’s only on the far right.

So:
- Left rectangle: 4 cm × 8 cm = 32 cm²
- Right bottom rectangle: 5 cm × 3 cm = 15 cm²
- Right top rectangle: 2 cm × 5 cm = 10 cm²

But wait — the right bottom and right top are both on the right side.

But the top rectangle is only 2 cm wide — so it’s on the rightmost part.

But the bottom rectangle is 9 cm wide — so the left 4 cm is the left bar, and the right 5 cm is the bottom of the right part.

Then the top rectangle is 2 cm wide and 5 cm high — so it sits on top of the rightmost 2 cm of the bottom rectangle.

So total area:

- Left bar: $ 4 \times 8 = 32 $
- Right bottom: $ (9 - 4) = 5 \text{ cm wide}, 3 \text{ cm high} = 5 \times 3 = 15 $
- Right top: $ 2 \times 5 = 10 $

But the right top is on top of the right bottom, so it’s fine.

Wait — but the right bottom is 5 cm wide, and the top is only 2 cm wide — so it’s okay.

But the left bar is 4 cm wide and 8 cm tall — so it includes the full height.

So total area = $ 32 + 15 + 10 = 57 $? But that counts overlapping?

No — the left bar is separate from the right parts.

But the bottom rectangle is 9 cm wide — so it includes the left 4 cm and right 5 cm.

But the left bar is also 4 cm wide and 8 cm tall — so it overlaps with the bottom rectangle.

Ah! So we must avoid double-counting.

Better: Divide the shape into two non-overlapping rectangles.

Option:

1. Large bottom rectangle: 9 cm × 3 cm = 27 cm²
2. Top rectangle: 2 cm × 5 cm = 10 cm²

But is the left side 4 cm wide? Yes — so the left part is 4 cm wide and 8 cm tall — but the bottom rectangle already covers the bottom 3 cm of it.

So the left bar is 4 cm × 8 cm, but the bottom rectangle covers the bottom 3 cm of it.

So the top part of the left bar is 4 cm × 5 cm — but we don’t have that.

Wait — perhaps the left bar is 4 cm wide and 8 cm tall — but the top rectangle is 2 cm wide and 5 cm high — so it’s on the right.

So the left bar is 4 cm × 8 cm = 32 cm²

The right part is:
- Bottom: 9 − 4 = 5 cm wide, 3 cm high → 15 cm²
- Top: 2 cm wide, 5 cm high → 10 cm²

But the bottom rectangle is 9 cm wide — so the left 4 cm is shared with the left bar.

So if we do:
- Left bar: 4 × 8 = 32
- Right bottom: (9 − 4) × 3 = 5 × 3 = 15
- Right top: 2 × 5 = 10

But the right bottom is only 5 cm wide — and the right top is 2 cm wide — so total right area = 15 + 10 = 25

But the left bar is 32

Total = 32 + 25 = 57 cm²

But is there overlap? No — the left bar is 4 cm wide, the right bottom is 5 cm wide, and the right top is 2 cm wide — but the right top is on top of the right bottom, so it’s fine.

But the left bar and right bottom share the bottom 3 cm of the left bar — but since we're adding, we’re not double-counting because the left bar includes the full 8 cm, and the right bottom is only on the right side.

Wait — the left bar is 4 cm wide and 8 cm tall — so it occupies x=0 to 4 cm, y=0 to 8 cm.

The bottom rectangle is 9 cm wide and 3 cm high — so x=0 to 9, y=0 to 3.

So the overlap is x=0 to 4, y=0 to 3 — which is counted in both.

So we cannot add them directly.

Better approach: Use subtraction or careful decomposition.

Let’s define the whole shape as:

- Bottom rectangle: 9 cm × 3 cm = 27 cm²
- Top rectangle: 2 cm × 5 cm = 10 cm²

But is the top rectangle attached to the right end of the bottom rectangle?

Yes — and the left side is 4 cm wide — so the left part of the bottom rectangle is 4 cm wide, and the right part is 5 cm wide.

But the top rectangle is only 2 cm wide — so it sits on the rightmost 2 cm of the bottom rectangle.

So the top rectangle is 2 cm × 5 cm = 10 cm²

And the bottom rectangle is 9 cm × 3 cm = 27 cm²

Are they adjacent? Yes — the top rectangle sits on the bottom rectangle.

So total area = 27 + 10 = 37 cm²

But then why is the left side labeled 4 cm? Because the left bar is 4 cm wide — but that is part of the bottom rectangle.

So the left bar is just the left 4 cm of the bottom rectangle — but it extends up to the top? No — the top rectangle is only 2 cm wide, so the left bar is only 3 cm high.

Wait — the left side is labeled 4 cm — but that’s the width, not the height.

The left side is 4 cm wide — meaning the vertical thickness on the left is 4 cm.

But the top rectangle is only 2 cm wide — so it’s on the right, and the left bar is 4 cm wide and 3 cm high (bottom) plus some?

No — the total height is 5 + 3 = 8 cm.

So the left bar must be 4 cm wide and 8 cm tall — but then it would extend to the top.

But the top rectangle is only 2 cm wide — so it’s on the right, and the left bar is 4 cm wide and 8 cm tall.

Then the bottom rectangle is 9 cm wide and 3 cm high — but it includes the left bar's bottom 3 cm.

So the left bar is 4 cm × 8 cm = 32 cm²

The bottom rectangle is 9 cm × 3 cm = 27 cm² — but this includes the left bar's bottom 3 cm.

So if we add them, we double-count the 4×3 = 12 cm² overlap.

So total area = (32 + 27) - 12 = 47 cm²

But then we haven't added the top rectangle yet.

Wait — the top rectangle is 2 cm × 5 cm = 10 cm² — and it's on top of the right side.

But the bottom rectangle is only 3 cm high — so the top rectangle is on top of it.

So the top rectangle is additional.

But the top rectangle is not part of the left bar.

So total area = (left bar) + (bottom rectangle) + (top rectangle) - overlap between left bar and bottom rectangle.

Overlap = 4 cm × 3 cm = 12 cm²

So:
- Left bar: 4 × 8 = 32
- Bottom rectangle: 9 × 3 = 27
- Top rectangle: 2 × 5 = 10
- Overlap (left bar and bottom): 4 × 3 = 12

But the top rectangle is not overlapping with left bar — it's on the right.

So total area = 32 + 27 + 10 - 12 = 57 cm²

But is that correct?

Let’s think differently.

The shape has:
- A left vertical rectangle: 4 cm wide, 8 cm high → 32 cm²
- A right horizontal rectangle: 5 cm wide (9−4), 3 cm high → 15 cm²
- A top rectangle: 2 cm wide, 5 cm high → 10 cm²

But the top rectangle is on top of the right horizontal rectangle, so it's fine.

But the right horizontal rectangle is only 5 cm wide — and the top rectangle is 2 cm wide — so it fits.

But the left vertical rectangle and right horizontal rectangle overlap in the bottom-left corner? No — the left vertical is 4 cm wide, the right horizontal is 5 cm wide — so together they make 9 cm wide.

So no overlap.

So total area = 32 (left) + 15 (right bottom) + 10 (top) = 57 cm²

But the top rectangle is only 2 cm wide — so it’s on the right end.

So yes.

But earlier I thought it was 37 — but that was wrong.

Let’s verify with a different method.

Alternative: Divide into:
1. Bottom rectangle: 9 cm × 3 cm = 27 cm²
2. Top rectangle: 2 cm × 5 cm = 10 cm²
3. Left vertical extension: 4 cm wide, 5 cm high (from y=3 to y=8) → 4 × 5 = 20 cm²

But the left vertical extension is not on the bottom — it's on the left, from y=3 to y=8, width 4 cm.

So it's separate.

So total area = 27 (bottom) + 10 (top) + 20 (left extension) = 57 cm²

Yes — and the bottom rectangle includes the bottom of the left bar.

So the left bar is: bottom 3 cm (included in bottom rectangle) + top 5 cm (left extension) = 8 cm tall.

So total area = 57 cm²

✔ Answer for (d): \boxed{57} \text{ cm}^2

---

## ✔ (e)

```
24cm
┌─────────────────────┐
│ │
│ │ 20cm
│ │
│ ┌─────┐ ┌─────┐
│ │ │ │ │
│ │ 8cm │ │ 8cm │
│ │ │ │ │
│ └─────┘ └─────┘
│
│ 6cm 6cm 2cm
│
└─────────────────────┘
```

This is a rectangle with two rectangles removed from the bottom.

So:
- Big rectangle: 24 cm × 20 cm = 480 cm²
- Two small rectangles removed:
- Each is 6 cm wide, 8 cm high → area = 6 × 8 = 48 cm²
- Two of them: 2 × 48 = 96 cm²
- But there is a third cut? No — the last one is 2 cm wide, but no height given.

Wait — the diagram shows:
- Three cuts at the bottom: 6cm, 6cm, 2cm — but the total width is 6+6+2 = 14 cm — but the big rectangle is 24 cm wide.

So the cuts are:
- First cut: 6 cm wide, 8 cm high
- Second cut: 6 cm wide, 8 cm high
- Third cut: 2 cm wide, 8 cm high

But the total width of cuts is 6+6+2 = 14 cm — but the big rectangle is 24 cm wide — so the cuts are not covering the whole width.

But the cut depth is 8 cm — so each cut removes a rectangle of 8 cm height.

So:
- Cut 1: 6 × 8 = 48
- Cut 2: 6 × 8 = 48
- Cut 3: 2 × 8 = 16
- Total removed = 48 + 48 + 16 = 112 cm²

So area = 480 − 112 = \boxed{368} \text{ cm}^2

✔ Answer for (e): \boxed{368} \text{ cm}^2

---

## ✔ (f)

```
8cm
┌─────────────┐
│ │
│ │ 7cm
│ │
│ ┌─────┐ │
│ │ │ │
│ │ 3cm │ │
│ │ │ │
│ └─────┘ │
│ │
│ 2cm 3cm │
│ │
└─────────────┘
```

This is a rectangle with a rectangular hole in the middle.

- Big rectangle: 8 cm × 7 cm = 56 cm²
- Hole: 2 cm × 3 cm = 6 cm²
- Area = 56 − 6 = \boxed{50} \text{ cm}^2

✔ Answer for (f): \boxed{50} \text{ cm}^2

---

## ✔ Final Answers:

| Shape | Area |
|------|------|
| (a) | 62 cm² |
| (b) | 110 cm² |
| (c) | 67 cm² |
| (d) | 57 cm² |
| (e) | 368 cm² |
| (f) | 50 cm² |

Let me know if you'd like a visual breakdown or explanation for any specific one!