Composite Shapes (Composite Figures) - Definition, Formulae ... - Free Printable

Let's solve the problem step by step.

We are given a composite shape made up of two parts:

1. A rectangle on the right.
2. A right trapezoid (or possibly a right triangle and rectangle combined) on the left.

But let’s analyze the figure carefully.

Step 1: Understand the Shape

The figure appears to be a composite polygon consisting of:
- A rectangle on the right with dimensions:
- Height = 10 in
- Width = 5 in
- On the left, there is an irregular shape that seems to be a trapezoid, or possibly a triangle attached to a rectangle.

Looking closely:

- The total base is 15 inches.
- The right part has a width of 5 inches, so the left part must span 15 - 5 = 10 inches horizontally.
- The height of the right rectangle is 10 inches.
- There's a vertical segment of 6 inches going up from the left side of the rectangle, and then a horizontal segment of 7 inches to the left.
- This suggests the left part is a right trapezoid or a triangle and rectangle.

Wait — actually, let's look at the angles and dimensions.

We see:
- A rectangle on the right: 5 in wide, 10 in tall.
- To the left of it, there is a vertical segment of 6 inches going upward from the base.
- Then a horizontal segment of 7 inches going left.
- Then a slanted line connecting back to the base.
- The total base is 15 in.

So here's how we can break it down:

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Step 2: Break into Simpler Shapes

We can divide the entire shape into two parts:

1. A rectangle on the right:
- Width = 5 in
- Height = 10 in
→ Area = $ 5 \times 10 = 50 $ in²

2. A trapezoid (or a right trapezoid) on the left:
- The bottom base is 15 in total, and the right part takes 5 in, so the left base is 10 in? Wait — not quite.

Actually, the left portion has:
- A horizontal base of 15 in
- But the top edge is broken: it goes up 6 in, then 7 in left, then 5 in right?

Wait — perhaps better to re-express the shape.

Let me reconstruct the shape based on the labels:

- Bottom: a straight line of 15 inches.
- On the far right: a vertical line of 10 inches up.
- At the top right: a horizontal line of 5 inches to the left.
- Then a vertical drop of 6 inches down?
- Then a horizontal line of 7 inches to the left?
- Then a slanted line down to the base?

Wait — this doesn't make sense.

Let’s label the points.

Assume the shape has these features:

- Right side: vertical line of 10 in (height).
- Top: horizontal line of 5 in (width), so the top right corner is 5 in from the right edge.
- From the top-left of that 5-in segment, go down 6 in vertically, then left 7 in, then a slanted line back to the bottom-left.

But the bottom is 15 in long.

So here’s a better way:

Let’s suppose the shape is composed of:
- A rectangle on the right: 5 in wide × 10 in high.
- To the left of it, a trapezoid or triangle + rectangle.

But notice: from the bottom-left, the base is 15 in.
From the right end of the base, go up 10 in to the top-right corner.
Then go left 5 in (top of the rectangle).
Then go down 6 in? But the rectangle is 10 in high, so if you go down 6 in from the top, you’re at 4 in above the base?

Wait — but there’s a smaller vertical segment labeled 6 in, and a square angle, so likely:

- The rectangle is 5 in wide and 10 in high.
- Attached to its left side, at the bottom, there is a vertical segment of 6 in going up from the base.
- Then from the top of that 6-in segment, go left 7 in to meet the top of the rectangle.

Wait — no, that would be inconsistent.

Let’s try to interpret the diagram correctly.

Looking at the image:

- The bottom is 15 in.
- The right side is 10 in.
- The top has a segment of 5 in.
- Between the top of the 5-in segment and the left, there is a vertical drop of 6 in, and a horizontal segment of 7 in to the left.
- And a right angle at the junction.

Ah! Here's the correct interpretation:

The shape is composed of:

1. A rectangle on the right:
- Width = 5 in
- Height = 10 in
→ Area = $ 5 \times 10 = 50 $ in²

2. A right trapezoid (or a rectangle + right triangle) on the left.

But more accurately: the entire shape can be split into:

- A rectangle on the right: 5 in × 10 in
- A trapezoid on the left:
- It has a height of 6 in (vertical leg)
- One base is 7 in (the horizontal leg)
- But wait — the bottom base is 15 in total, and the right rectangle takes 5 in, so the left part must have a base of 10 in?

Wait — let’s do coordinate geometry.

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Step 3: Use Coordinates

Let’s assign coordinates.

Place the bottom-left corner at the origin: $ (0, 0) $

- Bottom base: 15 in → point $ (15, 0) $
- Right side: 10 in up → point $ (15, 10) $
- Top: 5 in left → point $ (10, 10) $
- From $ (10, 10) $, go down 6 in → point $ (10, 4) $
- From $ (10, 4) $, go left 7 in → point $ (3, 4) $
- From $ (3, 4) $, go down to $ (3, 0) $? But that would make a rectangle, but the bottom is 15 in.

Wait — but the bottom is from $ (0,0) $ to $ (15,0) $, and the slanted side goes from $ (3,4) $ to $ (0,0) $? Or to $ (15,0) $?

No — the bottom is straight from $ (0,0) $ to $ (15,0) $

Now, what connects $ (3,4) $ to $ (0,0) $? That’s a diagonal.

But we also have a horizontal segment of 7 in from $ (10,4) $ to $ (3,4) $

And from $ (3,4) $, a slanted line down to $ (0,0) $

And from $ (0,0) $ to $ (15,0) $, then up to $ (15,10) $, then left to $ (10,10) $, down to $ (10,4) $, left to $ (3,4) $, then down to $ (0,0) $? That can’t be because $ (3,4) $ to $ (0,0) $ is a line.

But wait — the bottom is 15 in, so from $ (0,0) $ to $ (15,0) $

So the shape is bounded by:

- $ (0,0) $ → $ (15,0) $ → $ (15,10) $ → $ (10,10) $ → $ (10,4) $ → $ (3,4) $ → $ (0,0) $

Wait — that skips from $ (3,4) $ to $ (0,0) $, but $ (0,0) $ is the start.

But then the base is 15 in, so from $ (0,0) $ to $ (15,0) $, but we have a point at $ (3,4) $, which is above.

So the left side of the shape is from $ (0,0) $ to $ (3,4) $? No — that would be a slant.

But the only slanted line is from $ (3,4) $ to $ (0,0) $, and from $ (0,0) $ to $ (15,0) $, then up.

Wait — but the bottom is straight from $ (0,0) $ to $ (15,0) $

So the shape has vertices:

1. $ A = (0, 0) $
2. $ B = (15, 0) $
3. $ C = (15, 10) $
4. $ D = (10, 10) $
5. $ E = (10, 4) $
6. $ F = (3, 4) $
7. Back to $ A = (0, 0) $

Wait — but from $ F = (3,4) $ to $ A = (0,0) $ is a diagonal.

But the bottom is from $ (0,0) $ to $ (15,0) $, so the shape must include the area under that.

But in this case, the shape is not convex, and the line from $ (3,4) $ to $ (0,0) $ is one side.

So the full shape has:

- Bottom: $ (0,0) $ to $ (15,0) $
- Right: $ (15,0) $ to $ (15,10) $
- Top: $ (15,10) $ to $ (10,10) $
- Down: $ (10,10) $ to $ (10,4) $
- Left: $ (10,4) $ to $ (3,4) $
- Slant: $ (3,4) $ to $ (0,0) $

Yes — that makes sense.

So the shape is a polygon with 6 sides.

To find its area, we can split it into simpler shapes.

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Step 4: Split the Shape

Split into two parts:

#### Part 1: Rectangle on the right

- From $ (10,4) $ to $ (15,4) $ to $ (15,10) $ to $ (10,10) $
- Width = 5 in
- Height = 6 in (from y=4 to y=10)
- Area = $ 5 \times 6 = 30 $ in²

Wait — but earlier I thought it was 10 in high.

But now we see that from $ (10,10) $ down to $ (10,4) $, so only 6 in down.

And from $ (10,4) $ to $ (3,4) $, horizontal.

So the upper rectangle is only 6 in high.

But then from $ (15,0) $ to $ (15,10) $, so the right side is 10 in.

So there's a lower rectangle below the upper one.

Wait — let's re-analyze.

From $ (15,0) $ to $ (15,10) $: 10 in up.

From $ (15,10) $ to $ (10,10) $: 5 in left.

From $ (10,10) $ to $ (10,4) $: 6 in down.

From $ (10,4) $ to $ (3,4) $: 7 in left.

From $ (3,4) $ to $ (0,0) $: slanted line.

From $ (0,0) $ to $ (15,0) $: bottom.

But the bottom is from $ (0,0) $ to $ (15,0) $, so the shape includes all between.

But the right side is from $ (15,0) $ to $ (15,10) $, and the top is from $ (15,10) $ to $ (10,10) $, then down to $ (10,4) $, then left to $ (3,4) $, then down to $ (0,0) $.

So the only missing part is from $ (0,0) $ to $ (3,4) $, which is a triangle?

No — the bottom is straight, so the area is bounded.

Better idea: split the shape into two parts:

1. A rectangle from $ x=10 $ to $ x=15 $, $ y=0 $ to $ y=10 $:
- Width = 5 in
- Height = 10 in
→ Area = $ 5 \times 10 = 50 $ in²

2. The remaining part: from $ x=0 $ to $ x=10 $, but only up to $ y=4 $, and then from $ x=3 $ to $ x=10 $, $ y=4 $ to $ y=10 $? No.

Wait — no.

Actually, the left part is bounded by:
- Bottom: $ y=0 $ from $ x=0 $ to $ x=15 $
- But the top is not flat.

Let’s use coordinate geometry.

Vertices in order (clockwise):

1. $ (0,0) $
2. $ (15,0) $
3. $ (15,10) $
4. $ (10,10) $
5. $ (10,4) $
6. $ (3,4) $
7. Back to $ (0,0) $

Wait — but from $ (3,4) $ to $ (0,0) $ is a line.

So the shape has 6 vertices:

- $ A(0,0) $
- $ B(15,0) $
- $ C(15,10) $
- $ D(10,10) $
- $ E(10,4) $
- $ F(3,4) $
- Back to $ A(0,0) $

Now, to compute the area, we can use the shoelace formula.

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Step 5: Shoelace Formula

List the coordinates in order:

| x | y |
|---|---|
| 0 | 0 |
| 15| 0 |
| 15| 10 |
| 10| 10 |
| 10| 4 |
| 3 | 4 |
| 0 | 0 |

Now apply shoelace:

Sum of $ x_i y_{i+1} $:

- $ 0 \cdot 0 = 0 $
- $ 15 \cdot 10 = 150 $
- $ 15 \cdot 10 = 150 $
- $ 10 \cdot 4 = 40 $
- $ 10 \cdot 4 = 40 $
- $ 3 \cdot 0 = 0 $
- $ 0 \cdot 0 = 0 $

Wait — better to list pairs:

Shoelace:

$$
\text{Area} = \frac{1}{2} \left| \sum x_i y_{i+1} - \sum y_i x_{i+1} \right|
$$

List points in order:

1. $ (0,0) $
2. $ (15,0) $
3. $ (15,10) $
4. $ (10,10) $
5. $ (10,4) $
6. $ (3,4) $
7. $ (0,0) $

Now compute:

Sum of $ x_i y_{i+1} $:

- $ 0 \cdot 0 = 0 $
- $ 15 \cdot 10 = 150 $
- $ 15 \cdot 10 = 150 $
- $ 10 \cdot 4 = 40 $
- $ 10 \cdot 4 = 40 $
- $ 3 \cdot 0 = 0 $
- $ 0 \cdot 0 = 0 $ → but last is back to first

Wait, better:

For each $ i $, $ x_i y_{i+1} $:

- $ x_1 y_2 = 0 \cdot 0 = 0 $
- $ x_2 y_3 = 15 \cdot 10 = 150 $
- $ x_3 y_4 = 15 \cdot 10 = 150 $
- $ x_4 y_5 = 10 \cdot 4 = 40 $
- $ x_5 y_6 = 10 \cdot 4 = 40 $
- $ x_6 y_7 = 3 \cdot 0 = 0 $
- $ x_7 y_1 = 0 \cdot 0 = 0 $

Sum A = $ 0 + 150 + 150 + 40 + 40 + 0 + 0 = 380 $

Sum of $ y_i x_{i+1} $:

- $ y_1 x_2 = 0 \cdot 15 = 0 $
- $ y_2 x_3 = 0 \cdot 15 = 0 $
- $ y_3 x_4 = 10 \cdot 10 = 100 $
- $ y_4 x_5 = 10 \cdot 10 = 100 $
- $ y_5 x_6 = 4 \cdot 3 = 12 $
- $ y_6 x_7 = 4 \cdot 0 = 0 $
- $ y_7 x_1 = 0 \cdot 0 = 0 $

Sum B = $ 0 + 0 + 100 + 100 + 12 + 0 + 0 = 212 $

Now,

$$
\text{Area} = \frac{1}{2} |380 - 212| = \frac{1}{2} \times 168 = 84 \text{ in}^2
$$

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Step 6: Verify with Geometry

Alternative method: split into parts.

Part 1: Rectangle on the right

- From $ x=10 $ to $ x=15 $, $ y=0 $ to $ y=10 $
- Width = 5 in, height = 10 in
- Area = $ 5 \times 10 = 50 $ in²

Part 2: Trapezoid on the left

- From $ x=3 $ to $ x=10 $, $ y=4 $ to $ y=10 $? No.

Wait — the left part is:

- From $ x=0 $ to $ x=10 $, but the top is only from $ x=3 $ to $ x=10 $ at $ y=4 $, and then from $ x=3 $ to $ x=10 $ at $ y=10 $, but only for $ x=10 $ to $ x=10 $?

Better:

Split into:

1. Rectangle from $ x=3 $ to $ x=10 $, $ y=4 $ to $ y=10 $:
- Width = 7 in
- Height = 6 in
→ Area = $ 7 \times 6 = 42 $ in²

2. Triangle from $ x=0 $ to $ x=3 $, $ y=0 $ to $ y=4 $, with hypotenuse from $ (0,0) $ to $ (3,4) $

But wait — the bottom is from $ (0,0) $ to $ (15,0) $, and the left side is from $ (0,0) $ to $ (3,4) $, but what about the region between $ x=0 $ to $ x=3 $, $ y=0 $ to $ y=4 $?

Yes — the shape includes that.

But is there a rectangle from $ x=0 $ to $ x=3 $, $ y=0 $ to $ y=4 $? Only if the top is flat.

But from $ (3,4) $ to $ (0,0) $ is a straight line, so the area under it is a triangle.

Wait — no: the shape is bounded by:

- From $ (0,0) $ to $ (15,0) $: bottom
- From $ (15,0) $ to $ (15,10) $: right
- From $ (15,10) $ to $ (10,10) $: top
- From $ (10,10) $ to $ (10,4) $: down
- From $ (10,4) $ to $ (3,4) $: left
- From $ (3,4) $ to $ (0,0) $: slanted

So the left part consists of:

- A rectangle from $ x=3 $ to $ x=10 $, $ y=4 $ to $ y=10 $: $ 7 \times 6 = 42 $
- A rectangle from $ x=0 $ to $ x=3 $, $ y=0 $ to $ y=4 $? But the top is not flat — it's slanted.

Wait — no: the only boundary from $ (0,0) $ to $ (3,4) $ is the slanted line, so the area is under that line.

But the bottom is $ y=0 $, so the region from $ x=0 $ to $ x=3 $, $ y=0 $ to the line $ y = \frac{4}{3}x $

But also, from $ x=3 $ to $ x=10 $, $ y=0 $ to $ y=4 $? No — because from $ (3,4) $ to $ (10,4) $ is horizontal, and from $ (10,4) $ to $ (10,10) $, etc.

Wait — actually, the entire lower part from $ x=0 $ to $ x=15 $, $ y=0 $ is included.

But the upper part is only from $ x=3 $ to $ x=10 $ at $ y=4 $, and from $ x=10 $ to $ x=15 $ at $ y=10 $, and from $ x=10 $ to $ x=10 $ at $ y=10 $ to $ y=4 $, etc.

So the shape can be split as:

1. Right rectangle: $ x=10 $ to $ x=15 $, $ y=0 $ to $ y=10 $ → $ 5 \times 10 = 50 $

2. Middle rectangle: $ x=3 $ to $ x=10 $, $ y=4 $ to $ y=10 $ → $ 7 \times 6 = 42 $

3. Lower left triangle: from $ (0,0) $ to $ (3,0) $ to $ (3,4) $ to $ (0,0) $? But $ (3,4) $ to $ (0,0) $ is diagonal.

Wait — no: the region from $ x=0 $ to $ x=3 $, $ y=0 $ to $ y=\frac{4}{3}x $ is a triangle.

But the bottom is from $ (0,0) $ to $ (15,0) $, and the top is from $ (0,0) $ to $ (3,4) $, so the area between $ x=0 $ to $ x=3 $, $ y=0 $ to the line $ y = \frac{4}{3}x $ is a triangle.

But wait — is that correct?

Actually, the shape includes:

- From $ (0,0) $ to $ (3,4) $: diagonal
- From $ (3,4) $ to $ (10,4) $: horizontal
- From $ (10,4) $ to $ (10,10) $: vertical
- From $ (10,10) $ to $ (15,10) $: horizontal
- From $ (15,10) $ to $ (15,0) $: vertical
- From $ (15,0) $ to $ (0,0) $: horizontal

So the only missing part is the area above the diagonal from $ (0,0) $ to $ (3,4) $, but since the bottom is straight, the shape is bounded below by $ y=0 $, and above by the piecewise function.

So the area is:

- From $ x=0 $ to $ x=3 $: bounded below by $ y=0 $, above by line from $ (0,0) $ to $ (3,4) $: $ y = \frac{4}{3}x $
- From $ x=3 $ to $ x=10 $: bounded below by $ y=0 $, above by $ y=4 $
- From $ x=10 $ to $ x=15 $: bounded below by $ y=0 $, above by $ y=10 $

But wait — that would give a larger area than before.

But in our shoelace, we got 84.

Let’s compute this way.

Area by Integration (or summation)

1. From $ x=0 $ to $ x=3 $:
Height = $ y = \frac{4}{3}x $
Area = $ \int_0^3 \frac{4}{3}x \, dx = \frac{4}{3} \cdot \frac{1}{2} x^2 \big|_0^3 = \frac{2}{3} \cdot 9 = 6 $ in²

2. From $ x=3 $ to $ x=10 $:
Height = 4 in
Area = $ 7 \times 4 = 28 $ in²

3. From $ x=10 $ to $ x=15 $:
Height = 10 in
Area = $ 5 \times 10 = 50 $ in²

Total area = $ 6 + 28 + 50 = 84 $ in²

Matches shoelace!

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Final Answer:

$$
\boxed{84} \text{ square inches}
$$

This is the area of the given composite shape.