Area of an Irregular Polygon - Math Practice Worksheet

To find the area of each irregular polygon, we break them down into simpler shapes like rectangles, triangles, and trapezoids. We calculate the area of each part and then add them together.

Problem 1:
This shape can be split into three parts: a triangle on top, a rectangle in the middle, and another triangle on the bottom.
* Top Triangle: The base is the same as the width of the rectangle, which is $2.4\text{ m}$. The height is given as $1.2\text{ m}$.
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.4 \times 1.2 = 1.44\text{ m}^2$$
* Middle Rectangle: The width is $2.4\text{ m}$ and the height is $1.8\text{ m}$.
$$\text{Area} = \text{width} \times \text{height} = 2.4 \times 1.8 = 4.32\text{ m}^2$$
* Bottom Triangle: The base is $2.4\text{ m}$ and the height is $1.5\text{ m}$.
$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.4 \times 1.5 = 1.8\text{ m}^2$$
* Total Area: $1.44 + 4.32 + 1.8 = 7.56\text{ m}^2$

Problem 2:
This shape looks like a large rectangle with two triangles attached to the sides, or it can be seen as a central rectangle plus two side trapezoids/triangles. Let's split it vertically into a central rectangle and two side triangles.
Wait, looking at the dimensions:
The central vertical line divides it. Let's look at the horizontal segments.
The total width at the top is composed of a left part ($0.9\text{ m}$), a middle part ($1.4\text{ m}$), and a right part ($0.9\text{ m}$).
The height of the central rectangular section is $2.5\text{ m}$.
There are triangles on the left and right? No, let's look closer.
It looks like a central rectangle of width $1.4\text{ m}$ and height $2.5\text{ m}$.
On the left, there is a trapezoid or triangle? The label $0.9\text{ m}$ is the horizontal distance from the edge to the inner vertical line. The vertical side on the far left is not given directly, but the shape is symmetric.
Let's decompose it into:
1. Central Rectangle: Width $= 1.4\text{ m}$, Height $= 2.5\text{ m}$.
$$\text{Area} = 1.4 \times 2.5 = 3.5\text{ m}^2$$
2. Left Trapezoid/Triangle? Actually, the lines slant outwards. Let's assume the shape is composed of a central rectangle and two identical trapezoids on the sides? Or maybe two triangles on top/bottom?
Let's re-examine the diagram. It shows a hexagon-like shape.
Vertical center line height is $2.5\text{ m}$.
Horizontal top segment is $1.4\text{ m}$.
The side "wings" have a horizontal width of $0.9\text{ m}$.
The vertical height of the entire shape is not explicitly given as a single number, but the inner vertical line is $2.5\text{ m}$. The outer vertical edges seem shorter? No, usually in these problems, if not specified, we might assume the side shapes are triangles attached to the rectangle.
Let's look at the vertices. The top side is flat ($1.4\text{ m}$). The bottom side is flat ($1.4\text{ m}$).
The left side consists of a slanted line. The horizontal distance from the inner vertical line to the outermost point is $0.9\text{ m}$.
If we draw vertical lines down from the ends of the top $1.4\text{ m}$ segment, we get a central rectangle $1.4 \times 2.5$.
Then we have two side shapes. Are they rectangles? No, they are slanted.
Are they triangles? If the top and bottom widths are the same ($1.4\text{ m}$), and the shape bulges out, the side parts are likely trapezoids if the vertical sides are straight up and down. But here the sides are slanted.
Let's assume the standard decomposition: A central rectangle and two triangles on the left and right?
If the top and bottom are $1.4\text{ m}$ wide, and the "waist" is also $1.4\text{ m}$ wide (implied by the vertical lines), then the side parts are triangles with base $2.5\text{ m}$ (vertical) and height $0.9\text{ m}$ (horizontal)?
Let's check the orientation. The dimension $2.5\text{ m}$ is vertical. The dimension $0.9\text{ m}$ is horizontal.
So, we have a central rectangle of $1.4\text{ m}$ (width) $\times$ $2.5\text{ m}$ (height).
Plus two triangles on the sides. Each triangle has a vertical base of $2.5\text{ m}$ and a horizontal height of $0.9\text{ m}$.
Area of one side triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.5 \times 0.9 = 1.125\text{ m}^2$.
Two triangles $= 2 \times 1.125 = 2.25\text{ m}^2$.
Total Area $= \text{Rectangle} + \text{Triangles} = 3.5 + 2.25 = 5.75\text{ m}^2$.

Problem 3:
This shape is a kite or a quadrilateral divided into two triangles by a vertical diagonal.
* Top Triangle: Base $= 11\text{ cm}$ (the horizontal diagonal). Height $= 4.2\text{ cm}$.
$$\text{Area} = \frac{1}{2} \times 11 \times 4.2 = 23.1\text{ cm}^2$$
* Bottom Triangle: Base $= 11\text{ cm}$. Height $= 6.3\text{ cm}$.
$$\text{Area} = \frac{1}{2} \times 11 \times 6.3 = 34.65\text{ cm}^2$$
* Total Area: $23.1 + 34.65 = 57.75\text{ cm}^2$

Problem 4:
This shape is divided into a rectangle on the left and a trapezoid on the right? Or a rectangle and a triangle?
Let's look at the dimensions.
Left part is a rectangle with width $1.8\text{ m}$ and height $2.4\text{ m}$.
Right part is a trapezoid attached to the right side of the rectangle.
The vertical side shared with the rectangle is $2.4\text{ m}$.
The far right vertical side is $1.5\text{ m}$.
The horizontal width of this right part is $3.6\text{ m}$.
So, the right part is a trapezoid with parallel vertical sides of length $2.4\text{ m}$ and $1.5\text{ m}$, and a height (horizontal distance) of $3.6\text{ m}$.
* Area of Rectangle: $1.8 \times 2.4 = 4.32\text{ m}^2$.
* Area of Trapezoid: $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$
$$\text{Area} = \frac{1}{2} \times (2.4 + 1.5) \times 3.6$$
$$\text{Area} = \frac{1}{2} \times 3.9 \times 3.6$$
$$\text{Area} = 1.95 \times 3.6 = 7.02\text{ m}^2$$
* Total Area: $4.32 + 7.02 = 11.34\text{ m}^2$

Final Answer:
1) $7.56\text{ m}^2$
2) $5.75\text{ m}^2$
3) $57.75\text{ cm}^2$
4) $11.34\text{ m}^2$