If you have ever watched someone design a playground, paint a wall, or lay grass on a soccer field, they all have to answer the same question: How much space does this shape cover? That question is all about area. Today you will see how builders, artists, and game designers use a clever trick: they break shapes apart or fit shapes together to find area quickly and accurately.
You will learn how to find the area of right triangles, any triangles, special quadrilaterals, and other polygons by turning them into rectangles or triangles. These skills help you solve real-world problems, like figuring out how much carpet to buy or how much land a garden covers. 🌱
Area tells us how much flat space a shape covers. We measure area in square units like square centimeters \((\textrm{cm}^2)\), square meters \((\textrm{m}^2)\), or square feet \(\textrm{ft}^2\).
Imagine covering a rectangle with little 1-by-1 squares. The number of these squares is the area. As shown in [Figure 1], if the rectangle is 6 units long and 4 units wide, you could count the squares one by one, or multiply.
The area of a rectangle with length \(l\) and width \(w\) is
\[A = l \times w\]
If the rectangle is a square with side length \(s\), then
\[A = s \times s = s^2\]
Example 1: Rectangle area
A small garden is 7 meters long and 3 meters wide. What is its area?
Step 1: Identify length and width.
\(l = 7 \, \textrm{m}, \quad w = 3 \, \textrm{m}\)
Step 2: Use the rectangle area formula.
\[A = l \times w = 7 \times 3 = 21\]
The area is \(21 \, \textrm{m}^2\).
The most powerful idea in this lesson is the strategy of composing and decomposing shapes. 💡
When we work with area, we usually decompose a complicated shape into rectangles and triangles, because we know how to find their areas. Then we add or subtract those areas.
This is like doing a puzzle: you can break a picture into easy pieces or combine pieces to see the whole thing. Either way, the total area stays the same.
A right triangle has a right angle (an angle of \(90^\circ\)). The two sides that form the right angle are called the legs. If we call one leg the base and the other leg the height, we can connect this triangle to a rectangle.
As shown in [Figure 2], if you take two identical right triangles and put them together, they form a rectangle.
Suppose a right triangle has base \(b\) and height \(h\). The rectangle made by two copies of this triangle has:
So the rectangle’s area is
\[A_{\textrm{rectangle}} = b \times h\]
But that rectangle is made of two equal triangles. So the area of one triangle is
\[A_{\textrm{triangle}} = \frac{1}{2} \times b \times h\]
This is our formula for the area of any right triangle when you know a base and its matching height.
Example 2: Right triangle area
A right triangle has a base of 8 centimeters and a height of 5 centimeters. Find its area.
Step 1: Identify base and height.
\(b = 8 \, \textrm{cm}, \quad h = 5 \, \textrm{cm}\)
Step 2: Use the triangle area formula.
\[A = \frac{1}{2} \times b \times h = \frac{1}{2} \times 8 \times 5\]
First multiply \(8 \times 5 = 40\).
\[A = \frac{1}{2} \times 40 = 20\]
The area is \(20 \, \textrm{cm}^2\).
You can also think: the full \(8 \times 5\) rectangle has area \(40\); the triangle is half, so \(20\). This matches what we saw in [Figure 2].
What if the triangle is not a right triangle? It might be slanted, with no right angle. We can still use the same formula!
As seen in [Figure 3], if you make a copy of any triangle and flip it, you can slide it next to the first one to make a parallelogram.
A parallelogram has base \(b\) and height \(h\), just like a rectangle, and its area is
\[A_{\textrm{parallelogram}} = b \times h\]
Because the parallelogram is made of two identical triangles, one triangle has area
\[A_{\textrm{triangle}} = \frac{1}{2} \times b \times h\]
This is the same formula we used for right triangles:
\[A = \frac{1}{2} b h\]
For any triangle, \(b\) is the length of any side you choose as the base, and \(h\) is the height measured straight (perpendicularly) from the opposite vertex to that base (or its extension).
Example 3: Area of a non-right triangle
A triangle has a base of 10 meters and a height of 6 meters. Find its area.
Step 1: Identify base and height.
\(b = 10 \, \textrm{m}, \quad h = 6 \, \textrm{m}\)
Step 2: Use the triangle area formula.
\[A = \frac{1}{2} \times b \times h = \frac{1}{2} \times 10 \times 6\]
Compute \(10 \times 6 = 60\).
\[A = \frac{1}{2} \times 60 = 30\]
The area is \(30 \, \textrm{m}^2\).
This matches the idea that two such triangles form a parallelogram with area \(60 \, \textrm{m}^2\), so each triangle is half.
Now let’s look at some four-sided shapes (quadrilaterals) and see how composing and decomposing helps us find their areas. ⭐
Parallelograms
A parallelogram has two pairs of opposite sides that are parallel. A rectangle is a special parallelogram with right angles, but most parallelograms are slanted.
You can cut off a right triangle from one side of a parallelogram and slide it to the other side to make a rectangle. This does not change the area.
So a parallelogram with base \(b\) and height \(h\) has area
\[A = b \times h\]
The height is the perpendicular distance between the two bases.
Rhombuses
A rhombus is a parallelogram with all four sides the same length. Since it is still a parallelogram, its area formula is also
\[A = b \times h\]
Sometimes you might cut a rhombus into two congruent triangles and use the triangle area idea too.
Trapezoids
A trapezoid has one pair of parallel sides. We call these sides the bases. Let the lengths of the two bases be \(b_1\) and \(b_2\), and let \(h\) be the height (the distance between the bases).
There are two main ways to think about a trapezoid using composing and decomposing:
If you put two identical trapezoids together, you get a parallelogram with base \(b_1 + b_2\) and height \(h\). The area of that parallelogram is
\[A_{\textrm{para}} = (b_1 + b_2) \times h\]
But this parallelogram is made from two trapezoids, so one trapezoid has area
\[A_{\textrm{trap}} = \frac{1}{2} \times (b_1 + b_2) \times h\]
Example 4: Trapezoid area using composing
A trapezoid has bases \(b_1 = 5\) centimeters and \(b_2 = 9\) centimeters, and a height of \(4\) centimeters. Find its area.
Step 1: Add the bases.
\[b_1 + b_2 = 5 + 9 = 14\]
Step 2: Multiply by height.
\[14 \times 4 = 56\]
Step 3: Take half.
\[A = \frac{1}{2} \times 56 = 28\]
The area is \(28 \, \textrm{cm}^2\).
A polygon is any closed shape made of straight sides. For example: pentagons, hexagons, and irregular shapes you might get as a floor plan of a room. Many polygons can be broken into rectangles and triangles so their areas are easy to find.
A common shape in real life is an L-shaped figure, like a room that has a corner taken out. You can draw a line to split it into two rectangles. Then you find each rectangle’s area and add them.
Example 5: L-shaped room area (composing rectangles)
An L-shaped room looks like a big rectangle with a smaller rectangle missing from one corner. The total width is 9 meters, and the total length is 7 meters. A small 3 meter by 2 meter section is cut out of one corner. Use decomposing to find the area of the room.
There are two main strategies.
Strategy A: Subtract the missing rectangle
Step 1: Area of the big outer rectangle.
\[A_{\textrm{big}} = 9 \times 7 = 63\]
Step 2: Area of the missing rectangle.
\[A_{\textrm{missing}} = 3 \times 2 = 6\]
Step 3: Subtract.
\[A_{\textrm{room}} = A_{\textrm{big}} - A_{\textrm{missing}} = 63 - 6 = 57\]
The area of the room is \(57 \, \textrm{m}^2\).
Strategy B: Split into two rectangles (as in [Figure 4])
You could also draw a line dividing the L into two rectangles, find each area, and add them. You would get the same total, \(57 \, \textrm{m}^2\).
Example 6: Polygon decomposed into triangles
Suppose you have a pentagon (5-sided polygon) drawn on grid paper. By drawing two diagonals from one vertex, you split it into 3 triangles with areas \(6 \, \textrm{cm}^2\), \(8 \, \textrm{cm}^2\), and \(5 \, \textrm{cm}^2\). The pentagon’s area is the sum:
\[A = 6 + 8 + 5 = 19\]
The area of the pentagon is \(19 \, \textrm{cm}^2\).
These composing and decomposing strategies show up all the time in real life. 🎉
1. Flooring and Carpeting
When workers install tiles or carpet in a room shaped like an L, they must figure out the area to know how much material to buy. They usually treat the room as two rectangles or as one big rectangle minus a smaller one, just like Example 5.
Word Problem 1: Carpet for a classroom
A classroom floor is shaped like an L. One part is a 6 meter by 4 meter rectangle. The other part is a 3 meter by 4 meter rectangle attached along the 4 meter side. How much area needs carpeting?
Step 1: Find the area of the first rectangle.
\[A_1 = 6 \times 4 = 24\]
Step 2: Find the area of the second rectangle.
\[A_2 = 3 \times 4 = 12\]
Step 3: Add them together.
\[A_{\textrm{total}} = A_1 + A_2 = 24 + 12 = 36\]
The classroom floor area is \(36 \, \textrm{m}^2\).
2. Gardens and Parks 🌳
Gardeners and park planners design spaces with triangles and quadrilaterals. They use area to know how many plants or how much grass they need.
Word Problem 2: Triangular flower bed
A flower bed is shaped like a triangle with base 5 meters and height 4 meters. Each square meter can hold 3 plants. How many plants can fit?
Step 1: Find the area of the triangle.
\[A = \frac{1}{2} \times 5 \times 4 = \frac{1}{2} \times 20 = 10\]
The area is \(10 \, \textrm{m}^2\).
Step 2: Multiply by 3 plants per square meter.
\[\textrm{Number of plants} = 10 \times 3 = 30\]
So \(30\) plants can fit in the flower bed.
3. Sports Fields and Courts
Sports fields often include rectangles and half-circles, but some smaller practice areas might be triangular. By breaking a field into rectangles and triangles, designers can calculate how much artificial turf or line paint they will need.
Word Problem 3: School playground design
A school playground includes a trapezoid-shaped sand area. The two bases of the trapezoid are 6 meters and 10 meters. The height is 5 meters. The rest of the playground is a 10 meter by 8 meter rectangle. Find the total area of the playground.
Step 1: Area of the trapezoid.
\[A_{\textrm{trap}} = \frac{1}{2} (b_1 + b_2) h = \frac{1}{2} (6 + 10) \times 5\]
\(6 + 10 = 16\)
\[A_{\textrm{trap}} = \frac{1}{2} \times 16 \times 5 = 8 \times 5 = 40\]
The trapezoid has area \(40 \, \textrm{m}^2\).
Step 2: Area of the rectangle.
\[A_{\textrm{rect}} = 10 \times 8 = 80\]
Step 3: Add to find the total playground area.
\[A_{\textrm{total}} = 40 + 80 = 120\]
The playground covers \(120 \, \textrm{m}^2\).
Here are the key ideas you learned today. 🤔
