The area of a triangle
🎯 In this topic you will
- Derive the formula for the area of a triangle
- Use the formula to calculate the area of different triangles
🧠 Key Words
- perpendicular height
Show Definitions
- perpendicular height: The shortest distance between the base of a shape and its opposite vertex or side, forming a right angle with the base.
📐 Understanding the Area of a Triangle
The area of a triangle is always half of the area of the rectangle that surrounds it, as these diagrams show.
You find the area of a rectangle by multiplying the base by the height. So, the area of a triangle will be half of the base multiplied by the height.
You can use algebra to write the formula as: $\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height}$
Or simply: $A = \tfrac{1}{2}bh$
Note that the height measurement of the triangle must be the perpendicular height, from the base to the opposite vertex.
💡 Tip
The height must be at right angles (90°) to the base.
❓ EXERCISES
1. Copy and complete the workings to find the area of these triangles.
a.
$A=\tfrac12 bh=\tfrac12 \times 12 \times \square = \square\ \text{cm}^2$
b.
$A=\tfrac12 bh=\tfrac12 \times \square \times 7 = \square\ \text{m}^2$
👀 Show answer
b. $A=\tfrac12\times \mathbf{20}\times 7= \mathbf{70}\ \text{m}^2$
2. Work out the area of each triangle.
a.
b.
c.
d.
👀 Show answer
b. $A=\tfrac12\times 25\times 12=150\ \text{mm}^2$
c. $A=\tfrac12\times 16\times 11=88\ \text{m}^2$
d. $A=\tfrac12\times 12\times 9=54\ \text{cm}^2$
❓ EXERCISES
3. Arun, Marcus and Sofia are discussing the methods they use to find the area of a triangle.
Arun says: “I work out base times height, then divide the answer by $2$.”
Sofia says: “I work out half of the base, then times by the height.”
Marcus says: “I work out half of the height, then times by the base.”
Will they all get the same answer? Explain why.
👀 Show answer
Arun’s method: $(\text{base} \times \text{height}) \div 2 = \tfrac{1}{2} \times \text{base} \times \text{height}$. Sofia’s method: $\tfrac{1}{2} \times \text{base} \times \text{height}$. Marcus’s method: $\tfrac{1}{2} \times \text{height} \times \text{base} = \tfrac{1}{2} \times \text{base} \times \text{height}$.
All three expressions are equivalent, so they give the same result.
🧠 Think like a Mathematician
Task: Work individually to answer this question.
Look back at the methods that Arun, Marcus and Sofia used in Question 3. Whose method would it be best to use to find the areas of these triangles? Explain why.
Triangles:
- $\text{height} = 16 \text{ cm}$ and $\text{base} = 9 \text{ cm}$
- $\text{height} = 25 \text{ cm}$ and $\text{base} = 8 \text{ cm}$
- $\text{height} = 7 \text{ cm}$ and $\text{base} = 9 \text{ cm}$
- $\text{height} = 12 \text{ cm}$ and $\text{base} = 10 \text{ cm}$
👀 Show Answer
- a. Height $16$ and base $9$: Best to use Sofia’s method (half the base first) since $9$ is odd and halving the base is easier than halving the height.
- b. Height $25$ and base $8$: Best to use Marcus’s method (half the height first) because halving $25$ gives $12.5$, then multiply by $8$ easily.
- c. Height $7$ and base $9$: Either method works, but Sofia’s method is simpler because half of $9$ is $4.5$, then multiply by $7$.
- d. Height $12$ and base $10$: Any method works well, but Marcus’s method is simplest: half the height ($6$), then multiply by $10$.
Explanation: All methods are mathematically equivalent ($\tfrac{1}{2} \times \text{base} \times \text{height}$), but choosing which value to halve first can make the arithmetic simpler depending on the numbers.
❓ EXERCISES
5. Work out the area of each triangle, using the method shown. Look back at Question $3$ to check the method.
👀 Show answer
a. Base $12\ \text{cm}$, height $9\ \text{cm}$. Sofia halves the base, then multiplies by height:
$A=\tfrac{1}{2}\times b\times h=\tfrac{1}{2}\times 12\times 9=6\times 9=54\ \text{cm}^2$.
b. Base $14\ \text{cm}$, height $15\ \text{cm}$. Marcus halves the height, then multiplies by the base:
$A=\tfrac{1}{2}\times b\times h=14\times \tfrac{15}{2}=14\times 7.5=105\ \text{cm}^2$.
c. Base $7\ \text{cm}$, height $5\ \text{cm}$. Arun does base × height then divides by $2$:
$A=\tfrac{1}{2}\times 7\times 5=\tfrac{35}{2}=17.5\ \text{cm}^2$.
All three methods compute $A=\tfrac{1}{2}bh$, so they give the same results; only the order of operations differs.
❓ EXERCISES
6. This is part of Budi’s homework.
a. Explain the mistake that Budi has made.
b. Work out the correct answer.
👀 Show answer
a. Budi used $b=12\ \text{cm}$ with a slanted side of $9\ \text{cm}$. The height in $A=\tfrac{1}{2}bh$ must be the perpendicular height, not a slanted side. The perpendicular height shown is $8\ \text{cm}$.
b.$A=\tfrac{1}{2}\times b\times h=\tfrac{1}{2}\times 12\times 8=6\times 8=48\ \text{cm}^2$.
7. Work out the area of this triangle.
👀 Show answer
$A=\tfrac{1}{2}bh=\tfrac{1}{2}\times 40\ \text{mm}\times 20\ \text{mm}=20\times 20=400\ \text{mm}^2$.
8. Copy and complete the workings to find the area of this compound shape.
Area ① = $\dfrac{1}{2}bh$ = $\dfrac{1}{2}\times 10 \times \square$ = $\square\ \text{cm}^2$
Area ② = $bh$ = $10 \times \square$ = $\square\ \text{cm}^2$
Total area = $\square + \square$ = $\square\ \text{cm}^2$
👀 Show answer
The full height is $6\ \text{cm}$ and the lower rectangle height is $5\ \text{cm}$, so the triangle’s perpendicular height is $1\ \text{cm}$.
Area ①: $\tfrac{1}{2}\times 10\times 1=5\ \text{cm}^2$.
Area ②: $10\times 5=50\ \text{cm}^2$.
Total area: $5+50=55\ \text{cm}^2$.
9. Work out the areas of these compound shapes.
a.
b.
👀 Show answer
a.Not enough information provided. The diagram does not state all required perpendicular heights or parallel side lengths to determine a unique area without additional assumptions.
b.Not enough information provided. We need either the perpendicular height(s) of the triangular part or the lengths of the parallel sides to decompose and compute the area uniquely.
Method note: Once a shape is decomposed into rectangles/triangles with known perpendicular heights, use $A_{\text{rect}}=bh$ and $A_{\triangle}=\tfrac{1}{2}bh$, then add the parts.
❓ EXERCISES
10. The diagram shows a farmer’s field.
a. Work out the area of the field.
The farmer adds fertiliser to the field.
b. He uses $180$ kg of fertiliser per hectare. How many kilograms of fertiliser does he use for this field?
c. Fertiliser costs $80$ cents per kg. How much does it cost the farmer for the fertiliser for this field? Give your answer in dollars.
👀 Show answer
a. Treat the field as a trapezium with parallel sides $25\ \text{m}$ and $150\ \text{m}$, separated by $200\ \text{m}$:
$A=\tfrac{1}{2}(a+b)h=\tfrac{1}{2}(25+150)\times 200=\tfrac{1}{2}\times 175\times 200=17\,500\ \text{m}^2$$=1.75$ hectares.
b. Fertiliser needed: $1.75\times 180=315\ \text{kg}$.
c. Cost: $315\times \$0.80=\$252$.
Tip
shaded area = area of rectangle − area of triangle
11. Natasha works out that the shaded area in this diagram is $219\ \text{cm}^2$.
Is Natasha correct? Show your working.
👀 Show answer
Rectangle: $A_R=20\times 15=300\ \text{cm}^2$.
Triangle: $A_T=\tfrac{1}{2}\times 18\times 9=81\ \text{cm}^2$.
Shaded: $A=300-81=219\ \text{cm}^2$. ✔️ Natasha is correct.
12. In this diagram, the rectangle has a base length of $55\ \text{mm}$ and a height of $10\ \text{mm}$.
The triangle has a base length of $120\ \text{mm}$ and a height of $40\ \text{mm}$.
Work out the shaded area, in square centimetres.
👀 Show answer
Triangle: $A_T=\tfrac{1}{2}\times 120\times 40=2\,400\ \text{mm}^2$.
Rectangle: $A_R=55\times 10=550\ \text{mm}^2$.
Shaded (triangle − rectangle): $2\,400-550=1\,850\ \text{mm}^2=18.5\ \text{cm}^2$ (since $100\ \text{mm}^2=1\ \text{cm}^2$).
Tip
Try different values for the base length and height.
13. Marcus says:
Is Marcus correct? Explain your answer.
👀 Show answer
No. For a triangle $A=\tfrac{1}{2}bh$. Doubling both gives $A'=\tfrac{1}{2}(2b)(2h)=4\left(\tfrac{1}{2}bh\right)=4A$, so the area is quadrupled, not doubled. Example: $b=3,\ h=4\Rightarrow A=6$; doubling gives $b=6,\ h=8\Rightarrow A'=24$.