Area of Triangle
Every question with a fully worked solution.
- Area of Triangle - Video - Area of a Triangle Watch
Theory
The area of a triangle is half the area of a rectangle with the same base and the same perpendicular height: \(A = \dfrac{1}{2} \times b \times h\), also written \(A = \dfrac{1}{2}bh\). This Year 7 maths page shows where the height goes, what to do with sloping sides and obtuse triangles, and how to find a missing base or height.
Cut a rectangle along its diagonal. You get two identical triangles, so each triangle is half of the rectangle: \(A = \dfrac{1}{2} \times b \times h\).
\(b\) is the base length. \(h\) is the perpendicular height: the distance from the opposite vertex (corner) to the base, meeting the base at a right angle (shown by a small square). The sloping sides are not used.
Any side can be the base. The height is then drawn to that side at a right angle. In an obtuse triangle the height can fall outside the triangle, down to the base line made longer. The extension is not part of the base.
Same base and same height give the same area, whatever the shape of the triangle.
Missing base or height: substitute, double both sides to remove the half, then divide. Area is in square units; change the base and height to the same unit first.
The formula:
How to find the area of a triangle
- Find the base and the perpendicular height (at a right angle to the base).
- Change them to the same unit.
- Work out \(\dfrac{1}{2} \times b \times h\) (halve whichever number is even).
- Write the answer in square units.
How to find a missing base or height
- Substitute the area and the length you know.
- Double both sides to remove the half.
- Divide by the length you know.
| \(\dfrac{1}{2} \times 3 \times 8\) | \(=\) | \(\dfrac{1}{2} \times 8 \times 3\) |
| \(\) | \(=\) | \(4 \times 3\) |
| \(\) | \(=\) | \(12\) |
\(12~\text{cm}^2\) (halve the even number first)
| \(15\) | \(=\) | \(\dfrac{1}{2} \times 3 \times h\) |
| \(15 \times 2\) | \(=\) | \(3h\) |
| \(h\) | \(=\) | \(30 \div 3 = 10\) |
\(h = 10\) cm
| \(\dfrac{1}{2} \times 16 \times 15\) | \(=\) | \(8 \times 15\) |
| \(\) | \(=\) | \(120\) |
Only the dashed \(15\) cm line meets the base at a right angle. The area is \(120~\text{cm}^2\).
| \(\dfrac{1}{2} \times 6 \times 9\) | \(=\) | \(3 \times 9\) |
| \(\) | \(=\) | \(27\) |
The \(4\) cm is the base line made longer, not part of the base. The area is \(27~\text{cm}^2\).
| \(250 \div 100\) | \(=\) | \(2.5\) |
| \(\dfrac{1}{2} \times 4 \times 2.5\) | \(=\) | \(5\) |
| \(5 \times 12\) | \(=\) | \(60\) |
Area \(5~\text{m}^2\); the turf costs \(\$60\).
(i) Find its area.
(ii) Use a \(25\) cm side as the base. Find the perpendicular height to it.
| \(\dfrac{1}{2} \times 30 \times 20\) | \(=\) | \(300\) |
| \(300\) | \(=\) | \(\dfrac{1}{2} \times 25 \times h\) |
| \(600\) | \(=\) | \(25h\) |
| \(h\) | \(=\) | \(600 \div 25 = 24\) |
(i) \(300~\text{cm}^2\)
(ii) Same triangle, same area: \(h = 24\) cm. Check: \(\dfrac{1}{2} \times 25 \times 24 = 300\).
| \(200 \div 100\) | \(=\) | \(2\) |
| \(3\) | \(=\) | \(\dfrac{1}{2} \times 2 \times h\) |
| \(h\) | \(=\) | \(3\) |
\(h = 3\) m \(= 300\) cm
Common pitfalls
Frequently asked questions
What is the formula for the area of a triangle?
\(A = \dfrac{1}{2} \times b \times h\): half of base times perpendicular height.
Why is the area of a triangle half of base times height?
Two identical triangles make a rectangle (or parallelogram) with that base and height.
What is the perpendicular height?
The distance from the top vertex to the base, meeting the base at a right angle.
Can the height be outside the triangle?
Yes, in an obtuse triangle. It meets the base line made longer.
How do you find the height from the area?
Double the area, then divide by the base. Area \(15\), base \(3\): \(h = 30 \div 3 = 10\).