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Area of a triangle

Measurement · level 4, lesson 2 of 7

A question to start

Here are three triangles. Each one has a base 3 cm long at the bottom, and each one has its top corner on the same dashed line, 4 cm above the base. One is neat and pointy, one has a right angle, and one leans so far over that its top is not even above its base.

ABC

Which triangle has the biggest area? The leaning one looks biggest, because its long side stretches so far. But all three have exactly the same area: 6 cm². This lesson shows why, and gives one rule that works for every triangle there is.

The idea

Two triangles make a parallelogram

Take any triangle and make an exact copy of it. Turn the copy a half turn (upside down) and fit it against the original, long side to long side. Together they make a parallelogram.

a triangle → base b h

Why is it a parallelogram? The copy was turned a half turn, so each of its sides runs in exactly the same direction as the matching side of the original, just the other way round. So the top side of the shape (from the copy) is parallel to the base and the same length, and the two slanted sides are parallel and the same length too. Opposite sides equal and parallel: that is a parallelogram.

The parallelogram has the same base \(b\) as the triangle and the same height \(h\). From the last lesson, its area is \(b \times h\). It is made of two identical triangles, so each triangle is exactly half:

\[\text{area of a triangle} = \tfrac12 \times \text{base} \times \text{height}, \qquad A = \tfrac12 b h\]

This works for every triangle: pointy, flat, right-angled or leaning. Any triangle and its turned copy always make a parallelogram.

The height of a triangle

The height of a triangle (also called the altitude) is the distance from the base to the opposite corner, measured at a right angle to the base. Imagine the triangle standing on its base: the height is how tall it is, measured straight up from the base line to the top corner.

  • In a right triangle, the two sides that make the right angle are a base and its height. That makes right triangles the easiest: area = half of one short side × the other short side. A right triangle is plainly half a rectangle, cut corner to corner, just like the half squares in level 3.
  • In a pointed (acute) triangle, the height falls inside the triangle.
  • In a triangle with an obtuse angle (bigger than a right angle) at the base, the height falls outside.

Obtuse triangles: the height outside

base b h big right triangle minus small right triangle = the green triangle

When the triangle leans past its base, the height drawn from the top corner meets the base line outside the triangle. Extend the base with a dashed line and measure the height to it. The rule is still \(\tfrac12 b h\).

You might doubt it, since the height is not even touching the triangle. Here is a second reason. Put the green triangle together with the small right triangle in the gap on the right (between the dashed lines). Together they make a big right triangle. In the drawing, call the extra stretch of base line \(x\). Then:

  • the big right triangle has base \(b + x\) and height \(h\), so area \(\tfrac12 (b + x) h\);
  • the small right triangle has base \(x\) and height \(h\), so area \(\tfrac12 x h\);
  • the green triangle is the difference: \(\tfrac12 (b + x) h - \tfrac12 x h = \tfrac12 b h\).

The \(x\) cancels out. However far the triangle leans, its area depends only on \(b\) and \(h\). That answers the opening question: the three triangles share a base length of 3 cm and a height of 4 cm (their tops are all on the line 4 cm up), so each has area \(\tfrac12 \times 3 \times 4 = 6\) cm².

Three bases, three heights

A triangle has three sides, and any of them can be the base. Each base has its own height: the distance from that side to the corner opposite it. The three products are different, but each gives the same area, because it is the same triangle.

For example, a right triangle with short sides 6 cm and 8 cm has long side 10 cm. Using the short sides, \(A = \tfrac12 \times 6 \times 8 = 24\) cm². Using the long side as the base, \(\tfrac12 \times 10 \times h = 24\), so \(5h = 24\) and the height onto the long side is \(h = 4.8\) cm.

A grid trick: the box method

When a triangle's corners sit on grid points but none of its sides lie along the grid, there is no easy height to measure. Instead, draw the smallest rectangle (a box) around it, along the grid lines. The box is the triangle plus some right triangles in the corners. Find the box's area and subtract the corner pieces.

Worked examples

1. The basic rule. A triangle has base 10 cm and height 6 cm. Area \(= \tfrac12 \times 10 \times 6 = 30\) cm².

2. A right triangle. A right triangle has short sides 5 m and 12 m, and long side 13 m. Its area?

The short sides are a base and its height: \(\tfrac12 \times 5 \times 12 = 30\) m². The 13 m side is not needed.

3. An obtuse triangle. A triangle has base 5 cm. Its top corner is 4 cm above the base line, past the end of the base. Area \(= \tfrac12 \times 5 \times 4 = 10\) cm², exactly as for any other triangle.

4. Missing height. A triangle has area 36 cm² and base 8 cm. How tall is it?

\(\tfrac12 \times 8 \times h = 36\), so \(4h = 36\) and \(h = 9\) cm. (Check: \(\tfrac12 \times 8 \times 9 = 36\). ✓)

5. The box method. A triangle on squared paper has corners at \((0, 0)\), \((6, 2)\) and \((2, 5)\), where \((6, 2)\) means 6 squares right and 2 squares up. Find its area.

The box around it runs from 0 to 6 across and 0 to 5 up: area \(6 \times 5 = 30\). The three corner triangles are all right triangles:

  • from \((0, 0)\) to \((6, 2)\): base 6, height 2, area \(\tfrac12 \times 6 \times 2 = 6\);
  • from \((6, 2)\) to \((2, 5)\): base 4, height 3, area \(\tfrac12 \times 4 \times 3 = 6\);
  • from \((2, 5)\) to \((0, 0)\): base 2, height 5, area \(\tfrac12 \times 2 \times 5 = 5\).

Triangle \(= 30 - 6 - 6 - 5 = 13\) squares.

6. A sail. A sail is a right triangle 3 m along the boom (the bottom) and 5 m up the mast. Area \(= \tfrac12 \times 3 \times 5 = 7.5\) m².

Try it

  1. Find the area: (a) base 8 cm, height 7 cm; (b) base 12 m, height 2.5 m; (c) a right triangle with short sides 9 cm and 4 cm.
  2. A triangle has sides 13 cm, 13 cm and 10 cm. The height onto the 10 cm side is 12 cm. Find its area.
  3. A triangle has area 45 m² and height 9 m. How long is its base?
  4. A triangle has base 6 cm and height 6 cm, and leans so far that its height falls 10 cm past the end of its base. Find its area.
  5. A right triangle has short sides 15 cm and 20 cm, and long side 25 cm. Find its area, then the height onto the long side.
  6. Use the box method: a triangle has corners at \((0, 0)\), \((4, 1)\) and \((1, 3)\) on squared paper. Find its area.
  7. A square with side 10 cm is cut along one diagonal. What is the area of each triangle? Why is this the same as \(\tfrac12 \times 10 \times 10\)?
  8. Two triangles have the same base. The first is twice as tall as the second. How do their areas compare?
Answers
  1. (a) \(\tfrac12 \times 8 \times 7 = 28\) cm². (b) \(\tfrac12 \times 12 \times 2.5 = 15\) m². (c) \(\tfrac12 \times 9 \times 4 = 18\) cm².
  2. \(\tfrac12 \times 10 \times 12 = 60\) cm². (The 13 cm sides are not used.)
  3. \(\tfrac12 \times b \times 9 = 45\), so \(4.5b = 45\) and \(b = 10\) m.
  4. \(\tfrac12 \times 6 \times 6 = 18\) cm². How far it leans does not matter.
  5. Area \(= \tfrac12 \times 15 \times 20 = 150\) cm². On the long side: \(\tfrac12 \times 25 \times h = 150\), so \(12.5h = 150\) and \(h = 12\) cm.
  6. Box: 4 across by 3 up, area 12. Corner triangles: \((0,0)\) to \((4,1)\): \(\tfrac12 \times 4 \times 1 = 2\); \((4,1)\) to \((1,3)\): \(\tfrac12 \times 3 \times 2 = 3\); \((1,3)\) to \((0,0)\): \(\tfrac12 \times 1 \times 3 = 1.5\). Triangle \(= 12 - 2 - 3 - 1.5 = 5.5\) squares.
  7. The square has area \(10 \times 10 = 100\) cm², and the diagonal cuts it into two identical triangles, so each is \(50\) cm². Each is a right triangle with short sides 10 and 10, so the rule gives \(\tfrac12 \times 10 \times 10 = 50\) cm² too. The rule is "half the rectangle".
  8. The taller one has twice the area: \(\tfrac12 \times b \times 2h = 2 \times (\tfrac12 \times b \times h)\).

Common mistake

Forgetting the half. Base × height is the area of the whole parallelogram (or rectangle). The triangle is only half of it. If your triangle's area equals the area of the box around it, something is wrong.

Using a slanted side as the height. The height must meet the base line at a right angle. In worked example 2 the height is 12 m, not the 13 m side. Only in a right triangle is one of the sides itself a height.

Where it leads