Why is the area of a triangle base × height ÷ 2?

Because a triangle is exactly half its bounding rectangle. Box the triangle in a rectangle with the same base and height, then draw the diagonal: it cuts the rectangle into two matching triangles. The rectangle is base × height, so one triangle is half of that, base × height ÷ 2. A triangle with base 20 cm and perpendicular height 6.3 cm has a rectangle of 20 × 6.3 = 126 cm², so the triangle is 126 ÷ 2 = 63 cm². The answer 126 forgets the half and reports the whole rectangle.

Which height do I use for a triangle's area?

The perpendicular height: the straight distance from the base to the opposite point, measured at a right angle to the base. It is not a slanted side. If a triangle is drawn with base 12 cm, a marked perpendicular height of 5 cm, and a slanted side of 13 cm, the area is 12 × 5 ÷ 2 = 30 cm², using the perpendicular 5, not 12 × 13. Use the formula and what each number means, not just the two numbers printed on the diagram.

How do I find a triangle's area when the height is outside the triangle?

The same rule holds for an obtuse triangle whose perpendicular height falls outside the shape: area = base × perpendicular height ÷ 2. With base 10 cm and a perpendicular height of 6 cm measured to the line of the base, the area is 10 × 6 ÷ 2 = 30 cm². The position of the height does not change the formula or the halving.

GCSE Maths Foundation · AQA · Geometry

Triangle area: why it's base × height ÷ 2, not base × height

Finding a triangle’s area trips students who multiply base by height and stop, forgetting the $\tfrac{1}{2}$ . For a triangle with base 20 cm and perpendicular height 6.3 cm they write $20 \times 6.3 = 126$ , but that is the area of the bounding rectangle, not the triangle. A triangle is exactly half its rectangle: fold the rectangle along its diagonal and the two halves match, so the triangle is $126 \div 2 = 63$ cm².

The thirty-second fix: a triangle is half its bounding rectangle, so area = $\text{base} \times \text{perpendicular height} \div 2$ . Base 20, height 6.3 gives $20 \times 6.3 \div 2 = 63$ cm², never $20 \times 6.3 = 126$ .

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How to spot it in your own work

You worked out base × height and stopped, writing $20 \times 6.3 = 126$ instead of $20 \times 6.3 \div 2 = 63$ .
Your answer is the bounding rectangle, double the triangle: a sure sign the $\tfrac{1}{2}$ was dropped.
You multiplied a slanted side instead of the perpendicular height. The height in the formula is the perpendicular height, not whichever side is drawn.
A downstream answer (a cost, or a length back-solved from an area) came out twice what it should: the dropped half doubles everything after it.

An exam question that triggers it

Here is the canonical AQA Foundation trigger (Jun24 P1 Q18, non-calculator):

A triangle has base 20 cm and perpendicular height 6.3 cm.

Work out the area of the triangle.

The misconception answer is $20 \times 6.3 = 126$ cm²: base × height with no halving. But that 126 cm² is the area of the rectangle that boxes the triangle in.

A triangle is half its rectangle, so the area is $20 \times 6.3 \div 2 = 126 \div 2 = 63$ cm². The diagonal of the rectangle splits it into two matching triangles, and yours is one of them.

Why students fall for this

The formula $A = \tfrac{1}{2} b h$ is learned as a recipe, and the $\tfrac{1}{2}$ is the easiest ingredient to drop because $b \times h$ already feels like a finished area: it is, after all, exactly how you find a rectangle. With no picture in mind of why the half is there, the student multiplies the two numbers and reports the result.

The meaning fixes it: $b \times h$ is the area of the rectangle that just boxes the triangle in, and a triangle is exactly half that rectangle. Draw the diagonal of the rectangle and it cuts into two congruent triangles, so one triangle is $b \times h \div 2$ . The same half survives every kind of triangle: right-angled, with a dashed altitude for the height, or obtuse with the height falling outside the shape.

AQA Foundation papers exploit this directly (Jun24 P1 Q18, base 20 and height 6.3, non-calculator) and indirectly through back-solving (Jun23 P3 Q16 shape: find the length of a rectangle that has the same area as a triangle). When the triangle area is taken as $b \times h$ with no half, every downstream value, a turfing cost or a back-solved length, comes out doubled.

The fix: A triangle is half its bounding rectangle: area = base × perpendicular height ÷ 2

Box the triangle in a rectangle. The rectangle on the same base and perpendicular height has area $b \times h$ . For base 20 and height 6.3 that is $20 \times 6.3 = 126$ cm².

Halve it. The diagonal cuts the rectangle into two matching triangles, so the triangle is $126 \div 2 = 63$ cm², i.e. $20 \times 6.3 \div 2 = 63$ . If your answer equals the rectangle, you forgot the half.

Use the perpendicular height, not a slanted side. The height is the straight, right-angled distance from the base to the opposite point. With base 12, perpendicular height 5, and slant side 13, the area is $12 \times 5 \div 2 = 30$ cm²: the formula and its meaning, not just the two numbers shown.

Worked example

A triangle has base 12 cm and perpendicular height 8 cm. A rectangle has the same area as the triangle and a width of 6 cm. Find the length of the rectangle. This is the back-solve form of the trap: dropping the half doubles the final length.

Find the triangle’s area, then halve. The rectangle that boxes the triangle in is $12 \times 8 = 96$ cm², so the triangle is half:
$12 \times 8 \div 2 = 96 \div 2 = 48 \text{ cm}^2$
Use the equal area. The rectangle has the same area, 48 cm², and a width of 6 cm.
Back-solve the length.
$\text{length} = \text{area} \div \text{width} = 48 \div 6 = 8 \text{ cm}$
Spot the dropped-half trap. Taking the triangle area as $12 \times 8 = 96$ with no half gives a length of $96 \div 6 = 16$ cm, exactly double. The dropped $\tfrac{1}{2}$ doubles every value that follows.

The same halving settles the trigger question: base 20, perpendicular height 6.3, so the area is $20 \times 6.3 \div 2 = 63$ cm², not the rectangle’s 126 cm².

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Common questions

Why is the area of a triangle base × height ÷ 2?: Because a triangle is exactly half its bounding rectangle. Box the triangle in a rectangle with the same base and height, then draw the diagonal: it cuts the rectangle into two matching triangles. The rectangle is $b \times h$ , so one triangle is half of that, $b \times h \div 2$ . A triangle with base 20 cm and perpendicular height 6.3 cm has a rectangle of $20 \times 6.3 = 126$ cm², so the triangle is $126 \div 2 = 63$ cm². The answer 126 forgets the half and reports the whole rectangle.
Which height do I use for a triangle's area?: The perpendicular height: the straight distance from the base to the opposite point, measured at a right angle to the base. It is not a slanted side. If a triangle is drawn with base 12 cm, a marked perpendicular height of 5 cm, and a slanted side of 13 cm, the area is $12 \times 5 \div 2 = 30$ cm², using the perpendicular 5, not $12 \times 13$ . Use the formula and what each number means, not just the two numbers printed on the diagram.
How do I find a triangle's area when the height is outside the triangle?: The same rule holds for an obtuse triangle whose perpendicular height falls outside the shape: area = $\text{base} \times \text{perpendicular height} \div 2$ . With base 10 cm and a perpendicular height of 6 cm measured to the line of the base, the area is $10 \times 6 \div 2 = 30$ cm². The position of the height does not change the formula or the halving.

Related misconceptions

Perimeter vs area: computing the wrong measureThe neighbouring area trap: naming whether the question wants the boundary length or the space inside before you compute.
Area & volume don't scale like lengthDouble the lengths and the area goes up by 4 and the volume by 8, not by 2: measures in different dimensions scale differently.
Ratio: additive vs multiplicative scalingAnother recipe-without-meaning slip: scaling by multiplying, not adding, when a relationship must stay in proportion.

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