Area of a Triangle

Find the area of a triangle with half base times perpendicular height, find the height of an isosceles triangle first, and work back from an area to draw or find a side.

In the exam for Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024, Edexcel Further Pure Maths 4PM1, Edexcel IGCSE Maths 4MA1 · 37 real past-paper questions below

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Half a Rectangle

Draw a rectangle round a triangle, with the same base and height. The triangle fills exactly half of it.

Formula

A = \frac{1}{2} \times b \times h

$b$ is the base. $h$ is the perpendicular height: straight up from the base, at a right angle to it.

Worked example

1
$b = 10, h = 6$ — The base is 10 cm. The height is 6 cm, measured at a right angle to the base.
2
$10 \times 6 = 60$ — Base × height gives the rectangle drawn round the triangle: 60 cm².
3
$\frac{1}{2} \times 60 = 30 cm^{2}$ — The triangle is half of that rectangle, so halve 60. Area is in square units, cm².

Common mistake

Forgetting the half. 60 cm² is the whole rectangle, not the triangle.

Isosceles: Find the Height First

The 4-mark question: an isosceles triangle with only its three sides given. The height is not there, so find it.

The sides are 13 cm, 13 cm and 10 cm.

Worked example

1
$10 \div 2 = 5$ — The height runs from the top straight down to the middle of the base, so it cuts the 10 cm base into two 5s.
2
$h^{2} + 5^{2} = 1 3^{2}$ — Each half is a right-angled triangle. Its longest side is the 13 cm slope, so Pythagoras gives this line.
3
$h^{2} = 1 3^{2} - 5^{2}$ — Take $5^{2}$ from both sides, so $h^{2}$ is on its own.
4
$h^{2} = 169 - 25$ — Square each number: 13² = 169 and 5² = 25.
5
$h^{2} = 144$ — Subtract: 169 − 25 = 144.
6
$h = 12$ — Square root: √144 = 12. A length cannot be negative, so ignore −12. This is the height.
7
$A = \frac{1}{2} \times 10 \times 12$ — Now the formula. Use the whole base, 10, not the half of 5.
8
$A = 60 cm^{2}$ — Half of 10 is 5, and 5 × 12 = 60.

Common mistake

Using 13 as the height. 13 is a sloping side; the height goes straight up the middle.

Key idea

Write every line. Marks are given for halving the base, for the Pythagoras line and for the height, so a correct height alone already earns about half the marks.

Use the Right-Angle Height

The 3-mark question: a diagram shows three lengths. Use the base and the height at right angles to it. Give the units.

Worked example

1
$base 12, height 5, slant 7$ — The diagram gives three lengths. Look for the small square: it marks the one at a right angle to the base.
2
$\frac{1}{2} \times 12 \times 5$ — Use the base and the right-angle height. The 7 cm slant is left out.
3
$= 30 cm^{2}$ — Half of 12 is 6, and 6 × 5 = 30. Give the units, cm².

Common mistake

Using the slant: $\frac{1}{2} \times 12 \times 7 = 42$ is wrong. The slant side is longer than the height.

A right-angled triangle has sides 9 cm, 12 cm and 15 cm.

Worked example

1
$9, 12, 15$ — The longest side, 15 cm, is opposite the right angle. It is the slope, so leave it out.
2
$\frac{1}{2} \times 9 \times 12$ — The two shorter sides meet at the right angle, so one is the base and the other is the height.
3
$= 54 cm^{2}$ — 9 × 12 = 108, and half of 108 is 54.

Key idea

In a triangle with an angle over 90°, the height can fall outside the triangle. The next card shows one.

Triangles on a Grid

Grid questions draw the triangle on squares. Pick a side that runs along a grid line as the base, then count squares.

Find the area of the triangle with corners (1, 1), (1, 7) and (6, 9) on a 1 cm grid.

Worked example

1
$base = 7 - 1 = 6$ — (1, 1) and (1, 7) have the same x, so that side runs straight up a grid line. Count up: 6 squares.
2
$height = 6 - 1 = 5$ — The height goes straight across from the third corner, (6, 9), to the line x = 1. Across is at a right angle to the base: 5 squares.
3
$A = \frac{1}{2} \times 6 \times 5$ — Now the formula, with the counted base and height.
4
$A = 15 cm^{2}$ — 6 × 5 = 30, and half of 30 is 15. Each square is 1 cm², so the units are cm².

Key idea

The height meets the line x = 1 at (1, 9), above the top of the base and outside the triangle. That is fine: the height only has to be at a right angle to the base line.

Common mistake

Measuring the sloping side as the height. On a grid, only a straight-up or straight-across distance is at a right angle to a grid-line base.

Draw One, or Find a Missing Side

The 2-mark question: draw a right-angled triangle with a given area, or find a missing height. Double the area first.

Worked example

1
$A = 12$ — The question: draw a right-angled triangle with an area of 12 cm².
2
$b \times h = 2 \times 12 = 24$ — The formula halves base × height, so undo the half: base × height must be double the area.
3
$6 \times 4 or 8 \times 3$ — Pick any two numbers that multiply to 24. Draw the base along a grid line and the height straight up from one end.

Worked example

1
$A = 35, b = 10$ — This time the area and base are given. The height is the unknown.
2
$10 \times h = 2 \times 35 = 70$ — Double the area to undo the half: base × height = 70.
3
$h = 70 \div 10 = 7 cm$ — Divide both sides by the base, 10. Check: half of 10 × 7 is 35.

Common mistake

Drawing a 4 by 3 triangle for 12 cm². That makes 6 cm². The base times the height must be 24.

Sides Written With x

Some questions write the sides with x, and an earlier part finds x. Put x in first, then use the formula.

A right-angled triangle has sides (3x − 2) cm and (x + 4) cm next to the right angle. You found x = 5.

Worked example

1
$3 \times 5 - 2 = 13$ — Replace x with 5 in the first side: 3 × 5 = 15, then take 2. It is 13 cm.
2
$5 + 4 = 9$ — Replace x with 5 in the other side: 9 cm.
3
$A = \frac{1}{2} \times 13 \times 9$ — These two sides meet at the right angle, so they are the base and the height.
4
$A = 58.5 cm^{2}$ — 13 × 9 = 117, and half of 117 is 58.5.

Common mistake

Putting the longest side, the hypotenuse, into the formula. Only the two sides next to the right angle go in.

Key idea

Not sure your x is right? Use it anyway and finish. You still get the marks for the method with your own x.

Area Makes an Equation (Extended/Higher)

Harder questions give the area and write the sides with x. Put them into the formula and solve the equation.

A triangle has base (x + 3) cm, height x cm and area 20 cm². Find x.

Worked example

1
$\frac{1}{2} \times x \times (x + 3) = 20$ — Put the height, the base and the area into the formula.
2
$x (x + 3) = 40$ — Double both sides to undo the half: 2 × 20 = 40.
3
$x^{2} + 3 x = 40$ — Multiply out the bracket: x × x = x² and x × 3 = 3x.
4
$x^{2} + 3 x - 40 = 0$ — Take 40 from both sides, so the quadratic equals 0.
5
$(x + 8) (x - 5) = 0$ — Factorise: 8 × (−5) = −40 and 8 + (−5) = 3.
6
$x = - 8 or x = 5$ — One bracket must be 0: x + 8 = 0 or x − 5 = 0.
7
$x = 5$ — Reject −8: the height would be −8 cm, and a length cannot be negative.
8
$\frac{1}{2} \times 5 \times 8 = 20$ — Check: the height is 5, the base is 5 + 3 = 8, and half of 5 × 8 is 20.

Common mistake

Leaving both answers. Say which one you reject and why: a length cannot be negative.

Key idea

If the quadratic will not factorise, use the formula from The Quadratic Formula and Completing the Square, and give x to the accuracy asked.

Which Method? Read the Question

Check which lengths the diagram gives you.

You see	You do
Base and perpendicular height	use $\frac{1}{2} bh$
Isosceles, three sides only	half the base, Pythagoras, then area
A slant side as well	ignore it
Area given, find a side	double the area, divide
Triangle on a grid	base along a grid line, count the height straight across
Sides written with x	find x, put it in, then use $\frac{1}{2} bh$
Two sides and the angle between	Area of a Triangle Using Sine (Extended)
Four sides, one pair parallel	Area of a Parallelogram and Trapezium

Key idea

Always write the units: area is in square units, such as cm².

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