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A short class with Sarah and her students. The other ways are under More.
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We know the sides are 7, 8 and 9, but not the height. Which formula fits?
Heron's formula, because it only needs the three sides.
Step one is the semi-perimeter.
7 + 8 + 9 is 24, and half of that is 12.
Now the three brackets.
12 − 7 = 5, 12 − 8 = 4 and 12 − 9 = 3. So 12 × 5 × 4 × 3, which is 720.
And √720 is about 26.8. Exactly, it's 12√5, because 720 is 144 × 5.
Try the isosceles triangle with sides 5, 5 and 6.
s is 8, so 8 × 3 × 3 × 2 = 144, and the area is 12.
Can we check that another way?
The height splits the base 6 into 3 and 3, so the height is √(25 − 9) = 4. Half of 6 times 4 is 12.
Both methods agree. Now the plot of land with sides 20, 21 and 29 metres.
s is 35, so 35 × 15 × 14 × 6 = 44 100, and the area is 210 m². It's right-angled, so ½ × 20 × 21 also gives 210.
What happens with sides 2, 3 and 6?
s is 5.5, but s − 6 is −0.5. The product is negative, so we can't take the square root.
What is the geometry telling us?
2 + 3 is less than 6, so the short sides can't reach each other. There's no triangle.
Good. Now, how could Heron's formula help with a four-sided field?
Measure a diagonal, split the field into two triangles, find each area and add them.
Heron's formula finds a triangle's area from a, b and c without any height or angle.
s = (a + b + c)/2, then work out s − a, s − b and s − c.
the three brackets always add up to s.
simplify the surd for an exact answer (√720 = 12√5) and round only at the end.
a zero or negative bracket means the lengths fail the triangle inequality.
split a polygon into triangles with diagonals and add their Heron areas.
Sides 7, 8, 9: s = 12, area = √(12 × 5 × 4 × 3) = √720 = 12√5 ≈ 26.8 square units.
Sides 13, 14, 15: s = 21, area = √(21 × 8 × 7 × 6) = √7056 = 84 square units, so the height onto the side 14 is 2 × 84 ÷ 14 = 12.
Sides 20 m, 21 m, 29 m: s = 35, area = √(35 × 15 × 14 × 6) = √44 100 = 210 m², which matches ½ × 20 × 21 for this right-angled triangle.
Sides 2, 3, 6: s = 5.5 and s − 6 = −0.5, so the product is negative and these lengths cannot form a triangle.
Add the sides and cut in two, take each side from s, it's true; multiply the four and then, square root gives the area again.
Imagine Heron of Alexandria with only a rope to measure a triangular garden. He measured its three sides, halved the total, and by multiplying four numbers together and taking one square root, he knew how much seed to buy.
"Half, Subtract, Multiply, Root" - the four steps of Heron's formula, in order.
S.S.M.R. - Semi-perimeter, Subtract each side, Multiply all four, Root.
Term
What is the semi-perimeter of a triangle with sides 7, 8 and 9?
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Term
State Heron's formula.
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Term
What is the exact area of the 7, 8, 9 triangle?
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What is the area of a triangle with sides 13, 14 and 15?
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What does a negative value of s − c tell you?
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Area = √[s(s − a)(s − b)(s − c)], the area of a triangle from its three side lengths.
Half the perimeter of a triangle, s = (a + b + c)/2.
The total length around a shape; for a triangle, a + b + c.
In any triangle, the sum of any two sides is greater than the third side.
An answer written without rounding, such as 12√5 rather than 26.8.
A triangle with two equal sides.
A line segment joining two non-adjacent vertices of a polygon.
This lesson works through examples of Heron's formula, starting with the triangle of sides 7, 8 and 9: s = 12, so the area is √(12 × 5 × 4 × 3) = √720 = 12√5 ≈ 26.8 square units. It shows how to find a triangle's area from its three sides alone, check the answer and use it in real problems.
Heron's formula gives the area of a triangle from its side lengths a, b and c: first find the semi-perimeter s = (a + b + c)/2, then Area = √[s(s − a)(s − b)(s − c)]. For sides 7, 8 and 9, s = 12 and the area is √720 = 12√5, about 26.8 square units. This lesson sets out that calculation step by step, then works through a 13-14-15 triangle (area 84), an isosceles triangle checked against ½ × base × height, a right-angled plot of land, an equilateral triangle, decimal sides and a set of lengths that cannot form a triangle. It finishes by splitting a quadrilateral into two triangles and using Heron's formula twice.
The earlier lessons in this section gave three ways to find the area of a triangle: ½ × base × height, ½ab sin C, and Heron's formula. Heron's formula is the one to use when you know all three sides but no height or angle.
Heron's formula. For a triangle with sides a, b and c:
Example: sides 7, 8 and 9 units.
A useful check: the three brackets always add up to s, because 3s − (a + b + c) = 3s − 2s = s. Here 5 + 4 + 3 = 12.
More worked examples.
When the lengths do not make a triangle. For 2, 3 and 6, s = 5.5 and s − c = 5.5 − 6 = −0.5. The product under the root is negative, which is the formula's way of saying the triangle cannot exist: 2 + 3 is less than 6, so the two short sides cannot meet. Any bracket that is zero or negative means there is no triangle.
Areas of other shapes. A quadrilateral can be split along a diagonal into two triangles. If ABCD has AB = 9, BC = 40, CD = 28, DA = 15 and diagonal AC = 41, triangle ABC (9, 40, 41) has s = 45 and area √32 400 = 180, and triangle ACD (41, 28, 15) has s = 42 and area √15 876 = 126. The quadrilateral's area is 180 + 126 = 306 square units.
Accuracy tips. Keep s exact (it may be a decimal such as 5.5), work out each bracket before multiplying, and only round the final answer. Give exact surd answers such as 12√5 when asked for exact values.