Area of a triangle and Heron’s formula

A triangle with sides of 13, 14 and 15 m has an area of exactly 84 m², and you can reach that number by two quite different routes.

A plot you know only by its three sides

Here is the problem. A small triangular corner plot is to be sown with grass. Its three edges, measured with a tape, are 13 m, 14 m and 15 m. The seed packet says how many square metres one packet covers. How many square metres is the plot? You know the area of a triangle is half of base times height, but nobody measured a height, and walking a tape straight out from one edge to the far corner at exactly 90° is harder than it sounds. Try to find the area before reading on. One way uses only things from earlier units.

One route goes through the Pythagorean theorem. Take the 14 m side as the base and drop the height h from the opposite corner. It splits the base into two parts, say x and 14 − x, and makes two right triangles. From the left one, h² = 13² − x². From the right one, h² = 15² − (14 − x)². Set them equal: 169 − x² = 225 − 196 + 28x − x², so 28x = 140 and x = 5. Then h² = 169 − 25 = 144 and h = 12. The area is 14 × 12 ÷ 2 = 84 m². The two right triangles are 5, 12, 13 and 9, 12, 15, both with whole-number sides.

Why half of base times height

OpenStax Prealgebra starts from a rectangle with base b and height h, area b × h, and cuts it along a diagonal into two identical triangles, each with area half of bh. That settles right triangles. For any other triangle, draw the rectangle around it with the same base and height. The triangle takes up half of it, because the two pieces left over are copies of the two right triangles that make up the triangle itself. Either way, A = ½ × b × h. The height is measured at right angles to the chosen base, from the base line to the opposite corner, and every side can serve as the base if you use its own height.

For an obtuse triangle the height can fall outside the triangle, as on the right of the figure. That is not a mistake. Extend the base line, drop the perpendicular from the far corner to the extended line, and use that length as h. The formula still works, because the triangle is the difference between a big right triangle and a smaller one with the same height.

One consequence is worth knowing. Triangles with the same base and the same height have the same area, however slanted they are. Slide the top corner along a line parallel to the base and the area does not change. Euclid uses this idea inside his proof of the Pythagorean theorem, where a parallelogram is double a triangle on the same base between the same parallels.

Heron’s formula needs only the three sides

The second route needs no height at all. OpenStax Precalculus gives Heron’s formula for a triangle with sides a, b and c. First find s, half the perimeter: s = (a + b + c) ÷ 2. Then Area = √(s(s − a)(s − b)(s − c)). For 13, 14 and 15, s = 42 ÷ 2 = 21, and the area is √(21 × 8 × 7 × 6) = √7056 = 84. Same answer, no height. For the book’s own example, sides 10, 15 and 7, s = 16 and the area is √(16 × 6 × 1 × 9) = √864, about 29.4 square units.

A quick test of any formula is to feed it a case you already know. For the 3, 4, 5 right triangle, s = 6 and √(6 × 3 × 2 × 1) = √36 = 6, which is exactly 3 × 4 ÷ 2. For 5, 5, 8, s = 9 and √(9 × 4 × 4 × 1) = √144 = 12; the height to the side of 8 is 3, since 3² + 4² = 5², and 8 × 3 ÷ 2 = 12. Notice also what happens with sides that break the triangle inequality, such as 3, 4, 7: s = 7, one factor s − c is 0, and the area is 0. The formula agrees with unit 3.