Math by First Principles

Heron's Formula Advanced

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Learning goals

  • Check whether three side lengths form a nondegenerate triangle
  • Calculate exact and rounded areas from the semiperimeter and three side lengths
  • Explain how factoring the squared area produces Heron's formula
  • Use side-based area to find a height or compare two triangles

Check that the sides can close

Start with three positive lengths in the same unit. A nondegenerate triangle encloses positive area. In a flat arrangement, all three vertices lie on one line. To form one, the two shorter lengths must add to more than the longest. This checks all three triangle inequalities at once. Each of the other sums already contains the longest side and adds a positive length to it.

Worked example 1 Decide which sets can enclose a region

Problem. Each case gives three proposed side lengths in centimeters. Decide which sets can form a nondegenerate triangle and explain why.

Case A. The proposed lengths are 9, 16, and 25.

Case B. The proposed lengths are 7, 13, and 22.

Case C. The proposed lengths are 10, 17, and 24.

Case D. The proposed lengths are 12, 12, and 0.

Solution.

Case A. The shorter lengths reach exactly across the longest side.

9+16=25.9+16=25.

The frame is flat.

Case B. The shorter lengths leave a gap even when stretched straight.

7+13<22.7+13<22.

These lengths cannot close.

Case C. The shorter lengths can bend at their meeting point and still reach both ends of the longest side.

10+17>24.10+17>24.

All three lengths are positive, so they form a nondegenerate triangle.

Case D. A zero length cannot be a side of a nondegenerate triangle.

Check your understanding

Three rods have lengths 12, 19, and 30 units. What shows that they can form a nondegenerate triangle?

Find area from the semiperimeter

The semiperimeter is half the perimeter. Write it as ss. Heron’s calculation starts with ss and the three amounts left when you subtract one side at a time.

Worked example 2 Get an exact whole-number area

Problem. A triangular plate has sides 10 cm, 17 cm, and 21 cm. Check that the lengths can form a triangle and use Heron’s calculation to find the exact area.

Solution. The sides are positive and the shorter two reach farther than 21 cm.

10+17>21.10+17>21.

Add the lengths and halve the total.

s=10+17+212.s=\frac{10+17+21}{2}.s=24 cm.s=24\text{ cm}.

Use a,b,ca,b,c for the sides in the order listed. Subtract each length from 24 cm.

s−a=14 cm,s−b=7 cm,s−c=3 cm.\begin{aligned} s-a&=14\text{ cm},\\ s-b&=7\text{ cm},\\ s-c&=3\text{ cm}. \end{aligned}

Multiply those three differences and the semiperimeter, then take the positive square root.

K=24⋅14⋅7⋅3 cm2,K=7056 cm2,K=84 cm2.\begin{aligned} K&=\sqrt{24\cdot14\cdot7\cdot3}\text{ cm}^2,\\ K&=\sqrt{7056}\text{ cm}^2,\\ K&=84\text{ cm}^2. \end{aligned}

For positive side lengths a,b,ca,b,c that satisfy the strict triangle inequalities, Heron’s formula gives the area KK.

s=a+b+c2.s=\frac{a+b+c}{2}. K=s(s−a)(s−b)(s−c).K=\sqrt{s(s-a)(s-b)(s-c)}.

The length check keeps the differences positive. For instance, the perimeter is more than twice cc because the other two sides together exceed cc. Half the perimeter therefore exceeds cc.

Check your understanding

A triangle has sides 11, 18, and 22 units. What is the factor s−22s-22?

Worked example 3 Keep the exact value until rounding

Problem. A triangular garden has fence lengths 9 m, 14 m, and 19 m. Give its exact area and its area rounded to the nearest tenth of a square meter.

Three measured sides fix the enclosed areaA dashed altitude drops from the corner where the two shorter fences meet to the line of the longest fence.14 m9 m19 mh
The three fence lengths are known. The dashed altitude to the 19 m fence is a perpendicular distance that has not been measured.

Solution. The positive lengths can close because the two shorter sides add to 23 m, more than 19 m. Half the total length is 21 m. Subtract the sides in order to get 12 m, 7 m, and 2 m.

K=21⋅12⋅7⋅2 m2.K=\sqrt{21\cdot12\cdot7\cdot2}\text{ m}^2.K=3528 m2.K=\sqrt{3528}\text{ m}^2.

Factor out the perfect square 42242^2 to give the exact area.

K=422 m2.K=42\sqrt{2}\text{ m}^2.

Round the area after evaluating the radical.

K≈59.4 m2.K\approx59.4\text{ m}^2.

Check your understanding

An exact area calculation gives 15515\sqrt{5} square units. Which report correctly gives the area to the nearest tenth?

Why four factors appear

Use the sine area formula Advanced and the Law of Cosines Advanced on the same triangle. Both use the angle between the same two sides. Squaring the area lets us replace the sine squared with an expression involving only the cosine. The cosine law then replaces the cosine with an expression involving only side lengths.

Use the same included angle in both formulasThe angle arc at C identifies the included angle of sides a and b. The dashed altitude from A meets a at a right angle.ABCabch
Angle C lies between sides a and b. If CB is the base, the height from A to the line CB is the perpendicular distance h.

First use side lengths 10, 13, and 15, with angle CC opposite the 15-unit side. The positive lengths can close because 23 exceeds 15. The sine formula gives the area.

K=12⋅10⋅13sin⁡C.K=\tfrac12\cdot10\cdot13\sin C.

The Law of Cosines gives the cosine without finding the angle.

260cos⁡C=102+132−152,260cos⁡C=44.\begin{aligned} 260\cos C&=10^2+13^2-15^2,\\ 260\cos C&=44. \end{aligned}

Square the area formula and multiply by 16. The identity relating sine squared and cosine squared lets us subtract the square of 260cos⁡C260\cos C from 2602260^2.

16K2=2602sin⁡2C,16K2=2602−(260cos⁡C)2,16K2=2602−442,16K2=(260−44)(260+44).\begin{aligned} 16K^2&=260^2\sin^2 C,\\ 16K^2&=260^2-(260\cos C)^2,\\ 16K^2&=260^2-44^2,\\ 16K^2&=(260-44)(260+44). \end{aligned}

Each factor hides another difference of squares.

260−44=152−(13−10)2,260−44=(15−3)(15+3),260+44=(13+10)2−152,260+44=(23−15)(23+15).\begin{aligned} 260-44&=15^2-(13-10)^2,\\ 260-44&=(15-3)(15+3),\\ 260+44&=(13+10)^2-15^2,\\ 260+44&=(23-15)(23+15). \end{aligned}

The four factors are 12, 18, 8, and 38. With semiperimeter 19, these are twice 6, 9, 4, and 19. The four twos contribute the factor of 16 that cancels the 16 multiplying K2K^2.

K2=6⋅9⋅4⋅19,K=6114.\begin{aligned} K^2&=6\cdot9\cdot4\cdot19,\\ K&=6\sqrt{114}. \end{aligned}

The calculation used the two formulas and the difference of squares identity. The side lengths can be replaced with any valid positive triple.

Derive Heron's formula from the two triangle formulas#

Let a,b,ca,b,c be the sides of a nondegenerate triangle. Angle CC is opposite side cc and lies between aa and bb. Let DD stand for the expression that the cosine law gives for 2abcos⁡C2ab\cos C.

2abcos⁡C=a2+b2−c2,D=a2+b2−c2.\begin{aligned} 2ab\cos C&=a^2+b^2-c^2,\\ D&=a^2+b^2-c^2. \end{aligned}

Start with the sine area formula. Square it, then multiply by 16.

K=12absin⁡C,16K2=(2ab)2sin⁡2C.\begin{aligned} K&=\tfrac12 ab\sin C,\\ 16K^2&=(2ab)^2\sin^2 C. \end{aligned}

On the unit semicircle, the squared coordinates sum to one. Use sin⁡2C=1−cos⁡2C\sin^2 C=1-\cos^2 C, then replace 2abcos⁡C2ab\cos C with DD.

16K2=(2ab)2(1−cos⁡2C),16K2=(2ab)2−(2abcos⁡C)2,16K2=(2ab)2−D2.\begin{aligned} 16K^2&=(2ab)^2(1-\cos^2 C),\\ 16K^2&=(2ab)^2-(2ab\cos C)^2,\\ 16K^2&=(2ab)^2-D^2. \end{aligned}

A difference of squares is the product of a difference and a sum.

16K2=(2ab−D)(2ab+D).16K^2=(2ab-D)(2ab+D).

Expand DD in the first factor. The terms 2ab−a2−b22ab-a^2-b^2 combine to give the negative of (a−b)2(a-b)^2. Factor the resulting difference of squares.

2ab−D=2ab−a2−b2+c2,2ab−D=c2−(a−b)2,2ab−D=(c−a+b)(c+a−b).\begin{aligned} 2ab-D&=2ab-a^2-b^2+c^2,\\ 2ab-D&=c^2-(a-b)^2,\\ 2ab-D&=(c-a+b)(c+a-b). \end{aligned}

Expand DD in the second factor. The terms 2ab+a2+b22ab+a^2+b^2 combine to give (a+b)2(a+b)^2. Factor again.

2ab+D=2ab+a2+b2−c2,2ab+D=(a+b)2−c2,2ab+D=(a+b−c)(a+b+c).\begin{aligned} 2ab+D&=2ab+a^2+b^2-c^2,\\ 2ab+D&=(a+b)^2-c^2,\\ 2ab+D&=(a+b-c)(a+b+c). \end{aligned}

Twice the semiperimeter is the full sum. To get twice a difference such as s−as-a, subtract two copies of that side from the full sum.

a+b+c=2s,c−a+b=2(s−a),c+a−b=2(s−b),a+b−c=2(s−c).\begin{aligned} a+b+c&=2s,\\ c-a+b&=2(s-a),\\ c+a-b&=2(s-b),\\ a+b-c&=2(s-c). \end{aligned}

Substitute these factors. Each contributes one factor of 2.

2ab−D=4(s−a)(s−b),2ab+D=4s(s−c).\begin{aligned} 2ab-D&=4(s-a)(s-b),\\ 2ab+D&=4s(s-c). \end{aligned}16K2=16s(s−a)(s−b)(s−c),K2=s(s−a)(s−b)(s−c).\begin{aligned} 16K^2&=16s(s-a)(s-b)(s-c),\\ K^2&=s(s-a)(s-b)(s-c). \end{aligned}

The triangle encloses positive area, so take the positive square root.

K=s(s−a)(s−b)(s−c).K=\sqrt{s(s-a)(s-b)(s-c)}.

Check your understanding

Set D=a2+b2−c2D=a^2+b^2-c^2. After factoring (2ab)2−D2(2ab)^2-D^2, what does the factor 2ab+D2ab+D become in terms of ss?

Use area to recover a height or compare regions

Once the sides give you the area, you can use that area with a chosen base to recover the matching height. You can also compare the areas of two triangles without measuring either height.

Worked example 4 Find the height to a chosen base

Problem. A triangular sign has sides 16 cm, 17 cm, and 17 cm. Find the perpendicular height to its 16 cm side.

Area and the chosen base determine a heightThe dashed height splits the isosceles triangle down the middle and meets its base at a right angle.17 cm17 cm16 cmh
The height h belongs to the 16 cm base. The two 17 cm sides are slanted.

Solution. The sides are positive, and 16 cm plus 17 cm is more than 17 cm. The semiperimeter is 25 cm, so the three differences are 9 cm, 8 cm, and 8 cm.

K=25⋅9⋅8⋅8 cm2,K=120 cm2.\begin{aligned} K&=\sqrt{25\cdot9\cdot8\cdot8}\text{ cm}^2,\\ K&=120\text{ cm}^2. \end{aligned}

Using 16 cm as the base and working in centimeters, half the base times the height must give 120 square centimeters.

120=12⋅16h,120=8h,h=15.\begin{aligned} 120&=\tfrac12\cdot16h,\\ 120&=8h,\\ h&=15. \end{aligned}

The height is 15 cm.

Worked example 5 Compare areas with the same perimeter

Problem. Two triangular stencils each have perimeter 40 cm. Which stencil has more area? Use an exact comparison.

Stencil A. The side lengths are 12 cm, 12 cm, and 16 cm.

Stencil B. The side lengths are 11 cm, 14 cm, and 15 cm.

Solution. All lengths are positive. In A, 24 exceeds 16. In B, 25 exceeds 15. Both semiperimeters are 20 cm. Compare the squares of their areas, with all lengths expressed in centimeters.

Stencil A. The three differences are 8, 8, and 4.

KA2=20⋅8⋅8⋅4,KA2=5120.\begin{aligned} K_A^2&=20\cdot8\cdot8\cdot4,\\ K_A^2&=5120. \end{aligned}

Stencil B. The three differences are 9, 6, and 5.

KB2=20⋅9⋅6⋅5,KB2=5400.\begin{aligned} K_B^2&=20\cdot9\cdot6\cdot5,\\ K_B^2&=5400. \end{aligned}

Both areas are positive, so the larger squared area belongs to the larger area. Stencil B has more area.

Check your understanding

Heron's formula gives a triangle an area of 154 square units. What is the perpendicular height to its 22-unit side?

Common mistakes

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A bit of history (optional)

In the first book of the Metrica, Heron of Alexandria gives a proof of the triangle area rule that now bears his name. He also gives a way to estimate square roots by hand. To estimate the square root of 720, he starts with 27, whose square is 729. He divides 720 by 27 and averages the quotient with 27. When he squares that estimate to see how close he came, the result exceeds 720 by just one thirty-sixth. MacTutor’s account of Heron’s Metrica

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