Heron's Formula
Definition
Heron's formula is a formula that can be used to find the area of a triangle, when given its three side lengths. It can be applied to any shape of triangle, as long as we know its three side lengths. The formula is as follows:
The area of a triangle whose side lengths are \(a, b,\) and \(c\) is given by
\[A=\sqrt{s(s-a)(s-b)(s-c)},\]
where \(s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}\), semi-perimeter of the triangle.
Other useful forms are
\[\begin{align} A&=\frac 1 4\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}\\ \\ A&=\frac 1 4\sqrt{ \big[(a+b+c)(a+b-c) \big] \times \Big[\big(+(a-b)+c\big)\big(-(a-b)+c\big) \Big]}\\
A&=\frac 1 4\sqrt{\Big[(a+b)^2-c^2\Big] \times \ \Big[c^2-(a-b)^2\Big] }\\ \\
A&=\frac{1}{4}\sqrt{4a^2b^2-\big(a^2+b^2-c^2\big)^2}\\ A&=\frac{1}{4}\sqrt{2\left(a^2 b^2+a^2c^2+b^2c^2\right)-\left(a^4+b^4+c^4\right)} \\ A&=\frac{1}{4}\sqrt{\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)}. \end{align}\]
Although this seems to be a bit tricky (in fact, it is), it might come in handy when we have to find the area of a triangle, and we have no other information other than its three side lengths.
Proof of Heron's Formula
This formula follows from the area formula \(A=\frac{1}{2}ab\sin C\).
By the law of cosines, \(\cos C=\frac{a^2+b^2-c^2}{2ab}\).
Substituting into the Pythagorean identity \(\sin C=\sqrt{1-\cos^2 C}\) yields Heron's formula (after a series of algebraic manipulations). \(_\square\)
Examples
Find the area of the triangle below.
Imgur
Since the three side lengths are all equal to 6, the semiperimeter is \(s=\frac{6+6+6}{2}=9\). Therefore the area of the triangle is
\[A=\sqrt{9\times(9-6)\times(9-6)\times(9-6)}=9\sqrt{3}.\ _\square\]
Find the area of the triangle below.
Imgur
Since the three side lengths are 4, 5, and 7, the semiperimeter is \(s=\frac{4+5+7}{2}=8\). Therefore the area of the triangle is
\[A=\sqrt{8\times(8-4)\times(8-5)\times(8-7)}=4\sqrt{6}.\ _\square\]
What is the area of a triangle with side lengths 13, 14, and 15?
Since the three side lengths are 13, 14, and 15, the semiperimeter is \(s=\frac{13+14+15}{2}=21\). Therefore the area of the triangle is
\[A=\sqrt{21\times(21-13)\times(21-14)\times(21-15)}=84.\ _\square\]
Find the area of the triangle below.
Imgur
Since the three side lengths are 6, 8, and 10, the semiperimeter is \(s=\frac{6+8+10}{2}=12\). Therefore the area of the triangle is
\[A=\sqrt{12\times(12-6)\times(12-8)\times(12-10)}=24.\ _\square\]
Find the area of a triangle with side lengths \(4,13\) and \(15\).
We have \(a=4, b=13, c=15\) and \(s=\frac{4+13+15}{2}=16\). Hence,
\[A = \sqrt{16(16-4)(16-13)(16-15)} = 24. \ _\square\]
Find the area of the triangle outlined in black.
Image
We can use the Pythagorean theorem to find that the side lengths are \( 5, \sqrt{ 29}, 2 \sqrt{10} \).
If we used the direct form of \( A = \sqrt{ s (s-a)(s-b)(s-c) } \), we will quickly get into a huge mess because these lengths are not integers.Instead, we will use an alternate form of Heron's formula:
\[\begin{align} A & = \frac{1}{4}\sqrt{2\big(a^2 b^2+a^2c^2+b^2c^2\big)-\big(a^4+b^4+c^4\big)} \\ & = \frac{1}{4} \sqrt{ 2 ( 25 \times 29 + 25 \times 40 + 29 \times 40) - 25^2 - 29^2 - 40^2 } \\ & = \frac{1}{4} \sqrt{ 2704 } \\ & = 13. \ _\square \end{align}\]
Note: This triangle appears in Composite Figures, which is an easier approach.
Find the area of the triangle below.
Since the three side lengths are all equal to 6, the semiperimeter is \(s=\frac{6+6+6}{2}=9\). Therefore the area of the triangle is
\[A=\sqrt{9\times(9-6)\times(9-6)\times(9-6)}=9\sqrt{3}.\ _\square\]
Find the area of the triangle below.
Since the three side lengths are 4, 5, and 7, the semiperimeter is \(s=\frac{4+5+7}{2}=8\). Therefore the area of the triangle is
\[A=\sqrt{8\times(8-4)\times(8-5)\times(8-7)}=4\sqrt{6}.\ _\square\]
What is the area of a triangle with side lengths 13, 14, and 15?
Since the three side lengths are 13, 14, and 15, the semiperimeter is \(s=\frac{13+14+15}{2}=21\). Therefore the area of the triangle is
\[A=\sqrt{21\times(21-13)\times(21-14)\times(21-15)}=84.\ _\square\]
Find the area of the triangle below.
Since the three side lengths are 6, 8, and 10, the semiperimeter is \(s=\frac{6+8+10}{2}=12\). Therefore the area of the triangle is
\[A=\sqrt{12\times(12-6)\times(12-8)\times(12-10)}=24.\ _\square\]
Additional Problems
