Heron's Formula Calculator

This page provides some insight into Heron's Formula and allows you to find the area of triangle using Heron's Formula!
CalculationInputs﻿﻿a
:5m​
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﻿﻿b
:5m​
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﻿﻿c
:9m​
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Output﻿﻿s
:9.5m​
﻿
﻿﻿A
:9.808m2​
﻿
﻿
ExplanationHeron’s formula is credited to Heron of Alexandria (c. 62 CE) and used for finding the area of a triangle in terms of the lengths of its sides. In symbols, if a, b, and c are the lengths of the sides:
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 arc length calculation
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Then the area can be calculated by:
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A=(s(s−a)(s−b)(s−c)A = \sqrt{(s(s−a)(s−b)(s−c)}A=(s(s−a)(s−b)(s−c)​
Where:
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s=a+b+c2s = \frac{a+b+c}{2}s=2a+b+c​
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🕵️‍♀️ Heron of Alexandria
Heron of Alexandria, also called Hero (flourished c. AD 62, Alexandria, Egypt), was a Greek geometer and inventor whose writings preserved for posterity a knowledge of the mathematics and engineering of Babylonia, ancient Egypt, and the Greco-Roman world.
Heron’s most important geometric work, Metrica, was lost until 1896. It is a compendium, in three books, of geometric rules and formulas that Heron gathered from various sources, some of them going back to ancient Babylon, on areas and volumes of the plane and solid figures.
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17th-century German depiction of Heron
☝️ As well as his famous formula, Heron is credited with inventing the first vending machine. A set amount of holy water was dispensed when a coin was introduced via a slot on the top of the machine. This was included in his list of inventions in his book Mechanics and Optics. When the coin was deposited, it fell upon a pan attached to a lever. The lever opened up a valve which let some water flow out. The pan continued to tilt with the weight of the coin until it fell off, at which point a counterweight would snap the lever back up and turn off the valve. Cool!
✍️ Algebraic proof using the Pythagorean theorem
The following proof is very similar to the one given by Raifaizen. By the Pythagorean theorem, we have
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b2=h2+d2b^2 = h^2 + d^2b2=h2+d2
and
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a2=h2+(c−d)2a^2 = h^2 + (c-d)^2a2=h2+(c−d)2
According to the figure on the right, subtracting these yields 
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a2−b2=c2−2cda^2 - b^2 = c^2 -2cda2−b2=c2−2cd
This equation allows us to express d in terms of the sides of the triangle:
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d=−a2+b2+c22cd = \frac{-a^2+b^2+c^2}{2c}d=2c−a2+b2+c2​
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For the height of the triangle, we have: 
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h2=b2−d2h^2=b^2 - d^2h2=b2−d2
By replacing d with the formula given above and applying the difference of squares identity, we get
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h2=b2−(−a2+b2+c22c)2h^2 = b^2 -( \frac{-a^2+b^2+c^2}{2c})^2 h2=b2−(2c−a2+b2+c2​)2
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=(2bc−a2+b2+c2)(2bc+a2−b2−c2)4c2   = \frac{(2bc-a^2+b^2+c^2)(2bc+a^2-b^2-c^2)}{4c^2} =4c2(2bc−a2+b2+c2)(2bc+a2−b2−c2)​
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=((b+c)2−a2)(a2−(b−c)2)4c2   = \frac{((b+c)^2-a^2)(a^2-(b-c)^2)}{4c^2} =4c2((b+c)2−a2)(a2−(b−c)2)​
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=(b+c−a)(b+c+a)(a+b−c)(a−b+c)4c2   = \frac{(b+c-a)(b+c+a)(a+b-c)(a-b+c)}{4c^2} =4c2(b+c−a)(b+c+a)(a+b−c)(a−b+c)​
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=2(s−a) × 2s × 2(s−c) × 2(s−b)4c2   = \frac{2(s-a)\space ×\space 2s\space ×\space2(s-c)\space ×\space2(s-b)}{4c^2} =4c22(s−a) × 2s × 2(s−c) × 2(s−b)​
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=4s(s−a)(s−b)(s−c)c2   = \frac{4s(s-a)(s-b)(s-c)}{c^2} =c24s(s−a)(s−b)(s−c)​
We now apply this result to the formula that calculates the area of a triangle from its height:
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A=ch2  A = \frac{ch}{2} A=2ch​
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=c24.4s(s−a)(s−b)(s−c)c2= \sqrt{\frac{c^2}{4}.\frac{4s(s-a)(s-b)(s-c)}{c^2}}=4c2​.c24s(s−a)(s−b)(s−c)​​
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=s(s−a)(s−b)(s−c)= \sqrt{s(s-a)(s-b)(s-c)}=s(s−a)(s−b)(s−c)​
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