Herons Formula for Triangular Area
by Christy Williams, Crystal Holcomb, and Kayla Gifford
Heron of Alexandria
n n n
Physicist, mathematician, and engineer Taught at the museum in Alexandria Interests were more practical (mechanics, engineering, measurement) than theoretical He is placed somewhere around 75 A.D. (150)
Herons Works
n n n n n
Automata Mechanica Dioptra Metrica Pneumatica
n n n n n n
Catoptrica Belopoecia Geometrica Stereometrica Mensurae Cheirobalistra
The Aeolipile
Herons Aeolipile was the first recorded steam engine. It was taken as being a toy but could have possibly caused an industrial revolution 2000 years before the original.
Metrica
n n
Mathematicians knew of its existence for years but no traces of it existed In 1894 mathematical historian Paul Tannery found a fragment of it in a 13th century Parisian manuscript In 1896 R. Schne found the complete manuscript in Constantinople. Proposition I.8 of Metrica gives the proof of his formula for the area of a triangle
How is Herons formula helpful?
How would you find the area of the given triangle using the most common area formula?
A= 1 2 bh
Since no height is given, it becomes quite difficult
25
17
26
Herons Formula
Herons formula allows us to find the area of a triangle when only the lengths of the three sides are given. His formula states:
K = s (s a )(s b )(s c )
Where a, b, and c, are the lengths of the sides and s is the semiperimeter of the triangle.
The Preliminaries
Proposition 1
Proposition IV.4 of Euclids Elements. The bisectors of the angles of a triangle meet at a point that is the center of the triangles inscribed circle. (Note: this is called the incenter)
Proposition 2
Proposition VI.8 of Euclids Elements. In a right-angled triangle, if a perpendicular is drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle and to one another.
Proposition 3
In a right triangle, the midpoint of the hypotenuse is equidistant from the three vertices.
Proposition 4
If AHBO is a quadrilateral with diagonals AB and OH, then if HOB and HAB are right angles (as shown), then a circle can be drawn passing through the vertices A , O, B, and H.
Proposition 5
Proposition III.22 of Euclids Elements. The opposite angles of a cyclic quadrilateral sum to two right angles.
Semiperimeter
The semiperimeter, s, of a triangle with sides a, b, and c, is
s=
a +b +c 2
Herons Proof
Herons Proof
n n
The proof for this theorem is broken into three parts. Part A inscribes a circle within a triangle to get a relationship between the triangles area and semiperimeter. Part B uses the same circle inscribed within a triangle in Part A to find the terms s-a, s-b, and s-c in the diagram. Part C uses the same diagram with a quadrilateral and the results from Parts A and B to prove Herons theorem.
Restatement of Herons Formula
For a triangle having sides of length a, b, and c and area K, we have
K = s (s a )(s b )(s c )
where s is the triangles semiperimeter.
Herons Proof: Part A
Let ABC be an arbitrary triangle such that side AB is at least as long as the other two sides. Inscribe a circle with center O and radius r inside of the triangle. Therefore, OD = OE = OF .
Herons Proof: Part A (cont.)
Now, the area for the three triangles ? AOB, ? BOC, and ? COA is found using the formula (base)(height). Area ? AOB = 1 2 ( AB )(OD ) = 1 2 cr
1 1 Area ? BOC = 2 ( BC )(OE) = 2 ar
Area ? COA = 12 ( AC)(OF ) = 1 2 br
Herons Proof: Part A (cont.)
We know the area of triangle ABC is K. Therefore
K = Area(ABC ) = Area(AOB ) + Area(BOC ) + Area(COA)
If the areas calculated for the triangles ? AOB, ? BOC, and ? COA found in the previous slides are substituted into this equation, then K is
K =
1 2
cr +
ar +
a + b+ c br = r = rs 2
10
Herons Proof: Part B
When inscribing the circle inside the triangle ABC, three pairs of congruent triangles are formed (by Euclids Prop. I.26 AAS).
AOD AOF BOD BOE COE COF
Herons Proof: Part B (cont.)
n
Using corresponding parts of similar triangles, the following relationships were found:
AD = AF BD = BE CE = CF
AOD= AOF BOD= BOE COE = COF
11
Herons Proof: Part B (cont.)
n
The base of the triangle was extended to point G where AG = CE. Therefore, using construction and congruence of a triangle:
1 1 1 1 1
BG = BD + AD + AG = BD + AD + CE
BG = BG = BG = BG = BG =
2 2 2 2 2
(2BD + 2 AD + 2CE ) [(BD + BE )+ (AD + AF )+ (CE + CF )] [(BD + AD )+ (BE + CE )+ (AF + CF )] (AB + BC + AC )
(c + a + b ) = s
Herons Proof: Part B (cont.)
n
Since BG = s , the semi-perimeter of the triangle is the long segment straighten out. Now, s-c, s-b, and s-a can be found.
s c = BG AB = AG
Since AD = AF and AG = CE = CF,
s b = BG AC = BD + AD + AG AF + CF
) ( = (BD + AD + CE ) (AD + CE )
= BD
12
Herons Proof: Part B (cont.)
Since BD = BF and AG = CE,
s a = BG BC = BD + AD + AG BE + CE
) ( = (BD + AD + CE ) (BD + CE )
= AD
Herons Proof: Part B (cont.)
n
In Summary, the important things found from this section of the proof.
BG
(c
+ a + b
)=
s c = AG s b = BD s a = AD
13
Herons Proof: Part C
n
n n
The same circle inscribed within a triangle is used except three lines are now extended from the diagram. The segment OL is drawn perpendicular to OB and cuts AB at point K. The segment AM is drawn from point A perpendicular to AB and intersects OL at point H. The last segment drawn is BH. The quadrilateral AHBO is formed.
Herons Proof: Part C (cont.)
n
Proposition 4 says the quadrilateral AHBO is cyclic while Proposition 5 by Euclid says the sum of its opposite angles equals two right angles.
AHB + AOB = 2 right angles
14
Herons Proof: Part C (cont.)
n
By congruence, the angles around the center O reduce to three pairs of equal angles to give:
2 + 2 + 2 = 4 rt angles
Therefore,
+ + = 2 rt angles
Herons Proof: Part C (cont.)
n
Since
+ = AOB , and
+ + = 2 rt angles + AOB = 2 rt angles = AHB + AOB
Therefore, = AHB .
15
Herons Proof: Part C (cont.)
n
Since = AHB and both angles CFO and BAH are right angles, then the two triangles ? COF and ? BHA are similar. This leads to the following proportion using from Part B that AG = CF and OH = r :
AB CF AG = = r AH OF
which is equivalent to the proportion
AB AH = r AG
(*)
Herons Proof: Part C (cont.)
n
Since both angles KAH and KDO are right angles and vertical angles AKH and DKO are equal, the two triangles ? KAH and ? KDO are similar. This leads to the proportion:
AH OD r = = AK KD KD
Which simplifies to
AH AK = r KD
(**)
16
Herons Proof: Part C (cont.)
n
The two equations
AB AG = AH (*) and AH AK (**) = r r KD
are combined to form the key equation:
AB AK = AG KD
(***)
Herons Proof: Part C (cont.)
n
By Proposition 2, ? KDO is similar to ? ODB where ? BOK has altitude OD=r. This gives the equation:
(****) (r is the mean proportional between magnitudes KD and BD)
2
(KD)(BD) = r
KD r = r BD which simplifies to
17
Herons Proof: Part C (cont.)
n
One is added to equation (***), the equation is simplified, then BG/BG is multiplied on the right and BD/BD is multiplied on the left, then simplified.
AB AK = AG KD
BG BG AD BD BG AG = KD BD
AB AK +1= +1 AG KD
Using the equation (KD)(BD) = r 2 (****) this simplifies to:
AB+ AG AK+ KD = AG KD
BG AD = AG KD
(BG) = (AD )(BD) (AG)(BG ) r
2 2
Herons Proof: Part C (cont.)
n
Cross-multiplication of
2
(BG ) (AD )(BD ) (AG )(BG ) = r
2 2
produced
r 2 BG = AG BG AD BD . Next, the results from
Part B are needed. These are:
( ) ( )( )( )( )
= s
BG
s b = BD s a = AD
s c = AG
18
Herons Proof: Part C (cont.)
n
The results from Part B are substituted into the equation:
r 2 BG = AG BG AD BD
( ) ( )( )( )( )
2
r 2 s 2 = (s c )(s )(s b)(s c )
n
We know remember from Part A that K=rs, so the equation becomes:
K = s(s a )(s b )(s c )
n
Thus proving Herons Theorem of Triangular Area
Application of Herons Formula
We can now use Herons Formula to find the area of the previously given triangle
25 17
s=
1 2
(17 + 25 + 26) = 34
26
K = 34(34 17 )(34 25 )(34 26 ) = 41616 = 204
19
Eulers Proof of Herons Formula
Leonhard Euler provided a proof of Herons Formula in a 1748 paper entitled Variae demonstrationes geometriae His proof is as follows
Eulers Proof (Picture)
For reference, this is a picture of the proof by Euler.
20
Eulers Proof (cont.)
Begin with ABC having sides a, b, and c and angles , and Inscribe a circle within the triangle Let O be the center of the inscribed circle with radius r = OS = OU From the construction of the incenter, we know that segments OA , OB, and OC bisect the OBA = angles of ABC with OAB = 2, 2 , and OCA = 2
Eulers Proof (cont.)
Extend BO and construct a perpendicular from A intersecting this extended line at V Denote by N the intersection of the extensions of segment AV and radius OS Because AOV is an exterior angle of AOB , observe that
AOV = OAB + OBA = 2 + 2
21
Eulers Proof (cont.)
Because AOV is right, we know that AOV and OAV are complementary Thus, But
2 2
+ + OAV = 90 2
2
+ + 2 = 90 as well 2
= OCU
Therefore,OAV =
Eulers Proof (cont.)
Right triangles OAV and OCU are similar so we get AV / VO = CU / OU = z / r Also deduce that NOV and NAS are similar, as are NAS and BAV , as well as NOV and BAV Hence AV / AB = OV / ON
z This results in r
So, z SN = r ( x + y + z ) = rs
( )
AB ON
x+ y SN r
22
Eulers Proof (cont.)
Because they are vertical angles, BOS and VON are congruent, so
OBS = 90 BOS = 90 VON = ANS
NAS
and
BOS
are similar
Hence, SN / AS = BS / OS This results in
SN / x = y / r SN = ( xy ) / r
Eulers Proof (cont.)
Lastly, Euler concluded that
Area (ABC ) = rs = rs(rs ) = z SN (rs ) = z
( )rs =
xy r
( )
sxyz = s(s a )(s b )(s c )
23
Pythagorean Theorem
Herons Formula can be used as a proof of the Pythagorean Theorem
Pythagorean Theorem from Herons Formula
Suppose we have a right triangle with hypotenuse of length a, and legs of length b and c The semiperimeter is:
s=
a+b+c 2
24
Pythagorean Thm. from Herons Formula (cont.)
sa =
Similarly
a +b +c 2
a =
and
a + b +c 2
22a =
a + b+ c 2
s b =
a b+ c 2
sc =
a +b c 2
After applying algebra, we get
Pythagorean Thm. from Herons Formula (cont.)
(a + b + c )( a + b + c )(a b + c )(a + b c ) = [(b + c ) + a ][(b + c ) a ][a (b c )][a + (b c )] 2 2 = [(b + c ) a 2 ][a 2 (b c ) ] 2 2 2 2 = a 2 (b + c ) (b + c ) (b c ) a 4 + a 2 (b c )
= 2a 2b 2 + 2a 2c 2 + 2b 2 c 2 a 4 + b 4 + c 4
25
Pythagorean Thm. from Herons Formula (cont.)
Returning to Herons Formula, we get the area of the triangle to be
K = s (s a )(s b )(s c ) = =
+c a +b + c a b + c a + b c (a+ b )( 2 )( 2 )( 2 ) 2 2 a 2b 2 + 2 a 2 c 2 + 2 b 2 c 2 a 4 + b 4 + c 4 16
Pythagorean Thm. from Herons Formula (cont.)
Because we know the height of this triangle is c, we can equate our expression to the expression
1 K=1 2 bh = 2 bc
Equating both expressions of K and squaring both sides, we get
b 2c 2 4
2 a 2b 2 + 2 a 2c 2 + 2 b 2 c 2 a 4 +b 4 + c 4 16
Cross-multiplication gives us
4b 2 c 2 = 2 a 2 b 2 + 2a 2 c 2 + 2b 2 c 2 a 4 + b 4 + c 4
26
Pythagorean Thm. from Herons Formula (cont.)
Taking all terms to the left side, we have
(b + 2b c + c ) 2a b 2a c + a (b + c ) 2a (b + c ) + a = 0 [(b + c ) a ] = 0 (b + c ) a = 0
4 2 2 2 4 2 2 2 2 2 2 2 2 2 2 4 2 2 2 2 2 2
=0
a2 = b2 + c2
Thus, Herons formula provides us with another proof of the Pythagorean Theorem
27
