Chapter 3 : Area

For every polygon P we can associate to P a nonnegative real

number denoted area (P) such that

• (A1) Congruent polygons have equal areas.

• (A2) If P can be cut into two non-overlapping polygons and then area (P) = area ( ) + area ( ).

• (A3) If P is a square with sides of length 1, then

area (P) = 1.

1P2P 1P 2P

Area of Rectangles and Triangles

• Two rectangles are congruent if the bases are congruent and the heights are congruent. In general two polygons , and are congruent if corresponding sides are congruent and corresponding angles are congruent, that is, and .

• Theorem. Let ABCD and EFGH be rectangles such that , and , for some positive real number Then area (ABCD) = . area (EFGH).

nPPP ...21 nQQQ ...21

112121 ,..., QQPPQQPP nn nn QPQP ,...,11

EFAB FGBC .

• Proof. To prove this theorem we must first prove the special case in which is an integer, then the case in which is a rational number, and finally the general case. However, we will only do the first two cases now.

• If is a positive integer, then the length of divides the length of evenly. Hence (cf. Figure 3.2) we can find points between B and C on such that . At each point draw a line segment parallel to connecting with . This now breaks the rectangle ABCD into smaller non-overlapping rectangles. Each of these smaller rectangles has base congruent to and height congruent to and so, by (A1), each has area equal to area (EFGH).

BC

FG

121 ,...,, BBB BCCBBBBBFG 1211 ... iB

AB BC AD

EF FG

• But by (A2), the area of ABCD is the sum of the areas of these rectangles. So area (ABCD) = . area (EFGH). This proves the theorem for this special case.

• We next consider the case in which is a positive rational number. By definition, this means that where n and m are integers . Now, given ABCD and EFGH we construct a third rectangle R. The base of R will be congruent to and to . The height of R will be m times BC (Figure 3.3). But since , it follows that . So the height of R is also n times FG. Since n and m are integers, we may apply the previous case. We have

area (R) = m . area (ABCD)

and,

area (R) = n . area (EFGH).

Hence,

m . area (ABCD) = n . area (EFGH).

m

n

EFAB

FGm

nFGBC .. BCmFGn ..

• Or, area (ABCD) = area (EFGH) = . area (EFGH), which proves the theorem for rational .

• If is not a rational number we can find rational numbers that approximate very closely and apply the previous case.

m

n

m

n

• Corollary. If ABCD is a rectangle with AB = 1 and CB = h, then area (ABCD) = h.

• Proof. Let EFGH be a square, EF = GH =1. By the theorem, area (ABCD) = h . area (EFGH). But, by (A3), area (EFGH) = 1.

• Corollary. If ABCD is a rectangle with AB = b and BC = h, then area (ABCD) = bh.

• Proof. Let EFGH be a rectangle with

and FG = 1 as in Fig 3.4. By the theorem,

area (ABCD) = h . area (EFGH) and, by the previous corollary, area (EFGH) = b.

EFAB

• Triangles• Given we arbitrarily designate one side, say ,

as the base. Then the height will be the distance from A to . If D is a point on such that , then the line segment is called the altitude. Also the length of will be the height.

• Theorem. The area of a triangle is

• Proof. Let be a triangle with base . We will first prove the theorem in the special case in which is a right angle. In this case is the altitude. Draw lines through A parallel to and through C parallel to meeting at D (Fig. 3.5). The quadrilateral ABCD has

ABC BC

BC BC BCAD AD

AD

.2/1 heightbase

ABC BCB

AB

BC AB

opposite sides parallel and it has a right angle at B. Hence it is a rectangle. So area (ABCD) = AB . BC.

But, area (ABCD) is the sum of the areas of the two triangles, and . Now, the diagonal of a rectangle (or any parallelogram) divides it into two congruent triangles, so . In particular, area ABC = area ADC. Combining all this information, we see:

ABC ADC

ADCABC

AB . BC = area ABCD

= area ABC + area ADC

= area ABC + area ABC

= 2 . Area ABC.

Thus, area ABC = . AB . BC, as claimed for the case when .

2

1

.90B

• We now turn to the general case. Draw the altitude (Fig. 3.6) and compare the area of with the triangles and . These latter two triangles are right triangles, and by the case we already considered

area ABD = BD . AD,

and

area ACD = CD . AD.

• There are now two possibilities. If D is between B and C, then

area ABC = area ABD+ area ACD

= . CD . AD+ . BD . AD

= (CD+BD) . AD

= . BC . AD.

ADABC

ABD ACD

2

1

2

1

2

1

2

1

2

1

2

1

If D is not between B and C, then

area ABC = area ACD – area ABD

= . CD . AD - . BD . AD

= (CD – BD) . AD

= . BC . AD.

This completes the proof

2

1

2

1

2

1

2

1

The Pythagorean Theorem

Let be a right triangle at C. Denote the lengths of the sides by AB = c, AC = b, and BC = a. Since is the largest angle, will be the longest side. It is called the hypotenuse. The other two sides are called the legs. The Pythagorean theorem states that , or, conversationally, the square of the hypotenuse equals the sum of the squares of the legs.

• The Pythagorean theorem has many different proofs. We now present the one which is based on areas.

ABCC

AB

222 cba

• Theorem. Let be a right triangle with right angle at C. Denote AB = c, BC = a, and AC = b as in Fig. 3.7. Then .

• Proof. As in Fig. 3.7, we construct a square DEFG in which each side has length a + b. we next select points , and on the sides such that

and .

Then each of the four triangles , , , and is congruent to by SAS. Hence

, consequently, is a rhombus. We claim that it is actually a square. To prove this we need to show that all of its angles are .

ABC

222 cba

,,, 321 PPP 4P

bGPFPEPDP 4321 aDPGPFPEP 4321

41PDP 12PEP 23PFP

34PGP CABcPPPPPPPP 14433221 4321 PPPP

90

• Consider, for example :

• Therefore, is a square.

214 PPP

EPPDPPPPP 1214214 180

C.90

BA 180

4321 PPPP

• We now calculate the area of DEFG in two different ways. On the one hand, since DEFG is a square, it has area On the other hand, it has area equal to the sum of the areas of five pieces, the four triangles and the square. Each triangle has area and the square has area . Hence

or

The theorem now follows easily.

• Corollary (The converse of the Pythagorean Theorem). Let be any triangle and denote the lengths of the sides AB = c, BC = a, and AC = b. If , then .

2)( ba

ab.2

1 2c22

2

1.4)()( cabbaEFGHarea

.2)( 22 abcba

ABC222 cba

90C

• Proof. Construct a new triangle, , such that , and . Since is a right triangle we may apply the Pythagorean theorem to conclude that

Hence, EF = c, which implies that . Therefore , by SSS and .

EFG 90GACEG BCGF EFG

222 FGEGEF 22 ba

.2c

ABEF EFGABC 90 CG

• Corollary (SSA Theorem for Right Triangles). In triangles and assume that , and that and are right angles. Then .

• Proof. By the Pythagorean theorem , so . Likewise, . Since AB = EF and BC = FG, it follows that or AC = EG. Hence, by SSS.

ABC EFG EFAB FGBC C G

EFGABC

222 ABBCAC 222 BCABAC 222 FGEFEG

22 EGAC EFGABC

• Now, we want to present the geometric law of cosines which relates to even if is not a right angle.

• Theorem. Let be a triangle with base and altitude . Denote AB = c, BC = a, AC = b, and CD = x. If D is between B and C or if B is between C and D, then . If C is between B and D, then . .

2c 22 ba C

ABC BCAD

axbac 2222 axbac 2222

• Proof. Consider the case (a) of Fig. 3.10, in which D is on . Write h for AD. Since and are right triangles we have

, and .

Hence

The result now follows using a bit of algebra. The other two cases are similar.

BC ADC ADB

222 bxh 222 )( cxah

2222 )( xacxb 222 2 xaxac

Area of Triangles

• We prove two deeper results on areas of triangles. The first compares the areas of two different triangles sharing one equal angle and the second gives the area of a triangle in terms of the lengths of the three sides.

• Theorem. Let and be triangles in which , . Then

ABC DEFDA

..

.

DFDE

ACAB

DEFarea

ABCarea

• Proof. We first consider the special case in which If , then and the relation of the theorem reduces to 1 = 1. We lose no generality by supposing that . Choose a point G on such that . Now, by SAS. Let be the altitude from B to . Then area (ABC) = . BH . AC and area (ABG) = . BH . AG.

.DEAB DFAC DEFABC

DFAC ACDFAG DEFABG BH

AC2

1

2

1

• Hence

• Since we assume that AB = DE, this last ratio is also equal to . This completes the proof in this special case.

• The general case now follows easily from a clever construction. Given two triangles and such that , we construct a third triangle such that is congruent to and , , and .

)(

)(

)(

)(

ABGarea

ABCarea

DEFarea

ABCarea

AGBH

ACBH

..21

..21

AG

AC

.DF

AC

DEDF

ABAC

.

.

ABC DEF

GDA GHI

A D DEGH ACGI

• We may now apply the special case twice to the triangles , and and to and to yield the relations

and

• Multiplying the first equation by the second yields the result.

ABC GHI DEF GHI

GIGH

ACAB

GHIarea

ABCarea

.

.

)(

)(

DFDE

GIGH

DEFarea

GHIarea

.

.

)(

)(

• Heron’s Theorem. Let be any triangle and denote the lengths of the sides by AB = c, BC = a, and AC = b. Also, let s = the semiperimeter, . Then the area of .

• Proof. Let be the base and the altitude. As in the geometric law of cosines, let AD = h and CD = x (Fig. 3.13). The by that law,

so

(For the sake of simplicity we will assume that D is between B and C.) Also, by the Pythagorean theorem,

,

ABC

))()(( csbsass )(

2

1cbas

ABC

BC AD

axbac 2222

a

cbax

2

222

222 bxh

• So

Combining these two equations yields

• Now, simply take a square root and multiply by . The main tool for the simplification will be the formula for the difference of two squares, namely, .

.)2

( 2222

22

a

cbabh

222 xbh

a2

1

))((22 yxyxyx

• After one application of that rule we get

• Now multiply both sides by to see that

• Again, each factor is the difference of two squares, so

• Finally, we divide by 4:

2222

22 )2

(a

cbabh

)2

)(2

(222222

a

cbab

a

cbab

2a].)][()([

4

1)2)(2(

4

1 222222222222 cbabaccbaabcbaabha

).)()()((4

122 cbacbabacbacha

.2

.2

.2

.24

1 22 cbacbabacbacha

• But , and , so the right hand side of the equation equals . The left-hand side is , which is the square of the area. If we take the square root of both sides we get Heron’s formula.

• The proof of Heron’s formula given here uses a lot of algebra. Amazingly, Heron proved his formula long before symbolic algebra was invented. You can find Heron’s “non-algebraic” proof in B. M. Olive’s article “Heron’s remarkable triangle area formula,” in Mathematics Teacher, Feb. 1993.

scba

bsbac

asbac

2,

2,

2cs

cba

2

))()(( csbsass 2)

2

1( ah

Cutting and Pasting

• Informally, P and Q are equidecomposible if you can cut P into pieces and rearrange the pieces to form Q. Here is the more precise definition.

• Definition. Two polygons P and Q are said to be equidecomposible if P can be written as a union of non-overlapping polygons and Q can be written as a union of non-overlapping polygons such that , (Fig. 3.14). We will also say that P is equidecomposible with Q, or more simply that P is decomposed to Q.

nPP ,...,1

nQQ ,...,1

nn QPQP ,...,11

• It follows from the area axioms (A1) and (A2) that if two polygons are equidecomposible, then they have the same area.

• Theorem. If two polygons P and Q have equal area, then they are equidecomposible.

• The following lemma is the key result in proving the theorem. We will explain and illustrate it but not give a completely formal proof. It says that equidecomposibility is a transitive relation.

• Lemma. Given polygons P, Q, and R such that P and Q are equidecomposible and Q and R are equidecomposible, then P and R are equidecomposible.

• If we think of equidecomposiblity in terms of cutting up a polygon and rearranging the pieces, then the lemma is clear: In order to decompose P into R, we first cut P up and rearrange the pieces to form Q, and then cut Q up and rearrange the pieces to form R.

• In terms of the formal definition using congruent polygons, the situation is about the same. There are two decompositions of Q, one which can be rearranged to form R. If we superimpose these two decompositions of Q, we get one that can be rearranged to form either P or R (Fig. 3.15).

• For the rest of the proof of the theorem, pick out any line segment . Say for convenience that has length 1. We will prove that any polygon P is equidecomposible with a rectangle with base congruent to . This will imply that if P and Q have equal area, then they are each equidecomposible with the same rectangle and, in light of the lemma, this will imply that they are equidecomposible with each other.

• We first consider the case of triangles. We want to show that any triangle is equidecomposible with a rectangle of base 1. We do this using a series of constructions. The first three show how to do it in the special case that things work out nicely, and then the next two show what adjustments can be made if they don’t.

XY XY

XY

• Construction 1. Decompose any triangle with a parallelogram.

• Solution. Let be any triangle. Let D be the midpoint of parallel to and parallel to , as shown in Fig. 3.16. Then BEFC is a parallelogram, since opposite sides are parallel, and by ASA. So is equidecomposible with BEFC.

ABCDEAC, BC CF AB

DAEDCF ABC

• Construction 2. Decompose a parallelogram with a parallelogram with one side of length 1, if things work out nicely.

• Solution. Let ABCD be the parallelogram. If we’re lucky, then there is a point E on with AE = 1. In this case, let F on be such that (Fig. 3.17). Then AEFB must be a parallelogram because the sides and are parallel and congruent. Finally, by SAS:

because , , and

DE = DC – EC = EF – EC = CF.

DC

EC ABEF EF ABBCFADE

CFBDEA BFAE 1BFAE

• In the preceeding construction, there are two ways that things might not work out nicely. One such way would be if the height of ABCD is bigger than 1; then there will be no point E on with AE = 1. In this case we can use construction 4 (which follows) one or more times to decompose ABCD with a parallelogram with height less than or equal to 1.

• The other possible problem is if the points E on with AE =1 are all to the left or right of . In this case, we can use construction 5 (which follows) to decompose ABCD with a parallelogram with the same base and the same height, but with opposite side further to the right or left as required. So the missing step in this part of the program is to make our parallelogram into a rectangle.

DC

DCDC

AB

• Construction 3. Decompose a parallelogram with one side equal to 1 to a rectangle with one side equal to 1, if we’re lucky.

• Solution. Let ABCD be the parallelogram with AB = 1. Say that is obtuse, and let the altitude from A be and the altitude from B be . If things work out nicely, then E will be on . In this case we are done because , . If things don’t work out nicely, E will be to the right or left of . In this case, we can move using construction 5.

A AEBF

CDBFDAEC

CD CD

• Here now are the two constructions we need for the unlucky cases.

• Construction 4. Decompose a parallelogram with a parallelogram with half the height.

• Solution. See Fig. 3.19.

• Construction 5. Decompose any parallelogram ABCD with a parallelogram with or .

• Solution. We’ve already proven that the diagonal of a parallelogram divides it into two congruent triangles. With Fig. 3.20 for a hint you should be able to see how the decomposition works.

DCAB DC DC

• Proof of Main Theorem. Given any polygon P, we may cut P into non-overlapping triangles . Using constrictions 1-5, we can decompose each of into a rectangle , such that each of these rectangles has base 1. Then these rectangles can be stacked to form a bigger rectangle R, also of base 1.

• So, any polygon can be decomposed to a rectangle of base 1. As we remarked earlier, the transitive property of equidecomposibility now implies that any two polygons with equal area are equidecomposible.

nTT ,...,1

nTT ,...,1

nRR ,...,1