CHAPTER 5

AREA

5.1 Basic Properties

In this chapter we will use the word polygon to refer to a polygon together with its interior, even though properly we should use the term polygonal region. This should not cause any confusion.

Suppose one polygon is inside another. When treated as wire frames, the polygons would be considered as being disjoint; in the present context, they overlap. In general, if two figures share interior points, they will be considered as overlapping; otherwise, they will be considered as nonoverlapping.

5.1.1 Areas of Polygons

We will associate with each simple polygon a nonnegative number called its area, and we will assume that area has certain reasonable properties.

Postulates for polygonal areas:

Square brackets will be used to denote area. So, for example, the area of a quadrilateral ABCD will be denoted [ABCD].

Properties (ii) and (iii) certainly conform to our preconceived notions about area. We expect figures to have the same area if they have the same shape and size, and we also expect to be able to find the area of a large shape by summing the areas of the individual pieces making up the shape.

To develop a concrete theory of area, we must specify how it is to be measured, and our specification must satisfy (ii) and (iii). The easiest (and most useless) way would be to designate the area of every figure to be zero. To avoid this, a fundamental shape is chosen and defined to have a positive area, which is what statement (iv) accomplishes.

As an alternative to (iv), we could have used the unit square⁴ as the fundamental region, defined its area to be one square unit, and derived statement (iv) from it. This approach, although logical, presents some obstacles, especially when the rectangle has sides of irrational length.⁵ Because of this, we have chosen to use the set of all rectangles as a family of fundamental regions on which to base the computation of areas.

Exercise 5.1.1. The following figure is a regular five-pointed star. Which is larger, the area of the shaded part or the total area of the unshaded parts?

Since a square is a special case of a rectangle, we immediately have the following formula.

Theorem 5.1.2. (Area of a Square)

The area of an a × a square is a².

Now let us consider a parallelogram, of which a rectangle is a special case. We may choose any of its sides as the base. We will use the same symbol to designate both the length of the base and the base itself. The context will make it clear which meaning is intended. The distance from the base to the opposite side is called the altitude on that base.

Theorem 5.1.3. (Area of a Parallelogram)

The area of a parallelogram with altitude h on a base b is bh.

Proof. Let ABCD be a parallelogram with AB = b. Drop perpendiculars AE and BF to CD from A and B, respectively. Then ABFE is a rectangle with AE = h, so that [ABFE] = bh. Now, triangles ADE and BCF are congruent. Hence, [ADE] = [BCF] by the Invariance Property. By the Additivity Property,

which completes the proof.

Any side of a triangle may be designated as the base. The perpendicular from the opposite vertex to the base is called the altitude on that base. As with parallelograms, the word base has a dual meaning, referring to a specific side of a triangle and also to the length of that side. For triangles, the word altitude has a similar double meaning.

Theorem 5.1.4. (Area of a Triangle: Base-Altitude Formula)

The area of a triangle with altitude h on a base b is bh.

Proof. Let ABC be the triangle. Complete the parallelogram ABCD. Then the area of parallelogram ABCD is bh. Now, since ΔABC = ΔCDA, by the Invariance Property, we have

and, by the Additivity Property,

so that [ABC] = bh.

Theorem 5.1.5. (Area of a Triangle: Inradius Formula)

The area of a triangle with inradius r and semiperimeter s is rs.

(The semiperimeter is half the perimeter.)

Proof. Let I be the incenter of triangle ABC. Then

Example 5.1.6. Find the area of a trapezoid whose parallel bases have lengths b and t and whose altitude is h using:

Solution.

(1) Two copies of the trapezoid will form a parallelogram with base t + b and altitude h. Its area is (t + b)h, so the area of the trapezoid is (t + b)h.

(2) A diagonal divides the trapezoid into two triangles with altitudes h and respective bases t and b. The areas of the triangles are th and bh, from which the desired result follows.

The following example shows how area can be used to give a simple proof of Pythagoras’ Theorem.

Example 5.1.7. (Pythagoras’ Theorem)

Given a right triangle with sides of length a and b and with hypotenuse of length c,

Proof. Draw a square with side c and place four copies of the triangle around it, as shown below.

Since α + β = 90°, it follows that PTS is a straight line, and so PQRS is a square. Using the Additivity Property of area, we have

so that

and so

5.1.2 Finding the Area of Polygons

The area of an arbitrary polygon can be found by decomposing the polygon into triangular regions. If the polygon is convex, all diagonals are internal, so we can choose an arbitrary vertex and join it to all others by diagonals, as in figure (a) below, thereby dividing the polygon into triangles. Such a process is called triangulation. Since we can determine the area of each triangle, the Additivity Property yields the area of the polygon.

If the polygon is not convex, the following process can be used: extend the sides of all reflex angles until they meet the perimeter of the polygon. This will divide the polygon into convex polygons, each of which can be triangulated as in the preceding paragraph. Again, the Additivity Property yields the desired result.

The methods above show that the problem of finding the area of any polygon can always be reduced to the problem of finding the area of triangles.

However, from a computational point of view, triangulation methods are extremely clumsy, and we should find a better method. There are many different ways to triangulate a given polygon, and a good choice can ease the computation of the area. For example, we would likely prefer the triangulation of Figure (c) below rather than that of Figure (d), because the former decomposes the pentagon into congruent triangles.

Also, we should keep in mind the possibility of using other methods that do not involve triangulation. In the figure on the right, it is surely more convenient to view the area of the 7-gon as the difference in the areas of triangle ABG and the square CDEF.

5.1.3 Areas of Other Shapes

Is it possible to extend the notion of area to all bounded regions of the plane in a way that satisfies both the Invariance Property and the Additivity Property?

The answer to this question is yes—although the proof of this fact is well beyond the scope of this book. The positive answer means that the properties of polygonal area postulated earlier can therefore be applied to all bounded figures in the plane.

This does not mean, however, that the computation of areas of nonpolygonal regions is straightforward. In fact, one must resort to a limiting process even to find the area of such a basic region as a circular disk. The figure on the following page shows how to approximate a circle of radius r by polygons, with successive approximations coming closer and closer to the circle.

If the approximations in the figure above are cut along the dotted lines as shown below, the pie-shaped regions can be reassembled to form a parallelogram. As the approximations improve, the parallelogram comes closer and closer to becoming a rectangle whose altitude is the radius of the circle and whose base is half the circumference, which yields the following:

Theorem 5.1.8. The area of a circle is half the product of its radius and its circumference.

5.2 Applications of the Basic Properties

In this section we will develop several tools using the notion of area and give examples to show how these new tools can be used to solve a variety of problems. Some of the examples in this section may seem to have no connection with the notion of area … but they do.

Theorem 5.2.1. Let l and m be parallel lines. Let ABCD and EFGH be parallelograms with AB and EF on m and CD and GH on l. Then

Proof. The two parallelograms have equal altitude h on the respective bases AB and EF. Hence,

The same result holds for triangles.

Corollary 5.2.2. If ABC and DEF are triangles such that A and D are on a line l and BC and EF are on a parallel line m, then

Two special cases of this result that will be used frequently are:

Corollary 5.2.3. If AB and CD are parallel, then

Corollary 5.2.4. If B, C, and D are collinear points and if A is a point not collinear with them, then

The following is another proof that the three medians of a triangle are concurrent at the centroid.

Theorem 5.2.5. The medians of a triangle are concurrent.

Proof. Let D be the midpoint of the side BC of triangle ABC, and let G be a point on AD such that AG = 2GD. Suppose that the extension of BG meets CA at E. We will show that AE = EC and from this deduce that the three medians of ABC are concurrent at G.

Let [DEG] = x and [BDG] = y. Since

then by Corollary 5.2.4, we have

and

Hence,

so that AE = EC; that is, BE is a median.

Similarly, if we let the extension of CG meet AB at F, then AF = FB and AF is a median. Hence, the three medians are concurrent at G.

Example 5.2.6. Let D be a point on the side BC of triangle ABC. Construct a line through D which bisects the area of ABC.

Solution. If D is the midpoint of BC, then clearly AD is the desired line. Suppose D is between B and the midpoint M of BC. Draw a line through M parallel to AD, cutting CA at E.

We claim that DE is the desired line. By Corollary 5.2.3, [ADE] = [MAD], and so

Example 5.2.7. ABDE, BCFG, and CAHI are three parallelograms drawn outside ΔABC. The lines FG and HI meet at J. The extension of JC meets AB at K and the line DE at L. If JC = KL, prove that

Solution. Draw a line through B parallel to JL, cutting the line FG at M and the line DE at N. Draw a line through A parallel to JL, cutting the line HI at O and the line DE at P. By Theorem 5.2.1 and the Additivity Property of area, we get

which completes the proof.

Example 5.2.7 is due to Pappus and has Pythagoras’ Theorem as a corollary:

Corollary 5.2.8. (Pythagoras’ Theorem)

Let ABC be a triangle. If ∠C is a right angle, then

Proof. Suppose ∠ACB is a right angle. Draw squares ABDE, BCFG, and CAHI outside ABC. Let the extensions of GF and HI meet at J. Let the extension of JC meet AB at K and DE at L. Since triangles ABC and JCF are congruent, JC = AB = AE = KL. By Example 5.2.7,

The next example uses Corollary 5.2.3 to obtain another proof of the Angle Bisector Theorem. In the proof, we make use of the fact that an angle bisector is characterized by each point on the bisector being equidistant from the arms of the angle. Before stating the theorem, we recall that the notation refers to the directed distance from X to Y. This enables us to distinguish between interior and exterior angle bisectors.

Theorem 5.2.9. (The Angle Bisector Theorem)

Let D be a point on the line BC and A a point not on it. Then

(1) if and only if AD bisects angle A of triangle ABC and

(2) if and only if AD bisects the exterior angle A of triangle ABC.

Proof.

(1) Drop perpendiculars DE and DF from D to CA and AB, respectively. Suppose AD bisects ∠CAB. Since D is on the angle bisector, we have DE = DF. By Corollary 5.2.3,

Conversely, suppose . Reversing the argument, we have DE = DF, so D is equidistant from the arms of ∠BAC. It follows that AD bisects ∠CAB.

(2) Statement (2) can be proved in an analogous manner.

As the previous examples show, Corollary 5.2.3 is very useful. It can be generalized to the following result.

Theorem 5.2.10. Let D be a point on the line BC and A a point not on it. If E is a point on the line AD, then

Proof. The figure above shows that there are many possibilities for the location of the points D and E. We will consider one subcase of the situation where D is between B and C, namely, the case where E is between A and D as in the figure on the right.

Let BD/CD = t. Corollary 5.2.3 implies that

and that

Then

so that

We now apply Theorem 5.2.10 to solve the following, a special case of a result due to Routh.⁶

Example 5.2.11. Let D, E, and F lie respectively on the sides BC, CA, and AB of triangle ABC such that

Suppose also that AD meets BE at G and CF at H and that BE meets CF at I.

Determine

Solution.

Let the areas of triangles HAF, GBD, and ICE be x, y, and z, respectively. We will first show that [ABC] = 21x and that x = y = z.

By Corollary 5.2.3,

so that

Similarly, since DC = 2BD, Theorem 5.2.10 gives us

and so

Since BF = 2AF, from Corollary 5.2.3, we get [FBC] = 14x so that

In the same way, we can prove that [ABC] = 21y = 21z. Hence x = y = z and so

It follows that

so that

5.3 Other Formulae for the Area of a Triangle

The SAS congruency condition asserts that a triangle is uniquely determined given two of its sides and the included angle. All other congruency conditions also describe geometric data that determine a unique triangle. Since these conditions do not explicitly specify the altitude of the triangle, the formula for the area of a triangle cannot be used without some preliminary work. It would be much more convenient to be able to compute the area of a triangle directly from the data that describes it. In this section we develop a list of formulae for the area of a triangle based on the various congruency conditions. As you might expect, some of the formulae involve trigonometric functions.

In this section we will continue to follow the practice of denoting the sides of a triangle by the lowercase letters that correspond to the labels for the opposite vertices.

We start by proving a useful trigonometric tool.

Theorem 5.3.1. (Law of Sines)

In triangle ABC,

where R is the circumradius.

Proof.

Let BD be the diameter of the circumcircle. Then ∠BCD = 90° and

by Thales’ Theorem, Hence,

so that

Similarly,

are both equal to 2R.

Theorem 5.3.2. (SAS Case)

The area of a triangle with sides a, b and angle C is

Proof.

Let AD be the altitude on BC. Then AD = b sin C = b sin(180° − C). Hence,

Example 5.3.3. Prove that

where R is the circumradius of ΔABC.

Solution. We have

by the Law of Sines.

For the ASA case, we make use of all three angles.

Theorem 5.3.4. (ASA Case)

Given angles B and C and included side a of ΔABC, then

Proof. By the Law of Sines, a sin B = b sin A. Hence,

We remark that the Law of Sines is also useful when one needs to find the remaining dimensions of a triangle given a side and two angles.

The proof of the following uses the compound angle formula for the sine function, that is, sin(A + B) = sin A cos B + cos A sin B.

Theorem 5.3.5. (SSA Case)

Assuming that a > b, the area of ΔABC given sides a and b and angle A is

Proof. Since a > b, A > B, so that cos B > 0. By the Law of Sines, a sin B = b sin A. Hence,

In the following, s denotes the semiperimeter of a triangle; that is, for a triangle with sides a, b, and c,

Theorem 5.3.6. (SSS Case: Heron’s Formula)

For a triangle with sides a, b, and c,

Proof. Let BC be the longest side. Then the foot D of the altitude AD lies between B and C. Let CD = x and AD = y, so that BD = a − x. By Pythagoras’ Theorem, x² + y² = b² and (a − x)² + y² = c². Subtraction yields

Now,

Heron’s Formula was derived from Pythagoras’ Theorem, but it is possible to reverse directions and derive Pythagoras’ Theorem from Heron’s Formula, meaning that the two theorems are equivalent. (To derive Pythagoras’ Theorem, apply Heron’s Formula to a right triangle. The details are left as an exercise.)

The next result, which is far more familiar than Heron’s Formula, is also equivalent to Pythagoras’ Theorem.

Theorem 5.3.7. (The Law of Cosines)

In triangle ABC,

Proof. If ∠C = 90°, then cosC = 0 and the result is just Pythagoras’ Theorem. Suppose ∠C > 90°. Then the foot D of the altitude AD lies on the extension of BC. By Pythagoras’ Theorem,

since cos C < 0.

If ∠BCA < 90°, the argument is similar.

The Law of Cosines is not used to find the area of a triangle. It is indispensable, however, when one needs to find the remaining sides and angles in the SAS and SSS cases.

5.4 Solutions to the Exercises

Solution to Exercise 5.1.1

In the figure below, the shaded region has been divided into four parts, each of which is congruent to a part of the unshaded region. Therefore, the shaded region is equal in area to the total of the unshaded regions.

5.5 Problems

1. In the regular octagon shown below, is the area of the shaded region larger or smaller than the total area of the unshaded regions?

2. P is a point inside square ABCD. Show that

3. A square is divided into five nonoverlapping rectangles, with four of the rectangles completely surrounding the fifth rectangle, as shown in the diagram. The outer rectangles are the same area. Prove that the inner rectangle is a square.

4. ABCD is a quadrilateral with perpendicular diagonals inscribed in a circle with center O. Prove that [ADCO] = [ABCO].

5. A paper rectangle ABCD of area 1 is folded along a straight line so that C coincides with A. Prove that the area of the pentagon obtained is less than .

6. ABCD is a parallelogram. E is a point on BC and F a point on CD. AE cuts BF at G. AF cuts DE at H, and BF cuts DE at K, Prove that

7. ABCDEF is a convex hexagon in which opposite sides are parallel. Prove that [ACE] = [BDF].

8. G is a point inside triangle ABC such that [GBC] = [GCA] = [GAB]. Show that G is the centroid of ABC.

9. Seven children share a square pizza whose crust may be considered to consist only of the perimeter. Show how they make straight cuts to divide the pizza into seven pieces such that all pieces have the same amount of pizza and the same amount of crust.

10. AB and CD are four points such that AB = CD. Find the set of all points P such that [PAB] = [PCD].

11. ABCDEF is a convex hexagon with side AB parallel to CF, side CD parallel to BE, and side EF parallel to AD. Prove that [ACE] = [BDF].

12. ABCD is a convex quadrilateral, and F and H are the midpoints of BC and AD, respectively. If AC cuts FH at the midpoint K of FH, show that [ABC] = [ADC].

13. Given a convex quadrilateral ABCD, construct a point E on the extension of BC such that the area [ABCD] = [ABE].

14. ABCDE is a regular pentagon. The points P, Q, and R are the feet of the perpendiculars from A to BC, CD, and DE, respectively. The center O of the pentagon lies on AQ. If OQ = 1, compute AP + AQ + AR.

15. Given the figure below with 3BF = 2FC, AE = 2EF, and [DEF] = 1:

(a) Find [DFC].

(b) Find .

16. M is a point interior to the rectangle ABCD. Prove that AM · CM + BM · DM > ABCD.

17. Given a rectangle, construct a square having the same area.

18. A triangle is inside a parallelogram. Prove that the area of the triangle is at most half that of the parallelogram.

19. P is a point inside an equilateral triangle ABC. Perpendiculars PD, PE, and PF are dropped from P onto BC, CA, and AB, respectively. Prove that

20. Find the area of a triangle of sides 13, 18, and 31.

21. A triangle has sides 13, 14, and 15. Find its altitude on the base of length 14.

22. The sides of a triangle are 5, 7, and 8.

(a) Calculate its area.

(b) Calculate its inradius.

(c) Calculate its circumradius.

23. ABDE, BCGF, and CAHI are three squares drawn on the outside of triangle ABC, which has a right angle at C. Prove that

24. M is a point interior to the rectangle ABCD. Prove that

25. ABCDE is a convex pentagon such that

If M is the midpoint of BE, prove that CD = 2 AM.

⁴ The unit square is a 1 × 1 square.

⁵ A rational number is a real number that can be expressed as one integer divided by another nonzero integer. A number that cannot be expressed this way is called irrational.

⁶ A generalization of this example was given by E. J. Routh in 1891 (without proof) who needed this ratio