calculator that failed with Heron’s formula. What is amazing about this formula is that it
is just Heron’s formula rewritten! The use of parentheses is what forces the correct order of
operations for numerical stability.
9This is an issue even on modern computers. There is an excellent overview of this important subject in the
article What Every Computer Scientist Should Know About Floating-Point Arithmetic by D. Goldberg, available
at http://docs.sun.com/source/806-3568/ncg_goldberg.html
10Due to W. Kahan: http://www.eecs.berkeley.edu/~wkahan/Triangle.pdf
58
Chapter 2 • General Triangles
§2.4
Another formula11 for the area of a triangle given its three sides is given below:
For a triangle △ ABC with sides a ≥ b ≥ c, the area is:
2
Area = K = 1
a 2 c 2
(2.34)
2
− a 2+ c 2− b 2
2
For the triangle in Example 2.16, the above formula gives an answer of exactly K = 1 on
the same TI-83 Plus calculator that failed with Heron’s formula.
Exercises
For Exercises 1-6, find the area of the triangle △ ABC.
1. A = 70◦, b = 4, c = 12
2. a = 10, B = 95◦, c = 35
3. A = 10◦, B = 48◦, C = 122◦, c = 11
4. A = 171◦, B = 1◦, C = 8◦, b = 2
5. a = 2, b = 3, c = 4
6. a = 5, b = 6, c = 5
7. Find the area of the quadrilateral in Figure 2.4.3 below.
C
4
3.5
θ
6
B
D
2
5.5
A
Figure 2.4.3
Exercise 7
Figure 2.4.4
Exercise 8
8. Let ABCD be a quadrilateral which completely contains its two diagonals, as in Figure 2.4.4
above. Show that the area K of ABCD is equal to half the product of its diagonals and the sine of
the angle they form, i.e. K = 1 AC
2
· BD sin θ .
9. From formula (2.26) derive the following formula for the area of a triangle △ ABC:
a 2 sin B sin C
Area = K = 2 sin ( B+ C)
10. Show that the triangle area formula
Area = K = 1
( a
4
+ ( b + c))( c − ( a − b))( c + ( a − b))( a + ( b − c))
is equivalent to Heron’s formula. ( Hint: In Heron’s formula replace s by 1 ( a
2
+ b + c) . )
11. Show that the triangle area formula (2.34) is equivalent to Heron’s formula. ( Hint: Factor the
expression inside the square root. )
12. Find the angle A in Example 2.16, then use formula (2.23) to find the area. Did it work?
11Due to the Chinese mathematician Qiu Jiushao (ca. 1202-1261).
Circumscribed and Inscribed Circles • Section 2.5
59
2.5 Circumscribed and Inscribed Circles
Recall from the Law of Sines that any triangle △ ABC has a common ratio of sides to sines
of opposite angles, namely
a
b
c
=
=
.
sin A
sin B
sin C
This common ratio has a geometric meaning: it is the diameter (i.e. twice the radius) of
the unique circle in which △ ABC can be inscribed, called the circumscribed circle of the
triangle. Before proving this, we need to review some elementary geometry.
A central angle of a circle is an angle whose vertex is the center O of the circle and whose
sides (called radii) are line segments from O to two points on the circle. In Figure 2.5.1(a),
∠ O is a central angle and we say that it intercepts the arc BC.
A
A
O
O
D
B
B
B
C
C
C
(a) Central angle ∠ O
(b) Inscribed angle ∠ A
(c) ∠ A = ∠ D = 12 ∠ O
Figure 2.5.1
Types of angles in a circle
An inscribed angle of a circle is an angle whose vertex is a point A on the circle and
whose sides are line segments (called chords) from A to two other points on the circle. In
Figure 2.5.1(b), ∠ A is an inscribed angle that intercepts the arc BC. We state here without
proof12 a useful relation between inscribed and central angles:
Theorem 2.4. If an inscribed angle ∠ A and a central angle ∠ O intercept the same arc,
then ∠ A = 12 ∠ O . Thus, inscribed angles which intercept the same arc are equal.
Figure 2.5.1(c) shows two inscribed angles, ∠ A and ∠ D, which intercept the same arc BC
as the central angle ∠ O, and hence ∠ A = ∠ D = 12 ∠ O (so ∠ O = 2∠ A = 2∠ D ).
We will now prove our assertion about the common ratio in the Law of Sines:
Theorem 2.5. For any triangle △ ABC, the radius R of its circumscribed circle is given by:
a
b
c
2 R =
=
=
(2.35)
sin A
sin B
sin C
(Note: For a circle of diameter 1, this means a = sin A, b = sin B, and c = sin C.)
12For a proof, see pp. 210-211 in R.A. AVERY, Plane Geometry, Boston: Allyn & Bacon, 1950.
60
Chapter 2 • General Triangles
§2.5
To prove this, let O be the center of the circumscribed circle for a triangle △ ABC. Then
O can be either inside, outside, or on the triangle, as in Figure 2.5.2 below. In the first two
cases, draw a perpendicular line segment from O to AB at the point D.
C
C
C
b
a
c
c
b
a
2
D
2
A
B
b
O
a
c
R
R
A
B
R
R
O
R
O
R
A
c
D
c
B
2
2
(a) O inside △ ABC
(b) O outside △ ABC
(c) O on △ ABC
Figure 2.5.2
Circumscribed circle for △ ABC
The radii O A and OB have the same length R, so △ AOB is an isosceles triangle. Thus,
from elementary geometry we know that OD bisects both the angle ∠ AOB and the side AB.
So ∠ AOD = 1
. But since the inscribed angle
2 ∠ AOB and AD = c
2
∠ ACB and the central
angle ∠ AOB intercept the same arc AB, we know from Theorem 2.4 that ∠ ACB = 12 ∠ AOB.
Hence, ∠ ACB = ∠ AOD. So since C = ∠ ACB, we have
c
AD
c
c
sin C = sin ∠ AOD =
= 2 =
⇒
2 R =
,
O A
R
2 R
sin C
so by the Law of Sines the result follows if O is inside or outside △ ABC.
Now suppose that O is on △ ABC, say, on the side AB, as in Figure 2.5.2(c). Then AB is a diameter of the circle, so C = 90◦ by Thales’ Theorem. Hence, sin C = 1, and so 2 R = AB =
c = c
, and the result again follows by the Law of Sines.
QED
1 =
c
sin C
Example 2.17
Find the radius R of the circumscribed circle for the triangle △ ABC whose
B
sides are a = 3, b = 4, and c = 5.
5
Solution: We know that △ ABC is a right triangle. So as we see from Figure
3
2.5.3, sin A
O
= 3/5. Thus,
a
3
A
C
2 R
4
=
=
= 5
⇒
R = 2.5 .
sin A
3
5
Figure 2.5.3
Note that since R = 2.5, the diameter of the circle is 5, which is the same as
AB. Thus, AB must be a diameter of the circle, and so the center O of the
circle is the midpoint of AB.
Corollary 2.6. For any right triangle, the hypotenuse is a diameter of the circumscribed
circle, i.e. the center of the circle is the midpoint of the hypotenuse.
Circumscribed and Inscribed Circles • Section 2.5
61
For the right triangle in the above example, the circumscribed circle is simple to draw; its
center can be found by measuring a distance of 2.5 units from A along AB.
We need a different procedure for acute and obtuse triangles, since for an acute triangle
the center of the circumscribed circle will be inside the triangle, and it will be outside for
an obtuse triangle. Notice from the proof of Theorem 2.5 that the center O was on the
perpendicular bisector of one of the sides ( AB). Similar arguments for the other sides would
show that O is on the perpendicular bisectors for those sides:
Corollary 2.7. For any triangle, the center of its circumscribed circle is the intersection of
the perpendicular bisectors of the sides.
Recall from geometry how to create the perpendicular bisector
of a line segment: at each endpoint use a compass to draw an arc
d
d
with the same radius. Pick the radius large enough so that the
A
B
arcs intersect at two points, as in Figure 2.5.4. The line through
those two points is the perpendicular bisector of the line segment.
For the circumscribed circle of a triangle, you need the perpendic-
ular bisectors of only two of the sides; their intersection will be
the center of the circle.
Figure 2.5.4
Example 2.18
Find the radius R of the circumscribed circle for the triangle △ ABC from Example 2.6 in Section 2.2: a = 2, b = 3, and c = 4. Then draw the triangle and the circle.
Solution: In Example 2.6 we found A = 28.9◦, so 2 R = a
sin A =
2
sin 28.9◦ = 4.14, so R = 2.07 .
In Figure 2.5.5(a) we show how to draw △ ABC: use a ruler to draw the longest side AB of length c = 4, then use a compass to draw arcs of radius 3 and 2 centered at A and B, respectively. The
intersection of the arcs is the vertex C.
C
C
A
B
3
2
O
A
B
c = 4
(a) Drawing △ ABC
(b) Circumscribed circle
Figure 2.5.5
In Figure 2.5.5(b) we show how to draw the circumscribed circle: draw the perpendicular bisectors
of AB and AC; their intersection is the center O of the circle. Use a compass to draw the circle
centered at O which passes through A.
62
Chapter 2 • General Triangles
§2.5
Theorem 2.5 can be used to derive another formula for the area of a triangle:
Theorem 2.8. For a triangle △ ABC, let K be its area and let R be the radius of its circum-
scribed circle. Then
abc
abc
K =
(and hence R =
) .
(2.36)
4 R
4 K
To prove this, note that by Theorem 2.5 we have
a
b
c
a
b
c
2 R =
=
=
⇒
sin A =
, sin B =
, sin C =
.
sin A
sin B
sin C
2 R
2 R
2 R
Substitute those expressions into formula (2.26) from Section 2.4 for the area K :
a 2 sin B sin C
a 2 · b
abc
K =
=
2 R ·
c
2 R
=
QED
2 sin A
2 · a
4 R
2 R
Combining Theorem 2.8 with Heron’s formula for the area of a triangle, we get:
Corollary 2.9. For a triangle △ ABC, let s = 1 ( a
2
+ b + c). Then the radius R of its circum-
scribed circle is
abc
R =
.
(2.37)
4
s ( s − a)( s − b)( s − c)
In addition to a circumscribed circle, every triangle has an inscribed circle, i.e. a circle
to which the sides of the triangle are tangent, as in Figure 2.5.6.
C
E
F
a
b
O
r
A
B
c
D
Figure 2.5.6
Inscribed circle for △ ABC
Let r be the radius of the inscribed circle, and let D, E, and F be the points on AB, BC, and AC, respectively, at which the circle is tangent. Then OD ⊥ AB, OE ⊥ BC, and OF ⊥ AC.
Thus, △ OAD and △ OAF are equivalent triangles, since they are right triangles with the
same hypotenuse O A and with corresponding legs OD and OF of the same length r. Hence,
∠ OAD = ∠ OAF, which means that OA bisects the angle A. Similarly, OB bisects B and
OC bisects C. We have thus shown:
For any triangle, the center of its inscribed circle is the intersection of the bisectors of
the angles.
Circumscribed and Inscribed Circles • Section 2.5
63
We will use Figure 2.5.6 to find the radius r of the inscribed circle. Since O A bisects A, we see that tan 1 A
, and so r
A. Now,
2
= r
AD
= AD · tan 12
△ OAD and △ OAF are equivalent
triangles, so AD = AF. Similarly, DB = EB and FC = CE. Thus, if we let s = 1 ( a 2
+ b + c), we
see that
2 s = a + b + c = ( AD + DB) + ( CE + EB) + ( AF + FC)
= AD + EB + CE + EB + AD + CE = 2( AD + EB + CE)