Suppose that a quadrilateral ABCD be given with the following lengths of sides: AB = 3 cm, BC = 8 cm, CD = 6 cm and AD = 5 cm. (See Figure 1.) What is the radius of the circumcircle (r)? This article addresses the problem.

Figure 1

 

Let’s derive the formula first …

 

Let ABCD be a quadrilateral with the following lengths of sides: AB = a, BC = b, CD = c, AD = d. (See Figure 2.)

Figure 2

 

Draw the line segment \overline{BD} and let the length of \overline{BD} be p (i.e. BD = p).

In ΔBCD, let ∠BCD = C. By the Cosine Rule, we have:

p2 = b2 + c2 – 2bc cos C ………………………………………………………………………………………………………………………………………………………… (1)

In ΔBAD, let ∠BAD = A. By the Cosine Rule, we have:

p2 = a2 + d2 – 2ad cos A …………………………………………………………………………………………………………………………………………………………. (2)

Since the sum of the opposite angles of a cyclic quadrilateral is supplementary, it follows that A + C = 1800. This results in cos A = cos (1800 – C) = – cos C and (2) can be expressed as:

p2 = a2 + d2 + 2ad cos C ……………………………………………………………………………………………………………………………………………………. (2a)

From (1) and (2a), we have the following.

a2 + d2 + 2ad cos C = b2 + c2 – 2bc cos C

\cos C = \frac{b^2 + c^2 - a^2 - d^2}{2(ad+bc)} …………………………………………………………………………………………………………………………………………………………………….. (*)

1 + \cos C = \frac{(b+c)^2 - (a-d)^2}{2(ad+bc)} = \frac{(b+c+a-d)(b+c- a+d)}{2(ad+bc)}

Letting 2s = a + b + c + d, it can be easily proved that b + c + a – d = 2(s – d) and b + c – a + d = 2(s – a). It follows that:

1 + \cos C = \frac{2(s-d) \cdot 2(s-a)}{2(ad+bc)} = \frac{2(s-a)(s-d)}{ad+bc} …………………………………………………………………………………………………… (+)

From the double-angle formula \cos {2C} = 2 \cos^2 C - 1, we have \cos C = 2 \cos^2 {\frac{1}{2} C} - 1 and furthermore:

\cos {\frac{1}{2} C} = \sqrt{\frac{1+\cos C}{2}}  …………………………………………………………………………………………………………………………………………. (++)

(We have ignored the possibility that \cos {\frac{1}{2} C} = - \sqrt{\frac{1+\cos C}{2}} because 0 < \frac{1}{2} C < 90^0.)

Then, by combining (++) with (+), we have:

\cos {\frac{1}{2} C} = \sqrt{\frac{(s-a)(s-d)}{ad+bc}} ……………………………………………………………………………………………………………………………………………………………. (3)

 

To find \sin {\frac{1}{2} C}, we follow the steps below.

1 - \cos C = 1 - \frac{b^2 + c^2 - a^2 - d^2}{2(ad+bc)}

1 - \cos C = \frac{(a+d)^2 - (b-c)^2}{2(ad+bc)}

1 - \cos C = \frac{(a+d+b-c)(a+d-b+c)}{2(ad+bc)}

Since 2s = a + b + c + d, it holds that a+d+b–c = 2(s–c) and a+d–b+c = 2(s–b). Accordingly,

1 - \cos C = \frac{2(s-c) \cdot 2(s-b)}{2(ad+bc)} = \frac{2(s-b)(s-c)}{ad+bc}  ……………………………………………………………………………………………………….. (+++)

From the double-angle formula \cos {2C} = 1 - 2 \sin^2 C, we have \cos C = 1 - 2 \sin^2 {\frac{1}{2} C}. Moreover,

\sin {\frac{1}{2} C} = \sqrt{\frac{1 - \cos C}{2}}  ……………………………………………………………………………………………………………………………………………… (++++)

(We have ignored the possibility that \sin {\frac{1}{2} C} = - \sqrt{\frac{1 - \cos C}{2}} because 0 < \frac{1}{2} C < 90^0.)

Then, by combining (++++) with (+++), we have:

\sin {\frac{1}{2} C} = \sqrt{\frac{(s-b)(s-c)}{ad+bc}} ……………………………………………………………………………………………………………………………………………………… (4)

 

By (3), (4), and the identity \sin C = 2 \sin {\frac{1}{2} C} \cos {\frac{1}{2} C}, we have:

\sin C = 2 \sqrt{\frac{(s-a)(s-b)(s-c)(s-d)}{(ad+bc)^2}} = \frac{2 \sqrt{(s-a)(s-b)(s-c)(s-d)}}{ad+bc}

 

It can be proved that the area of a concyclic quadrilateral is L = \sqrt{(s-a)(s-b)(s-c)(s-d)}. Consequently,

\sin C = \frac{2L}{ad+bc} …………………………………………………………………………………………………………………………………………………………… (**)

 

Now, consider ΔBOD. By the Cosine Rule, we get:

p2 = r2 + r2 – 2r2 cos ∠BOD ………………………………………………………………………………………………………………………………………… (5)

Since ∠BOD = 2C, equation (5) can also be expressed as:

p2 = 2r2 – 2r2 cos 2C

p2 = 2r2(1 – cos 2C)

p2 = 2r2(2 sin2 C)     [using the double-angle formula cos 2C = 1 – 2 sin2 C]

p2 = 4r2 sin2 C ……………………………………………………………………………………………………………………………………………………………………….. (6)

Combining (2a) with (6), we obtain the following.

4r2 sin2 C = a2 + d2 + 2ad cos C

r = \frac{\sqrt{a^2+d^2+2ad \cos C}}{2 \sin C} …………………………………………………………………………………………………………………………………………………………. (7)

Substituting (*) and (**) into (7), we get:

r = \frac{\sqrt{(ad+bc)(ab+cd)(ac+bd)}}{4L} ……………………………………………………………………………………………………………. (8)

 

Going back to the problem at the beginning of this article, what is the radius of the circumcircle?

 

In this case, a = 3 cm, b = 8 cm, c = 6 cm and d = 5 cm.

First, we compute the area of ​​the rectangle.

s = \frac{3+8+6+5}{2} \: cm = 11 \: cm

L = \sqrt{(11-3)(11-8)(11-6)(11-5)} \: cm^2 = 12 \sqrt{5} \: cm^2

Now, we are ready to calculate the circumradius.

Substituting a = 3 cm, b = 8 cm, c = 6 cm, d = 5 cm, and L = 10√6 cm2 into (8), we get:

r = \frac{\sqrt{(3 \cdot 5 + 8 \cdot 6)(3 \cdot 8 + 6 \cdot 5)(3 \cdot 6 + 8 \cdot 5)}}{4 \cdot 12 \sqrt{5}} \: cm = \frac{3}{40} \sqrt{3045} \: cm.

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