Problem of the Week
Problem D and Solution
A Circle in a Trapezoid

Problem

A circle with centre \(O\) and radius \(15 \text{ m}\) is inside trapezoid \(ABCD\) such that each side of \(ABCD\) is tangent to the circle. In the trapezoid, \(AB \parallel CD\) and \(AD=BC\), so \(ABCD\) is an isosceles trapezoid.

If the area of \(ABCD\) is \(2025 \text{ m}^2\), determine the lengths of \(AD\) and \(BC\).

For this problem, you may want to use the following known results about circles:

1. If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of tangency.

2. A line drawn from the centre of a circle perpendicular to a tangent line meets the tangent line at the point of tangency.

Solution

Draw line segment \(RT\) through \(O\) perpendicular to \(AB\) and \(CD\) so that \(R\) lies on \(AB\) and \(T\) lies on \(CD\). From Result 2 given after the problem, \(R\) and \(T\) are the points of tangency on \(AB\) and \(CD\), respectively.

Since \(OR\) and \(OT\) are both radii of the circle, \(RT=OR+OT=15+15=30\). Since \(RT \perp CD\), \(RT\) is the height of trapezoid \(ABCD\).

We will now proceed with two different approaches.

Solution 1

Let \(y=AB\) and \(z=CD\). Then using the formula for the area of a trapezoid, \[\begin{align} \text{Area} &= (AB + CD) \times (RT) \div 2\\ 2025 &= (y + z) \times 30 \div 2\\ 2025 &= (y + z) \times 15\\ 135 &= y + z \tag{1} \end{align}\]

Let \(x=AD=BC\). Join \(O\) to each of the vertices of \(ABCD\), creating four triangles, \(\triangle AOB\), \(\triangle BOC\), \(\triangle COD\), and \(\triangle AOD\). Then join \(O\) to each of the points of tangency, \(R\), \(S\), \(T\), and \(U\), on \(AB\), \(BC\), \(CD\), and \(AD\), respectively. Each of these line segments is a radius so \(OR=OS=OT=OU=15\). Furthermore, from Result 1 given after the problem, these line segments are each perpendicular to their respective sides of the trapezoid.

We can now find the area of the trapezoid a second way by summing the areas of the four triangles \(\triangle AOB\), \(\triangle BOC\), \(\triangle COD\), and \(\triangle AOD\). \[\begin{aligned} \text{Area} &= \text{Area }\triangle AOB + \text{Area }\triangle BOC + \text{Area }\triangle COD + \text{Area }\triangle AOD\\ 2025&= \frac{AB \times OR}{2} + \frac{BC \times OS}{2} + \frac{CD \times OT}{2} + \frac{AD \times OU}{2}\\ 2025 &= \frac{15y}{2} + \frac{15x}{2} + \frac{15z}{2} + \frac{15x}{2}\\ 2025 &= 15x + \frac{15(y+z)}{2}\\ 2025 &= 15x + \frac{15(135)}{2} \qquad(\text{from }(1))\\ 2025 &= 15x + \frac{2025}{2}\\ 15x &= \frac{2025}{2}\\ x &= \frac{135}{2} = 67.5 \end{aligned}\] Therefore, \(AD\) and \(BC\) each have a length of \(67.5\) m.

Solution 2

Join \(O\) to each of the points of tangency, \(R\), \(S\), \(T\), and \(U\), on \(AB\), \(BC\), \(CD\), and \(AD\), respectively. From Result 1 given after the problem, these line segments are each perpendicular to their respective sides of the trapezoid.

Let \(x=AD=BC\), \(a=AU\) and \(b=BS\). Then \(DU=x-a\) and \(CS=x-b\).

Join \(O\) to \(A\), forming right-angled triangles \(\triangle AUO\) and \(\triangle ARO\). Using the Pythagorean theorem, \(AO^2=AU^2+OU^2\) and \(AO^2=AR^2+OR^2\). Since \(OU\) and \(OR\) are both radii, \(OU=OR\). Thus \(AR=AU=a\). Using a similar reasoning, we can conclude that \(BR=BS=b\), \(CT=CS=x-b\), and \(DT=DU=x-a\).

Then using the formula for the area of a trapezoid, \[\begin{aligned} \text{Area} &= (AB + CD) \times (RT) \div 2\\ 2025 &= ((AR+BR)+(DT+CT)) \times 30 \div 2\\ 2025 &= ((a+b) + (x-a + x-b) \times 15\\ 2025 &= 2x \times 15\\ 2025 &= 30x\\ x &= \frac{2025}{30} = 67.5 \end{aligned}\] Therefore, \(AD\) and \(BC\) each have a length of \(67.5 \text{ m}\).