In the following figure, a triangle
𝐴𝐵𝐶 is drawn to circumscribe a circle of radius three centimetres such that 𝐵𝐷
and 𝐷𝐶 are of lengths six centimetres and nine centimetres, respectively. If the area of the triangle is 54
centimetres squared, find the lengths of 𝐴𝐵 and 𝐴𝐶.
Our first step is to let 𝐸 and 𝐹
be the points, where the tangents 𝐴𝐶 and 𝐴𝐵 touch the circle. One of our circle theorems states
that two tangents to a circle from the same point will be of equal length. This means that the lengths 𝐵𝐹
and 𝐵𝐷 will be equal. They will both be six
centimetres. In the same way, the lengths 𝐶𝐸
and 𝐶𝐷 will both be equal to nine centimetres. Finally, the two tangents from 𝐴,
𝐴𝐸 and 𝐴𝐹, will also be of equal length. In this case, we will call them 𝑥
We were asked in the question to
find the lengths of 𝐴𝐵 and 𝐴𝐶. These will be equal to 𝑥 plus six
and 𝑥 plus nine, respectively. We are also told in the question
that the area of our triangle was 54 centimetres squared.
We can split the larger triangle
into three smaller triangles: triangle 𝑂𝐵𝐶, triangle 𝑂𝐴𝐶, and triangle
𝑂𝐴𝐵. We know that the sum of the areas
of these three triangles is equal to 54. The area of any right-angled
triangle is equal to the base multiplied by the height divided by two.
Let’s firstly consider triangle
𝑂𝐵𝐶. This triangle has a base of 15
centimetres and a height of three centimetres. Therefore, we can calculate its
area by multiplying 15 by three and dividing by two. 15 multiplied by three is equal to
45. Therefore, the area of triangle
𝑂𝐵𝐶 is 45 over two.
Next, let’s consider triangle
𝑂𝐴𝐶. This triangle has a base of 𝑥 plus
nine and a height once again of three, the radius of the circle. Therefore, the area is given by 𝑥
plus nine multiplied by three all divided by two. Expanding or multiplying out the
bracket gives us three 𝑥 plus 27 as three multiplied by 𝑥 is three 𝑥 and three
multiplied by nine is equal to 27. Therefore, the area of triangle
𝑂𝐴𝐶 is three 𝑥 plus 27 over two.
Our third triangle 𝑂𝐴𝐵 has a
base of 𝑥 plus six. It also has a height of three. Therefore, the area is given by 𝑥
plus six multiplied by three divided by two. Once again, expanding the bracket
on the numerator gives us three 𝑥 plus 18. Therefore, the area of triangle
𝑂𝐴𝐵 is three 𝑥 plus 18 divided by two.
We know that the sum of these three
expressions equals 54. This can be written as 45 over two
plus three 𝑥 plus 27 over two plus three 𝑥 plus 18 over two is equal to 54. We can get rid of the denominator
of the three fractions on the left-hand side by multiplying all four terms by
On the left-hand side, the twos
cancel and on the right-hand side we get 108 as 54 multiplied by two is 108. Collecting the like terms on the
left-hand side gives us six 𝑥 plus 90. Three 𝑥 plus three 𝑥 is equal to
six 𝑥 and 45 plus 27 plus 18 is equal to 90. We are, therefore, left with six 𝑥
plus 90 is equal to 108. Subtracting 90 from both sides of
this equation gives us six 𝑥 is equal to 18. And dividing both sides of this new
equation by six gives us a value of 𝑥 equal to three. 18 divided by six is equal to
Remember we wanted to work out the
length of 𝐴𝐵 and 𝐴𝐶. Well, 𝐴𝐵 was equal to 𝑥 plus six
and 𝐴𝐶 was equal to 𝑥 plus nine. Substituting in 𝑥 equals three
give us values of nine centimetres and 12 centimetres, respectively.
The line 𝐴𝐵 has length nine
centimetres and the line 𝐴𝐶 has length 12 centimetres.