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4.8 G) Chords
4.8 G) Chords
A chord is a straight line that goes from one point on the circumference of the circle to another point on the circumference of the circle. An example of a chord is shown below.
Many people tend to forget about the circle theorem that involves a chord, so it is worth spending a while on this circle theorem and worth getting it down on a revision card.
The circle theorem states that the perpendicular from the centre of the circle to a chord bisects the chord. The perpendicular means that the line going from the centre of the circle to the chord makes an angle of 90° with the chord. The word bisects means that the line going from the centre of the circle to the chord cuts the chord exactly in half. This circle theorem is shown on the diagram below.
The circle theorem states that the perpendicular from the centre of the circle to a chord bisects the chord. The perpendicular means that the line going from the centre of the circle to the chord makes an angle of 90° with the chord. The word bisects means that the line going from the centre of the circle to the chord cuts the chord exactly in half. This circle theorem is shown on the diagram below.
The chord on the above diagram is the straight line ABC. The line OB intersects the chord at point B and it is perpendicular to the chord, which means that angle OBC and OBA are 90° (right angles). Also, the line OB bisects the chord, which means that AB and BC are the same length.
This circle theorem creates two right angle triangles. On the diagram below, the two right angle triangles are ABO and BCO.
This circle theorem creates two right angle triangles. On the diagram below, the two right angle triangles are ABO and BCO.
Common questions that involve this circle theorem also involve either trigonometry or Pythagoras’ theorem. We are going to have a look at one example of each. It may be worth going through the sections on Pythagoras’ theorem and trigonometry before working through the examples in this section (click here to be taken to the Pythagoras’ theorem section and click here to be taken to the trigonometry section).
Example 1
O is the centre of the circle below. The radius of the circle is 9 cm and the chord AB has a length of 10 cm. How far is the midpoint of the chord away from the centre of the circle? Give your answer to 1 decimal place.
O is the centre of the circle below. The radius of the circle is 9 cm and the chord AB has a length of 10 cm. How far is the midpoint of the chord away from the centre of the circle? Give your answer to 1 decimal place.
The first step in answering this question is to draw on a line going from the centre of the circle to the midpoint of the chord. The line that we have drawn on will bisect the chord (it will cut the chord in half/ meet the chord at the midpoint of the chord). I am going to label the point where the line from the centre of the circle meets the chord as C. As the line going from the centre touches the midpoint of the chord, the length of AC and BC will be the same. The length of the whole chord is 10 cm (given at the start of the question), and this means that the length of AC and the length of BC will be 5 cm.
We can also add the length of the radius to the diagram. We are told in the question that the radius is 9 cm. I have drawn another radius going from O to A (the length of OA will be 9 cm).
The line from the centre meets the chord at 90°. We now have two right angle triangles; ACO and CBO. I am going to draw the triangle CBO out below; OB is the radius so it will be 9 cm and CB is 5 cm.
The line from the centre meets the chord at 90°. We now have two right angle triangles; ACO and CBO. I am going to draw the triangle CBO out below; OB is the radius so it will be 9 cm and CB is 5 cm.
We want to find the length of OC and we are going to find it by using Pythagoras’ theorem. His theorem states that the hypotenuse squared is the sum of the squares of the other two sides. The formula for his theorem is shown below:
c in the above formula is the hypotenuse, and a and b are the other two sides (it does not matter which side is a and b out of the two non-hypotenuse sides). We now need to label up the sides on the right angle triangle. The side that is 9 cm is the hypotenuse, so this side will be c. It does not matter which of the non-hypotenuse sides is a and which one is b; I am going to let a be the side that is 5 cm and b be the side that we are trying to work out.
We are now ready to sub the values into the formula.
The first step in finding out the value of b is to isolate all of the b’s; we get all of the b’s to one side of the equation and all of the numbers to the other side. As we have a b on the left and no b’s on the right, it makes sense to get all of the b’s to the left and all of the numbers to the right. Therefore, we need to move the 25 that is currently on the left to the right, which we can do by doing the opposite; we minus 25 from both sides of the equation.
We want to find the value of b and not b2. Therefore, we square root both sides of the equation.
The question asks us to give our answer to one decimal place.
Therefore, the distance of the midpoint from the centre of the circle is 7.5 cm.
Example 2
The radius of the circle below is 12 cm. The chord DE is 16 cm. What is the size of angle ODE? Give your answer to one decimal place.
The radius of the circle below is 12 cm. The chord DE is 16 cm. What is the size of angle ODE? Give your answer to one decimal place.
The circle theorem that we are looking at in this section states that the perpendicular from the centre of a circle to a chord bisects the chord. Therefore, we can draw a line going from the centre of the circle to the midpoint of the chord. I am going to label the point where this line meets the midpoint of the chord F.
Also, I am going to draw another radius going from O to E. The length of OD and OE will be 12 cm.
Also, I am going to draw another radius going from O to E. The length of OD and OE will be 12 cm.
The angle OFD and OFE will be 90° (right angles). Also, the line going from the centre of the circle to the chord bisects the chord. This means that DF and EF will be the same length and they will both be half of the length of the whole chord. The whole chord is 16 cm, which means that DF and EF are 8 cm.
We want to find the size of angle ODE and I am going to use the triangle ODF to find the size of this angle. This triangle is a right angle triangle and the right angle in the triangle is OFD. We know that OD is 12 cm (it is the radius and we were given this in the question) and we know that DF is 8 cm. I am going to label the angle that we are looking for x. A diagram of the triangle ODF is given below.
We want to find the size of angle ODE and I am going to use the triangle ODF to find the size of this angle. This triangle is a right angle triangle and the right angle in the triangle is OFD. We know that OD is 12 cm (it is the radius and we were given this in the question) and we know that DF is 8 cm. I am going to label the angle that we are looking for x. A diagram of the triangle ODF is given below.
We are able to use trigonometry to find the value of x. The trigonometry formula triangles are given below:
We find which of the three trigonometry formula triangles we use by labelling up the right angle triangle. The side OD is the hypotenuse and the side DF is the adjacent.
This means that we are looking for the formula triangle that contains the hypotenuse and the adjacent. Therefore, we will be using the CAH formula triangle.
We find the calculation that we need to undertake by covering up what we are looking for in the formula triangle. We are looking for the angle, so we cover up C in the formula triangle. This tells us that we need to complete the following calculation:
We can now sub the values in.
We want to find the value of x and not cos(x). Therefore, we need to get rid of the cos, which we do by taking the inverse of cos for both sides of the equation; we cos-1 both sides of the equation.
The question asks us to give our answer to one decimal place.
Angle ODE is 48.2°.
End Note
I personally believe that this is one of the trickier circle theorems to remember and I think it is because it doesn’t come up that often. Therefore, I think that it would be very beneficial to get this circle theorem down on a revision card.
I personally believe that this is one of the trickier circle theorems to remember and I think it is because it doesn’t come up that often. Therefore, I think that it would be very beneficial to get this circle theorem down on a revision card.