Perpendicular bisector
In the diagram below, AB is the chord of a circle with centre O.
OM is perpendicular to AB (meet at a right angle).
Look at triangles OAM and OBM.
The hypotenuses (OA and OB) are the same, as they are both the radius of the circle.
OM is common to both triangles.
OMA and OMB are both right angles.
Triangles OAM and OBM are congruent, so it follows that \(AM=MB.\)
Therefore, M is the midpoint of AB, and the chord has been bisected.
Since these properties allow us to form a right-angled triangle, often these questions use Pythagoras.
Question
The entrance to a tunnel is shown in the picture.
The road is nine metres wide and the radius is seven metres.
Calculate the height of the tunnel.
\(QP=9\div2=4.5m\)
Now we can use Pythagoras to calculate the length of OQ.
\(O{Q^2}={7^2}-{4.5^2}=28.75\)
\(OQ=\sqrt{28.75}=5.3619...\)
Since OR is the radius, then \(OR=7m\).
Therefore the height of the tunnel equals:
\(7+5.3619...\)
\(=12.3619...\)
\(=12.4m\,(to\,1\,d.p.)\)
More guides on this topic
- Determine the gradient of a straight line
- Circle geometry
- Calculating the volume of a standard solid
- Applying Pythagoras Theorem
- Using similarity
- Working with two-dimensional vectors
- Working with three-dimensional coordinates
- Using vector components
- Calculating the magnitude of a vector
- An Approximate History of Co-ordinates