When a triangle is formed inside a semicircle, two lines from either side of the diameter meet at a point on the circumference at a right angle.

The angle in a semicircle is a right angle of \(90^\circ\).

- Question
In the diagram PR is a diameter and \(\angle PRQ = 25^\circ\).

What is the size of \(\angle QPR\)?

\(\angle PQR = 90^\circ\) since it is the angle in a semicircle.

The three angles in the triangle add up to \(180^\circ\), therefore:

\[\angle QPR = 180^\circ - 90^\circ - 25^\circ\]

\[\angle QPR = 65^\circ\]

- Question
In the diagram KL is a diameter of the circle and is 8 cm long.

LM = 3 cm.

Calculate the size of KM.

KL is a diameter so we have an angle in a semicircle therefore \(\angle KML = 90^\circ\).

We have a right-angled triangle and so can use Pythagoras.

KM is not the hypotenuse so:

\[K{M^2} = {8^2} - {3^2} = 55\]

\[KM = \sqrt {55} = 7.461...\]

\[KM = 7.4cm\,(to\,1\,d.p.)\]