Chapter 21: Review

Return to the chapter problem

The chord could fit in the window. Its diameter is $2(7)=14$ cm, and a chord cannot be longer than the diameter. Since $13<14$, this chord is possible.

The two points on the circumference lie in the same segment and see the same chord. Angles in the same segment are equal, so $x=38^\circ$.

The radius meets the chord at right angles. A perpendicular from the centre bisects a chord, so each half is equal. If the chord length is $c$, each half is $c/2$. The radius and tangent theorem is not needed here because the line meets a chord, not a tangent.

Mixed practice

Work without notes first. For each question, state which method applies and why.

1. A circle has radius $6.5$ cm. Work out its diameter.

1. An angle at the circumference is $29^\circ$. Find the centre angle on the same arc.
2. A cyclic quadrilateral has one angle of $73^\circ$. Find its opposite angle.
3. Two tangents from one point have lengths $3x+2$ and $20$ cm. Find $x$.
4. A perpendicular from the centre meets a chord of length $22$ cm. Find each half.
5. A tangent-chord angle is $41^\circ$. Find the angle in the alternate segment.
6. An angle in a triangle standing on a diameter is $90^\circ$. Another angle is $33^\circ$. Find the third angle and give both reasons.

1. An isosceles triangle has equal angles of $52^\circ$. Find the third angle. Explain why you use the triangle angle sum rather than a circle theorem.

What can you now do?

Write one circle fact you can use confidently. Write one fact you want to practise again, and name a question that would help.