Angles in a Circle and Cyclic Quadrilaterals: Class 9

Understand Angles in a Circle and Cyclic Quadrilaterals

Angles in a Circle and Cyclic Quadrilaterals: formulas to remember

Central angle theorem

$\angle AOB=2\angle APB$

$P$ on the remaining circle; $O$ is the centre.

Angles in the same segment

$\angle APB=\angle AQB$

$P$, $Q$ on the same side of $AB$.

Angle in a semicircle

$\angle APB=90^{\circ }\text{ when }AB\text{ is a diameter}$

Converse: a right angle at $P$ means $AB$ is a diameter of the circle through $A,P,B$.

Cyclic quadrilateral

$\angle A+\angle C=180^{\circ },\angle B+\angle D=180^{\circ }$

Opposite angles are supplementary.

Exterior angle

$\text{exterior angle}=\text{interior opposite angle}$

Produce one side of a cyclic quadrilateral.

Circumradius of a right triangle

$R=\frac{\text{hypotenuse}}{2}$

The circumcentre is the midpoint of the hypotenuse.

How to solve Angles in a Circle and Cyclic Quadrilaterals questions

1. 1Mark the centre if given, and note which arc each given angle stands on.
2. 2For an angle at the centre and one on the circle on the same arc, use $\text{central}=2\times \text{inscribed}$.
3. 3Look for angles in the same segment (equal) and for a diameter (right angle).
4. 4If four points of a quadrilateral are on the circle, use opposite angles summing to $180^{\circ }$, or exterior angle = interior opposite angle.
5. 5Fill in the remaining angles with the angle sum of a triangle ($180^{\circ }$) and the linear pair.
6. 6Write a reason beside every line, and check the answer with a second route if one exists.

Angles in a Circle and Cyclic Quadrilaterals in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Tricks to remember Angles in a Circle and Cyclic Quadrilaterals

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Centre sees double

  The centre sees an arc at twice the angle that the circle sees it. Half the central angle gives any inscribed angle.

  Use it when: Finding an angle on the circle from the centre, or the reverse.

• Opposite angles make a straight line

  Opposite angles of a cyclic quadrilateral add to $180^{\circ }$, the angle of a straight line.

  Use it when: Cyclic quadrilateral numericals and proofs.

• Same side, same angle

  Two viewers on the same side of a chord see it at the same angle.

  Use it when: Angles in the same segment.

Angles in a Circle and Cyclic Quadrilaterals: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Where students lose marks in Angles in a Circle and Cyclic Quadrilaterals

The slips examiners see most often, and what to write instead.

• ✗Mistake: Applying the central angle theorem to the wrong arc

  Using the inscribed angle from the same side as the centre, on the minor arc.

  Wrong working

  $P$ on the minor arc $AB$ with $\angle AOB=100^{\circ }$, so $\angle APB=50^{\circ }$.

  Right working

  For $P$ on the minor arc, $\angle APB=\frac{1}{2}(360^{\circ }-100^{\circ })=130^{\circ }$.

  ✓Do this: Ask: is $P$ on the major arc or the minor arc? The reflex angle belongs with the minor-arc point.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Pairing the wrong angles in a cyclic quadrilateral

  Adding adjacent angles to $180^{\circ }$.

  Wrong working

  $\angle A+\angle B=180^{\circ }$.

  Right working

  $\angle A+\angle C=180^{\circ }$ (opposite).

  ✓Do this: Label the quadrilateral in order and add angles that do not share a side.

  Usually costs: The whole result.

• ✗Mistake: No reason beside each step

  Writing the angle without naming the theorem.

  Wrong working

  $\angle ACB=55^{\circ }$.

  Right working

  $\angle ACB=55^{\circ }$ (angle at the centre is twice the angle on the circle).

  ✓Do this: Write the theorem in brackets on every line.

  Usually costs: ½ mark per step.

Practice questions on Angles in a Circle and Cyclic Quadrilaterals

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   In the figure, $O$ is the centre and $\angle ACB=35^{\circ }$ with $C$ on the major arc $AB$. Then $\angle AOB$ is
   (a) $17.5^{\circ }$
   (b) $35^{\circ }$
   (c) $70^{\circ }$
   (d) $140^{\circ }$

   Hint

   Central angle is twice the angle on the circle.

   Answer

   (c) $70^{\circ }$, since $\angle AOB=2\times 35^{\circ }$.

2. Q2 · 3 marks · Short answerBoard level

   $ABCD$ is a cyclic quadrilateral. Side $AB$ is produced to $E$ and $\angle CBE=82^{\circ }$. Find $\angle ADC$. Then, if $\angle BAD=65^{\circ }$, find $\angle BCD$.

   Hint

   The exterior angle equals the interior opposite angle.

   Answer

   $\angle ADC=\angle CBE=82^{\circ }$ (exterior angle of a cyclic quadrilateral equals the interior opposite angle). $\angle BCD=180^{\circ }-65^{\circ }=115^{\circ }$ (opposite angles supplementary).

3. Q3 · 3 marks · Short answerHarder board level

   A right-angled triangle has legs 9 cm and 12 cm. Find the radius of the circle passing through its three vertices, and the distance of the circumcentre from the right-angle vertex. Justify with a circle theorem.

   Hint

   The hypotenuse is a diameter because the angle opposite it is $90^{\circ }$.

   Answer

   Hypotenuse $=\sqrt{81+144}=15$ cm. Angle in a semicircle is $90^{\circ }$, so the converse says the hypotenuse is a diameter. Radius $=7.5$ cm, and the circumcentre (midpoint of the hypotenuse) is $7.5$ cm from the right-angle vertex.

4. Q4 · 5 marks · Long answerHarder board level

   $AB$ is a diameter of a circle, and $C$ is a point on the circle with $\angle CAB=2x$ and $\angle ABC=3x-10$ (in degrees). Find $x$, then find the three angles of $△ABC$ and show that $AC<AB$.

   Hint

   Use angle in a semicircle for $\angle ACB$.

   Answer

   $\angle ACB=90^{\circ }$ (semicircle). Then $2x+3x-10=90$, so $5x=100$ and $x=20$. Angles: $\angle CAB=40^{\circ }$, $\angle ABC=50^{\circ }$, $\angle ACB=90^{\circ }$. $AB$ is a diameter, the longest chord of the circle, so $AC<AB$. (Equally, $AB$ is the hypotenuse of the right triangle $ABC$.)

5. Q5 · 4 marks · Case-basedHOTS / competency

   Case: four lamp posts $P,Q,R,S$ stand on a circular plaza in that order. The angles of quadrilateral $PQRS$ are $\angle P=(3y+6{)}^{\circ }$, $\angle Q=(4y{)}^{\circ }$, $\angle R=(6y-6{)}^{\circ }$ and $\angle S$. (i) Find $y$. (ii) Find $\angle S$. (iii) A surveyor claims that diagonal $PR$ is a diameter. What would $\angle PQR$ have to be? (iv) Is the claim true?

   Hint

   Opposite angles add to $180^{\circ }$; check against a diameter.

   Answer

   (i) $\angle P+\angle R=180^{\circ }$: $3y+6+6y-6=9y=180$, so $y=20$. (ii) $\angle P=66^{\circ }$, $\angle R=114^{\circ }$, $\angle Q=80^{\circ }$, so $\angle S=180^{\circ }-80^{\circ }=100^{\circ }$. (iii) If $PR$ were a diameter, $\angle PQR$ would be $90^{\circ }$ (angle in a semicircle). (iv) But $\angle Q=80^{\circ }\ne 90^{\circ }$, so the claim is false: $PR$ is not a diameter.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Angle at centre versus angle on the circle?Show answer

   The central angle is double.

2. 2.Angle in a semicircle?Show answer

   $90^{\circ }$.

3. 3.Opposite angles of a cyclic quadrilateral?Show answer

   They add up to $180^{\circ }$.

4. 4.Are angles in the same segment equal?Show answer

   Yes.

5. 5.Which parallelograms are cyclic?Show answer

   Only rectangles (including squares).

Angles in a Circle and Cyclic Quadrilaterals: frequently asked questions