Grade 11 Maths · Analytical Geometry

Use angle in a semicircle

Use the fact that an angle in a semicircle is 90°, so an angle standing on a diameter is a right angle.

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Circle Geometry Grade 11 : Angle in Semi CircleKevinmathscience

Circle Geometry Grade 11 Euclidean Geometry: Angle in semi-circleMiss Martins Maths and Science

Angle in semicircle worked examplesGraphicMaths

Angle in a Semicircle (Circle Theorems)StudyPass

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Question 6.2 · March 2022 · Provincial · 4 marks

In the diagram below, P, Q, R and S are points on the circle. QS and PR intersect at point T. The line from V is a tangent at P.

Q \hat{R} P = 5 6^{\circ}

and

R \hat{P} V = 7 0^{\circ}

6.2 If

\hat{Q}_{1} = 3 7^{\circ}

, then explain why QS is not a diameter of the circle.(4)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: use angle in a semicircle

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3. Question 5.2.4 · March 2022 · Provincial · 2 marks
5.2 In the diagram, O is the centre of the circle and L is a point on the circumference. RLM is a tangent at L.
5.2.4 Determine with reasons the size of $\hat{Q}_{1}$ .(2)
Show the memo (1 tick)Memo
1. ✓
  Angle in semi-circle $M \hat{L} P = 9 0^{\circ}$ (This reasoning in memo might refer to $P L Q$ being $90$ if $PQ$ diameter, but $PQ$ is diameter through O. Let's assume P,O,Q collinear). If $POQ$ is diameter, $\hat{L}_{2 + 3 + 4} = 9 0^{\circ}$ .
  Memo says: $M \hat{L} P = 9 0^{\circ}$ (angle in semi circle?) - wait, $R L M$ is tangent. $O L ⊥ R L M$ . So $M \hat{L} O = 90$ . If $P - O - Q$ is a straight line, $P \hat{L} Q = 90$ .
  Memo calculates $\hat{Q}_{1} = 9 0^{\circ} - \hat{P} = 5 8^{\circ}$ (Sum of angles in $Δ PQ L$ ).
  2 marks · Use angle in a semicircle
4. Question 3.3.2 · March 2022 · Provincial · 2 marks
Give reasons for your statements and calculations in Question 3.
3.3 In the diagram, O is centre of the circle. Points A, B, C and D lie on the circumference of the circle. BOD is a diameter. $\hat{A}_{1} = 5 1^{\circ}$ and $\hat{B}_{1} = 2 9^{\circ}$ . A line is drawn from O to F such that the length of BC = 24 m and radius of the circle centred O is 13 m.
3.3.2 Determine the size of $\hat{A}_{2}$ .(2)
Show the memo (2 ticks)Memo
1. ✓
  Statement $D \hat{A} B = 9 0^{\circ}$ , so $\hat{A}_{2} = 9 0^{\circ} - 5 1^{\circ} = 3 9^{\circ}$
  1 mark · Use angle in a semicircle
2. ✓
  Reason $∠$ in semi-circle (BD is a diameter)
  1 mark · Use angle in a semicircle
5. Question 5.2.1 · March 2020 · Provincial · 3 marks
5.2 AE is a diameter of the circle. AE is produced to D. DB is a tangent to the circle at B. DC is perpendicular to AD. CBA is a straight line. BE and EC are drawn. Let $\hat{A} = x .$
5.2.1 Prove that BCDE is a cyclic quadrilateral.(3)
Show the memo (2 ticks)Memo
1. ✓
  Identify right angle at B $\hat{B}_{3} = 9 0^{\circ}$ ( $∠$ in semi-circle)
  2 marks · Use angle in a semicircle
2. ✓
  Conclude Cyclic Quad $∴ \hat{B}_{3} = C \hat{D} E$ (both $= 9 0^{\circ}$ )
  BCDE is a cyclic quadrilateral (converse: ext. $∠$ of cyclic quad)
  1 mark · Prove a quadrilateral is cyclic
6. Question 8.2.1 · March 2022 · Provincial · 4 marks
In the diagram, $A B$ lies on circle with centre $O$ , and is produced to $C$ which lies on circle with centre $P$ . Similarly, $FE$ is produced to $D$ . $BE ∥ C D$ .
8.2 In the diagram, $A B$ is the diameter of circle with centre $P$ . $EP ⊥ A B$ .
8.2.1 Prove that $Δ BPE ∣∣∣ Δ B D A$ .(4)
Show the memo (3 ticks)Memo
1. ✓
  State common angle $\hat{B}$ is common
  1 mark · Prove triangles similar
2. ✓
  Identify 90 degree angles $\hat{D} = 9 0^{\circ}$ (Angle in semi-circle)
  $\hat{P}_{2} = 9 0^{\circ}$ (Given $EP ⊥ A B$ )
  2 marks · Use angle in a semicircle
3. ✓
  Conclusion $Δ BPE ∣∣∣ Δ B D A$ (AAA)
  1 mark · Prove triangles similar

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