Back to the lesson plan builder
Lesson length30 min45 min60 min
Questions🔀 New questions

Print the plan for yourself; the student sheet prints on its own page after it.

Math lesson plan · 10th grade (Year 11 in England) · 45 minutes

Name the circle theorem that justifies a step

Geometry

CCSS HSG.C.A.2CCSS HSG.C.A.3England: GCSE Higher Geometry

Objective and success criteria

Students will: Name the circle theorem that justifies a step.

  • I can answer questions on today's skill on my own, without help.
  • I can explain each step and say why it works.
  • I can check my answer and spot a mistake.

Before and after

Needs first: Circle theorems: angle at the centre and angle in a semicircle; Circle theorems: cyclic quadrilaterals and angles in the same segment

Leads to: the next skill in this area.

Materials: Rulers and protractors, Squared or dotted paper, Shape cut-outs, Mini whiteboards and pens, Printed student sheet (below).

Warm-up · 5 min

  • Quick retrieval from earlier learning (Circle theorems: angle at the centre and angle in a semicircle; Circle theorems: cyclic quadrilaterals and angles in the same segment). Students answer on mini whiteboards.
  • Show answers together and note who needs support in the main task.
  1. O is the centre of a circle. A, B and C are on the circle, with C on the major arc. ∠ACB = 39°. Find ∠AOB. → 78Prompt: Both angles stand on the same arc AB.
  2. ABCD is a cyclic quadrilateral. ∠DAB = 108°. Find ∠BCD. → 72Prompt: Are the two angles opposite each other?
  3. O is the centre. ∠AOB = 46° and C is on the major arc. Find ∠ACB. → 23Prompt: Angle at centre = 2 × angle at circumference.

Hook · 5 min

  • Show a photo of a building, a tile pattern or a road sign and ask what shapes and angles students can spot.
  • Share the objective: Name the circle theorem that justifies a step. Read the success criteria together.

I do: model it · 10 min

  • Work through the example out loud, saying why you do each step, not just what you do.
  • Example: O is the centre of the circle. Which theorem justifies this step? ∠OTP = 90°, where PT touches the circle at T
  • Step 1: A tangent is perpendicular to the radius
  • Answer: A tangent is perpendicular to the radius
  • Pause after each step and ask a student to say what comes next.

We do: guided practice · 10 min

  • Do these together. Students try each step on whiteboards before you reveal it.
  • Use the hints as prompts rather than giving the answer.
  1. O is the centre of the circle. Which theorem justifies this step? ∠AOB = 2 × ∠ACB (A tangent is perpendicular to the radius / Angles in the same segment are equal / The angle at the centre is twice the angle at the circumference / The angle in a semicircle is 90°) → The angle at the centre is twice the angle at the circumferencePrompt: Look at where the angle's vertex is: at the centre, on the circle, or outside it.
  2. O is the centre of the circle. Which theorem justifies this step? PA = PB, where PA and PB touch the circle (Angles in the same segment are equal / The angle at the centre is twice the angle at the circumference / Tangents from an outside point are equal in length / The angle in a semicircle is 90°) → Tangents from an outside point are equal in lengthPrompt: Look at where the angle's vertex is: at the centre, on the circle, or outside it.

You do: independent practice · 10 min

  • Students work on the student sheet. Circulate and check the first two answers of every student early.
  • Students who finish: write a question of their own for a partner, with the answer.

The 6 questions are on the student sheet below, with answers in the key.

Exit ticket · 5 min

  • Two questions on their own. Sort answers into got it / nearly / not yet to plan the next lesson.
  1. O is the centre of the circle. Which theorem justifies this step? ∠OTP = 90°, where PT touches the circle at T (The angle in a semicircle is 90° / A tangent is perpendicular to the radius / Opposite angles of a cyclic quadrilateral add up to 180° / The angle at the centre is twice the angle at the circumference) → A tangent is perpendicular to the radiusPrompt: Look at where the angle's vertex is: at the centre, on the circle, or outside it.
  2. O is the centre of the circle. Which theorem justifies this step? ∠ABC + ∠ADC = 180° in quadrilateral ABCD drawn inside the circle (The angle in a semicircle is 90° / Opposite angles of a cyclic quadrilateral add up to 180° / Tangents from an outside point are equal in length / A tangent is perpendicular to the radius) → Opposite angles of a cyclic quadrilateral add up to 180°Prompt: Look at where the angle's vertex is: at the centre, on the circle, or outside it.

Watch out for

  • Thinking a square turned on its corner is no longer a square.
  • Reading the wrong scale on a protractor.

Support and stretch

Support

  • Review first: Circle theorems: angle at the centre and angle in a semicircle.
  • Use concrete materials: rulers and protractors.
  • Give the first step of each question, or a partly worked example to finish.
  • Pair with a confident partner for the guided practice.

Stretch

  • Explain a common mistake in writing and how to fix it.
  • Write a word problem that uses today's skill, then solve a partner's.

Homework and more practice

Free online practice with hints and printable worksheets for this skill: eurekami.com/math/grade-10.

Free lesson plan from eurekami.com/lesson-plan-builder

eurekami

Name the circle theorem that justifies a step

Name

Date

Score/ 6

  1. Question 1

    O is the centre of the circle. Which theorem justifies this step? AB is a diameter, so ∠ACB = 90°

    • A. Angles in the same segment are equal
    • B. The angle in a semicircle is 90°
    • C. Opposite angles of a cyclic quadrilateral add up to 180°
    • D. The angle at the centre is twice the angle at the circumference

  2. Question 2

    O is the centre of the circle. Which theorem justifies this step? ∠OTP = 90°, where PT touches the circle at T

    • A. A tangent is perpendicular to the radius
    • B. Angles in the same segment are equal
    • C. Tangents from an outside point are equal in length
    • D. The angle at the centre is twice the angle at the circumference

  3. Question 3

    O is the centre of the circle. Which theorem justifies this step? ∠AOB = 2 × ∠ACB

    • A. The angle in a semicircle is 90°
    • B. The angle at the centre is twice the angle at the circumference
    • C. Tangents from an outside point are equal in length
    • D. A tangent is perpendicular to the radius

  4. Question 4

    O is the centre of the circle. Which theorem justifies this step? ∠ADB = ∠ACB, both standing on chord AB on the same side

    • A. A tangent is perpendicular to the radius
    • B. Opposite angles of a cyclic quadrilateral add up to 180°
    • C. Angles in the same segment are equal
    • D. Tangents from an outside point are equal in length

  5. Question 5

    O is the centre of the circle. Which theorem justifies this step? ∠ABC + ∠ADC = 180° in quadrilateral ABCD drawn inside the circle

    • A. A tangent is perpendicular to the radius
    • B. Opposite angles of a cyclic quadrilateral add up to 180°
    • C. The angle at the centre is twice the angle at the circumference
    • D. The angle in a semicircle is 90°

  6. Question 6

    O is the centre of the circle. Which theorem justifies this step? PA = PB, where PA and PB touch the circle

    • A. The angle in a semicircle is 90°
    • B. Opposite angles of a cyclic quadrilateral add up to 180°
    • C. Tangents from an outside point are equal in length
    • D. Angles in the same segment are equal

Free from eurekami.com

for the adult only

Answer key: Name the circle theorem that justifies a step

  1. 1. B. The angle in a semicircle is 90°

  2. 2. A. A tangent is perpendicular to the radius

  3. 3. B. The angle at the centre is twice the angle at the circumference

  4. 4. C. Angles in the same segment are equal

  5. 5. B. Opposite angles of a cyclic quadrilateral add up to 180°

  6. 6. C. Tangents from an outside point are equal in length