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Math lesson plan · 10th grade (Year 11 in England) · 30 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 · 4 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 = 50°. Find ∠AOB. → 100Prompt: Both angles stand on the same arc AB.
  2. ABCD is a cyclic quadrilateral. ∠ABC = 66°. Find ∠ADC. → 114Prompt: Are the two angles opposite each other?
  3. O is the centre. ∠AOB = 160° and C is on the major arc. Find ∠ACB. → 80Prompt: Angle at centre = 2 × angle at circumference.

Hook · 3 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 · 7 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 · 6 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? PA = PB, where PA and PB touch the circle (Tangents from an outside point are equal in length / Opposite angles of a cyclic quadrilateral add up to 180° / The angle in a semicircle is 90° / The angle at the centre is twice the angle at the circumference) → 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.
  2. O is the centre of the circle. Which theorem justifies this step? ∠AOB = 2 × ∠ACB (The angle in a semicircle is 90° / Angles in the same segment are equal / Opposite angles of a cyclic quadrilateral add up to 180° / The angle at the centre is twice the angle at the circumference) → 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.

You do: independent practice · 7 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 · 3 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? AB is a diameter, so ∠ACB = 90° (The angle in a semicircle is 90° / Angles in the same segment are equal / A tangent is perpendicular to the radius / The angle at the centre is twice the angle at the circumference) → The angle in a semicircle is 90°Prompt: 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? ∠ADB = ∠ACB, both standing on chord AB on the same side (Opposite angles of a cyclic quadrilateral add up to 180° / Angles in the same segment are equal / Tangents from an outside point are equal in length / The angle in a semicircle is 90°) → Angles in the same segment are equalPrompt: 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

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  1. Question 1

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

  2. Question 2

    O is the centre of the circle. Which theorem justifies this step? ∠ABC + ∠ADC = 180° in quadrilateral ABCD drawn inside 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. 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 at the centre is twice the angle at the circumference
    • B. Angles in the same segment are equal
    • C. The angle in a semicircle is 90°
    • 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. 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

  5. Question 5

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

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

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

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Answer key: Name the circle theorem that justifies a step

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

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

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

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

  5. 5. A. The angle in a semicircle is 90°

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