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Free Grade 3 Compound Shapes in Context Techniques Lesson Plans

Free Grade 3 lesson plans for Compound Shapes in Context Techniques: 12 real questions with a full answer key and worked solutions. Perimeter is the distance around a shape; area is the space inside. Rectangle area = length × width; triangle = ½ × base × height; circle = πr².

Price

Free

Difficulty

Core

Estimated time

30 min

Total marks

24

Learning objectives

Every objective is checkable, and each one maps to an activity and an assessment.

  • UnderstandExplain compound shapes in context techniques

    Explain compound shapes in context techniques accurately in a familiar context.

    Assessed by: Exit ticket of four short items

  • ApplyCalculate compound shapes in context techniques

    Calculate compound shapes in context techniques accurately in a familiar context.

    Assessed by: Exit ticket of four short items

  • AnalyseCompare compound shapes in context techniques

    Compare compound shapes in context techniques accurately in a familiar context.

    Assessed by: Exit ticket of four short items

  • EvaluateJustify compound shapes in context techniques

    Justify compound shapes in context techniques accurately in a familiar context.

    Assessed by: Exit ticket of four short items

Prerequisites

Close these gaps first — each has its own free lesson, worksheet and quiz.

How Compound Shapes in Context Techniques works

Perimeter is the distance around a shape; area is the space inside. Rectangle area = length × width; triangle = ½ × base × height; circle = πr².

Key wordsperimeterareasquare unitsradiuscircumference

12 practice questions

Easy to hard, 24 marks in total. Try them first, then open the answers.

  1. 1.A rectangle is 7 cm by 9 cm. Find its perimeter and area.[1]
  2. 2.A rectangle is 2 cm by 2 cm. Find its perimeter and area.[1]
  3. 3.A rectangle is 3 cm by 7 cm. Find its perimeter and area.[1]
  4. 4.A rectangle is 7 cm by 2 cm. Find its perimeter and area.[1]
  5. 5.A triangle has base 4 cm and height 13 cm. Find its area.[2]
  6. 6.A triangle has base 24 cm and height 10 cm. Find its area.[2]
  7. 7.A triangle has base 30 cm and height 6 cm. Find its area.[2]
  8. 8.A triangle has base 24 cm and height 2 cm. Find its area.[2]
  9. 9.Find the area and circumference of a circle with radius 7 cm (to 1 d.p.).[3]
  10. 10.Find the area and circumference of a circle with radius 4 cm (to 1 d.p.).[3]
  11. 11.Find the area and circumference of a circle with radius 3 cm (to 1 d.p.).[3]
  12. 12.Find the area and circumference of a circle with radius 2 cm (to 1 d.p.).[3]
Show answers and working
  1. 1. Perimeter 32 cm, area 63 cm²

    Perimeter = 7 + 9 + 7 + 9 = 32 cm. Area = 7 × 9 = 63 cm².

  2. 2. Perimeter 8 cm, area 4 cm²

    Perimeter = 2 + 2 + 2 + 2 = 8 cm. Area = 2 × 2 = 4 cm².

  3. 3. Perimeter 20 cm, area 21 cm²

    Perimeter = 3 + 7 + 3 + 7 = 20 cm. Area = 3 × 7 = 21 cm².

  4. 4. Perimeter 18 cm, area 14 cm²

    Perimeter = 7 + 2 + 7 + 2 = 18 cm. Area = 7 × 2 = 14 cm².

  5. 5. 26 cm²

    Area = ½ × base × height. ½ × 4 × 13 = 26 cm².

  6. 6. 120 cm²

    Area = ½ × base × height. ½ × 24 × 10 = 120 cm².

  7. 7. 90 cm²

    Area = ½ × base × height. ½ × 30 × 6 = 90 cm².

  8. 8. 24 cm²

    Area = ½ × base × height. ½ × 24 × 2 = 24 cm².

  9. 9. Area 153.9 cm², circumference 44 cm

    Area = πr² = π × 7² = 153.9 cm². Circumference = 2πr = 2 × π × 7 = 44 cm.

  10. 10. Area 50.3 cm², circumference 25.1 cm

    Area = πr² = π × 4² = 50.3 cm². Circumference = 2πr = 2 × π × 4 = 25.1 cm.

  11. 11. Area 28.3 cm², circumference 18.8 cm

    Area = πr² = π × 3² = 28.3 cm². Circumference = 2πr = 2 × π × 3 = 18.8 cm.

  12. 12. Area 12.6 cm², circumference 12.6 cm

    Area = πr² = π × 2² = 12.6 cm². Circumference = 2πr = 2 × π × 2 = 12.6 cm.

Worked examples

Easy example

A rectangle is 4 cm by 7 cm. Find its perimeter and area.

  1. Perimeter = 4 + 7 + 4 + 7 = 22 cm.
  2. Area = 4 × 7 = 28 cm².
  3. Answer: Perimeter 22 cm, area 28 cm²
Medium example

A triangle has base 26 cm and height 13 cm. Find its area.

  1. Area = ½ × base × height.
  2. ½ × 26 × 13 = 169 cm².
  3. Answer: 169 cm²
Hard example

Find the area and circumference of a circle with radius 10 cm (to 1 d.p.).

  1. Area = πr² = π × 10² = 314.2 cm².
  2. Circumference = 2πr = 2 × π × 10 = 62.8 cm.
  3. Answer: Area 314.2 cm², circumference 62.8 cm

Interactive practice

Free, no account required — answer on screen or print it.

Common mistakes

What learners typically get wrong, and how to fix it.

  • Mixes up area and perimeter.

    Correction: Perimeter is a fence (around), area is grass (inside).

  • Uses the slanted side as the height of a triangle.

    Correction: Height must be at right angles to the base.

Teacher tips

  • · Draw shapes on squared paper and count squares first.

Parent tips

  • · Measure a room and work out how much carpet you'd need.

Real-life applications

  • · Carpeting a room, fencing a garden.

Assessment objectives

How mastery is evidenced, and which standards it maps to.

EDEXCEL EDE-299.1 — Mapped statement covering Compound Shapes in Context Techniques.CBSE CBS-346.2 — Mapped statement covering Compound Shapes in Context Techniques.

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Frequently asked

Is this Grade 3 Compound Shapes in Context Techniques lesson plan really free?
Yes. Every lesson plans on FreeMathWorksheet is free to use, print and share — no account, no paywall and no watermark.
What should a learner already know before this?
Start with Compound Shapes in Context Essentials, Working with Compound Shapes. Each one has its own free lesson, worksheet and quiz.
How is the lesson plan sequenced?
Recognition first, then understanding, application, reasoning and finally mastery — the same ladder used across every free resource on the platform.
What comes next after Compound Shapes in Context Techniques?
Move on to Compound Shapes in Context Word Problems, Compound Shapes in Context Common Errors, Introducing Compound Shapes, Working with Compound Shapes, which build directly on this idea.

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