--- title: "Squares and square roots" subtitle: "Number Foundations · Squares, powers and primes · Lesson 4" --- # Squares and square roots **Name:** {{field-24}} **Date:** {{field-15}} **Class:** {{field-15}} > **Key question:** A square patio uses 49 tiles. How long is each side? Squares and square roots go back and forth between a side and an area. :::teacher-note **In one sentence:** Students grow a square on a grid to see a square number as an area, learn the perfect squares to 144, and use the square root as the inverse, the side of a square with a given area (AC9M7N01, nodes 1 and 2). **Students should be able to say:** I can work out the square of a number and list the perfect squares to 144. I can find the square root of a perfect square and explain what it means. I can find the side and the perimeter of a square from its area. **Width:** the unmarked parts are the floor every student does (predict, run, investigate and make). Typical adds the modify practice; wider adds the real-world contexts. They add contexts, not difficulty. Students work in the online tutorial alongside this sheet. ::: ## Predict Picture a square with side 5. How many unit squares does it cover? :::keep-together **What is $5^2$?** ☐ 7 ☐ 10 ☐ 25 ☐ 52 My prediction, and why: {{lines-3}} ::: :::teacher-note **What happens:** 25. 5² means 5 × 5: a square with side 5 has 5 rows of 5 unit squares. 10 is 5 × 2, double, not squared. ::: ## Run Try it on the board in the tutorial. Was your prediction right? What was different? {{lines-2}} ## Squaring and its inverse :::keep-together ### Undo the square The square root asks: what number, times itself, makes this? 1. $6^2 = 6 \times 6 = 36$ — Squaring: multiply the number by itself (a square with side 6). 2. $\sqrt{36} = 6$ — The square root goes back: 6 × 6 = 36, so the side is 6. 3. $1,\ 4,\ 9,\ 16,\ 25,\ 36,\ 49,\ 64,\ 81,\ 100,\ 121,\ 144$ — The perfect squares from 1² to 12². Learn them. Now try one: $\sqrt{81}$. **Work out $\sqrt{81}$.** Answer: {{field-15}} ::: :::teacher-note **Answer:** 9 9 × 9 = 81, so √81 = 9. ::: :::keep-together ### Find the mistake **Ava finds the area of a square with side 5 cm. Click the line where her working goes wrong.** ``` 1 A square has side 5 cm. Find its area. 2 Area = side squared. 3 5² = 5 × 2 = 10 4 So the area is 10 cm². ``` Line: {{field-15}} Why is it wrong? {{lines-2}} ::: :::teacher-note **Answer:** line 3 Line 3: 5² means 5 × 5 = 25, not 5 × 2. The area is 25 cm². ::: :::keep-together ### Why it works **Why is $\sqrt{64} = 8$?** ☐ Because $8 \times 8 = 64$ ☐ Because 64 is even ☐ Because $64 \div 8 = 8$ and $64 \div 2 = 32$ ::: :::teacher-note **Answer:** Because $8 \times 8 = 64$ The square root of 64 is the number that times itself makes 64: 8 × 8 = 64. ::: :::typical ## One change at a time **The biggest in the list.** Each task changes one thing. **A root.** **A side from an area.** **Two steps: the perimeter.** **Not a perfect square.** Work in the tutorial. Plan first: what will you change, and how will you test it? {{box-4}} ::: ## Squares and roots :::keep-together ### Square **Work out $9^2$.** Answer: {{field-15}} ::: :::teacher-note **Answer:** 81 9 × 9 = 81. ::: :::keep-together ### Root **Work out $\sqrt{100}$.** Answer: {{field-15}} ::: :::teacher-note **Answer:** 10 10 × 10 = 100. ::: :::keep-together ### A side **A square rug has area 9 m². How long is each side, in metres?** Answer: {{field-15}} ::: :::teacher-note **Answer:** 3 √9 = 3, so each side is 3 m. ::: :::keep-together ### Review: integers **Review: work out $-4 \times 3$.** Answer: {{field-15}} ::: :::teacher-note **Answer:** -12 4 × 3 = 12, and the signs differ, so −12. ::: ## Make: a bigger square Then write its square root, and check by multiplying. Work in the tutorial. Plan first: what will you change, and how will you test it? {{box-4}} ## Exit ticket :::keep-together **A square has area 81 cm². Find its side and its perimeter.** {{lines-3}} ::: :::teacher-note **Answer:** Side √81 = 9 cm; perimeter 4 × 9 = 36 cm. :::