This worksheet guides students through Pythagoras' Theorem by starting with basic square and square root calculations before applying the theorem to find missing sides.

Let’s work through each part of the worksheet step by step.

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Section A: Squares and Square Roots

We’ll calculate each one carefully.

1. 3² = 9
→ 3 × 3 = 9

2. 5² = 25
→ 5 × 5 = 25

3. 7² = 49
→ 7 × 7 = 49

4. 9² = 81
→ 9 × 9 = 81

5. 14² = 196
→ 14 × 14 = 196

6. 25² = 625
→ 25 × 25 = 625

7. 118² = ?
→ 118 × 118
Let’s compute:
(100 + 18)² = 100² + 2×100×18 + 18² = 10000 + 3600 + 324 = 13924

8. 6² + 2² = 36 + 4 = 40

9. 4.1² – 3.5² = ?
→ 4.1 × 4.1 = 16.81
→ 3.5 × 3.5 = 12.25
→ 16.81 – 12.25 = 4.56

10. √16 = 4
→ Because 4 × 4 = 16

11. √64 = 8
→ 8 × 8 = 64

12. √36 = 6
→ 6 × 6 = 36

13. √144 = 12
→ 12 × 12 = 144

14. √196 = 14
→ 14 × 14 = 196

15. √12 ≈ ?
→ Not a perfect square. √12 = √(4×3) = 2√3 ≈ 3.464 (but if exact form is needed, leave as 2√3 — but since others are decimals or integers, probably decimal to 3 places: 3.464)

*Wait — looking at context, maybe they want simplified radical? But other answers are integers. Let’s check next ones.*

16. √250 = ?
→ √(25×10) = 5√10 ≈ 5 × 3.162 = 15.811

Again, not integer. Maybe they accept simplified radical? But let’s see pattern.

Actually, looking back — some problems like √(12² + 5²) will give integers. So perhaps for √12 and √250, we leave as simplified radicals? But the instruction says “work out”, so likely decimal approximations.

However, in school worksheets, sometimes they expect exact forms. But let’s look at the last two:

17. √(12² + 5²) = √(144 + 25) = √169 = 13

18. √(30² – 24²) = √(900 – 576) = √324 = 18

So for √12 and √250, since they don’t simplify to integers, and the worksheet doesn’t specify, I’ll give decimal approximations to 3 decimal places.

But wait — let me double-check the worksheet layout. It has 6 rows per column. In third column:

Row 1: √144 → 12
Row 2: √196 → 14
Row 3: √12 → ?
Row 4: √250 → ?
Row 5: √(12²+5²) → 13
Row 6: √(30²–24²) → 18

So yes, √12 and √250 are meant to be calculated as decimals.

√12 ≈ 3.464
√250 ≈ 15.811

But let’s confirm with calculator-style precision:

√12 = 3.464101615... → 3.464
√250 = 15.8113883... → 15.811

Okay.

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Section B: Pythagoras’ Theorem

Recall: In a right-angled triangle,
a² + b² = c², where c is the hypotenuse (longest side, opposite right angle).

For each triangle, identify which side is missing, then apply the formula.

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Triangle a:

Given: legs = 12 and a, hypotenuse = 20
So:
12² + a² = 20²
144 + a² = 400
a² = 400 – 144 = 256
a = √256 = 16

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Triangle b:

Given: legs = b and 10, hypotenuse = 26
So:
b² + 10² = 26²
b² + 100 = 676
b² = 676 – 100 = 576
b = √576 = 24

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Triangle c:

Given: legs = 45 and c, hypotenuse = 117
So:
45² + c² = 117²
2025 + c² = 13689
c² = 13689 – 2025 = 11664
c = √11664

Let’s compute √11664:

Note: 100² = 10000, 110² = 12100 → too big.
Try 108² = (100+8)² = 10000 + 1600 + 64 = 11664 → YES!

So c = 108

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Triangle d:

Given: legs = d and 21, hypotenuse = 35
So:
d² + 21² = 35²
d² + 441 = 1225
d² = 1225 – 441 = 784
d = √784 = 28

(Check: 28×28 = 784 → correct)

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Now, compiling all answers:

Section A:

First column:
3² = 9
5² = 25
7² = 49
9² = 81
14² = 196
25² = 625

Second column:
118² = 13924
6² + 2² = 40
4.1² – 3.5² = 4.56
√16 = 4
√64 = 8
√36 = 6

Third column:
√144 = 12
√196 = 14
√12 ≈ 3.464
√250 ≈ 15.811
√(12² + 5²) = 13
√(30² – 24²) = 18

Section B:

a = 16
b = 24
c = 108
d = 28

All calculations verified.

Final Answer:
Section A:
3² = 9, 5² = 25, 7² = 49, 9² = 81, 14² = 196, 25² = 625
118² = 13924, 6² + 2² = 40, 4.1² – 3.5² = 4.56, √16 = 4, √64 = 8, √36 = 6
√144 = 12, √196 = 14, √12 ≈ 3.464, √250 ≈ 15.811, √(12² + 5²) = 13, √(30² – 24²) = 18

Section B:
a = 16, b = 24, c = 108, d = 28