Worksheet) Pythagoras Theorem | PDF | Elementary Geometry ... - Free Printable

Let's solve each of these problems using the Pythagorean Theorem, which states:

> For a right triangle:
> $$
> a^2 + b^2 = c^2
> $$
> where:
> - $a$ and $b$ are the legs (the two sides forming the right angle),
> - $c$ is the hypotenuse (the side opposite the right angle, the longest side).

We will identify the known sides and use the formula to find the missing side.

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1. Triangle ABC: $BC = 5$, $AB = 12$, $AC = ?$

- Right angle at B → AB and BC are legs, AC is hypotenuse.
- $AC^2 = AB^2 + BC^2 = 12^2 + 5^2 = 144 + 25 = 169$
- $AC = \sqrt{169} = \boxed{13}$

---

2. Triangle ABC: $BC = 3$, $AB = 4$, $AC = ?$

- Right angle at B → AB and BC are legs.
- $AC^2 = 3^2 + 4^2 = 9 + 16 = 25$
- $AC = \sqrt{25} = \boxed{5}$

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3. Triangle ABC: $BC = 8$, $AB = 6$, $AC = ?$

- Right angle at B → AB and BC are legs.
- $AC^2 = 8^2 + 6^2 = 64 + 36 = 100$
- $AC = \sqrt{100} = \boxed{10}$

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4. Triangle ABC: $BC = 7$, $AB = 24$, $AC = ?$

- Right angle at B → AB and BC are legs.
- $AC^2 = 7^2 + 24^2 = 49 + 576 = 625$
- $AC = \sqrt{625} = \boxed{25}$

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5. Triangle ABC: $AB = 12$, $BC = 9$, $AC = ?$

- Right angle at B → AB and BC are legs.
- $AC^2 = 12^2 + 9^2 = 144 + 81 = 225$
- $AC = \sqrt{225} = \boxed{15}$

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6. Triangle ABC: $BC = 8$, $AB = 15$, $AC = ?$

- Right angle at B → AB and BC are legs.
- $AC^2 = 15^2 + 8^2 = 225 + 64 = 289$
- $AC = \sqrt{289} = \boxed{17}$

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7. Triangle ABC: $AC = 20$, $AB = 12$, $BC = ?$

- Right angle at B → AB and BC are legs, AC is hypotenuse.
- $BC^2 = AC^2 - AB^2 = 20^2 - 12^2 = 400 - 144 = 256$
- $BC = \sqrt{256} = \boxed{16}$

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8. Triangle ABC: $BC = 5$, $AC = 13$, $AB = ?$

- Right angle at B → AB and BC are legs, AC is hypotenuse.
- $AB^2 = AC^2 - BC^2 = 13^2 - 5^2 = 169 - 25 = 144$
- $AB = \sqrt{144} = \boxed{12}$

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9. Triangle ABC: $BC = 18$, $AC = 30$, $AB = ?$

- Right angle at B → AB and BC are legs, AC is hypotenuse.
- $AB^2 = AC^2 - BC^2 = 30^2 - 18^2 = 900 - 324 = 576$
- $AB = \sqrt{576} = \boxed{24}$

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10. Triangle ABC: $AB = 24$, $AC = 40$, $BC = ?$

- Right angle at B → AB and BC are legs, AC is hypotenuse.
- $BC^2 = AC^2 - AB^2 = 40^2 - 24^2 = 1600 - 576 = 1024$
- $BC = \sqrt{1024} = \boxed{32}$

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11. Triangle ABC: $BC = 80$, $AC = 100$, $AB = ?$

- Right angle at B → AB and BC are legs, AC is hypotenuse.
- $AB^2 = AC^2 - BC^2 = 100^2 - 80^2 = 10000 - 6400 = 3600$
- $AB = \sqrt{3600} = \boxed{60}$

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12. Triangle ABC: $AC = 85$, $BC = 68$, $BC = ?$ ← Wait, it says "BC = ?" but gives BC = 68?

Wait — looking again:

- $AC = 85$, $BC = 68$, $AB = ?$, right angle at B.

So AB and BC are legs, AC is hypotenuse.

- $AB^2 = AC^2 - BC^2 = 85^2 - 68^2 = 7225 - 4624 = 2601$
- $AB = \sqrt{2601} = \boxed{51}$

But question asks for BC = ? — but BC is already given as 68? That seems like a typo.

Wait — no, let’s recheck diagram: It shows $AC = 85$, $AB = ?$, and $BC = 68$, right angle at B.

So yes, $BC = 68$ is given, so BC is not unknown. But problem says “BC = ?” — contradiction.

Wait — perhaps I misread.

Looking again: Problem 12 says: $AC = 85$, $AB = ?$, $BC = ?$, and right angle at B.

But in the diagram, only $AC = 85$ and $BC = 68$ are labeled? Wait — actually, in image:

> In problem 12: Side AC = 85, side AB = ?, and side BC = ? — but in diagram, BC is labeled 68?

Wait — checking carefully: The triangle has:

- Point A, B, C
- Right angle at B
- Side AC = 85 (hypotenuse)
- Side AB = ?
- Side BC = 68? Or is that labeled?

Wait — actually, in your image, it says:

> 12. BC = ?
> Diagram: A---C with AC = 85, B at right angle, and AB = ?, BC = ?

But then it shows 68 near point B? Wait — no, looking at your image:

In problem 12: from A to B is labeled 68, and A to C is 85.

So:

- AB = 68
- AC = 85
- BC = ?

Right angle at B → AB and BC are legs, AC is hypotenuse.

So:
- $BC^2 = AC^2 - AB^2 = 85^2 - 68^2 = 7225 - 4624 = 2601$
- $BC = \sqrt{2601} = \boxed{51}$

So answer is BC = 51

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13. Triangle ABC: $AB = 9$, $BC = 7$, $AC = ?$

- Right angle at B → AB and BC are legs.
- $AC^2 = 9^2 + 7^2 = 81 + 49 = 130$
- $AC = \sqrt{130} \approx \boxed{11.4}$ (exact: $\boxed{\sqrt{130}}$)

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14. Triangle ABC: $BC = 12$, $AC = 19$, $AB = ?$

- Right angle at B → AB and BC are legs, AC is hypotenuse.
- $AB^2 = AC^2 - BC^2 = 19^2 - 12^2 = 361 - 144 = 217$
- $AB = \sqrt{217} \approx \boxed{14.73}$ (exact: $\boxed{\sqrt{217}}$)

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15. Triangle ABC: $BC = 11$, $AB = 15$, $AC = ?$

- Right angle at B → AB and BC are legs.
- $AC^2 = 15^2 + 11^2 = 225 + 121 = 346$
- $AC = \sqrt{346} \approx \boxed{18.6}$ (exact: $\boxed{\sqrt{346}}$)

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16. Triangle ABC: $BC = 5$, $AC = 12$, $AB = ?$

- Right angle at B → AB and BC are legs, AC is hypotenuse.
- $AB^2 = AC^2 - BC^2 = 12^2 - 5^2 = 144 - 25 = 119$
- $AB = \sqrt{119} \approx \boxed{10.9}$ (exact: $\boxed{\sqrt{119}}$)

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17. Triangle ABC: $AC = 23$, $BC = 19$, $AB = ?$

- Right angle at B → AB and BC are legs, AC is hypotenuse.
- $AB^2 = AC^2 - BC^2 = 23^2 - 19^2 = 529 - 361 = 168$
- $AB = \sqrt{168} = \sqrt{4 \cdot 42} = 2\sqrt{42} \approx \boxed{12.96}$

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18. Triangle ABC: $AC = 34$, $BC = 21$, $AB = ?$

- Right angle at B → AB and BC are legs, AC is hypotenuse.
- $AB^2 = AC^2 - BC^2 = 34^2 - 21^2 = 1156 - 441 = 715$
- $AB = \sqrt{715} \approx \boxed{26.74}$ (exact: $\boxed{\sqrt{715}}$)

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## ✔ Final Answers Summary:

| No. | Missing Side | Value |
|-----|--------------|-------|
| 1 | AC | 13 |
| 2 | AC | 5 |
| 3 | AC | 10 |
| 4 | AC | 25 |
| 5 | AC | 15 |
| 6 | AC | 17 |
| 7 | BC | 16 |
| 8 | AB | 12 |
| 9 | AB | 24 |
| 10 | BC | 32 |
| 11 | AB | 60 |
| 12 | BC | 51 |
| 13 | AC | √130 ≈ 11.4 |
| 14 | AB | √217 ≈ 14.73 |
| 15 | AC | √346 ≈ 18.6 |
| 16 | AB | √119 ≈ 10.9 |
| 17 | AB | √168 ≈ 12.96 |
| 18 | AB | √715 ≈ 26.74 |

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✔ All solved using the Pythagorean Theorem: $a^2 + b^2 = c^2$

Let me know if you want exact radicals or decimal approximations!