Practice problems for applying the Pythagorean Theorem to find missing sides of right triangles.

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.
- $AC^2 = 3^2 + 4^2 = 9 + 16 = 25$
- $AC = \sqrt{25} = \boxed{5}$

---

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 BC is already given as 68?

Wait — actually, the triangle has:
- Right angle at B,
- $AC = 85$ (hypotenuse),
- $BC = 68$,
- $AB = ?$

So we're solving for AB.

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

But the question asks for BC = ?, but BC is already labeled as 68? That seems inconsistent.

Wait — look again:
The diagram shows:
- Right angle at B,
- $AC = 85$,
- $BC = 68$,
- But asks: BC = ?

This is likely a typo in labeling. Probably meant to ask for AB = ?

But since it says BC = ?, and BC is marked as 68, maybe it's asking to verify or there's confusion.

Wait — no, perhaps the label "BC = ?" is correct, and the number 68 is on side AB?

Let’s re-analyze:

In triangle ABC:
- Right angle at B,
- Side from A to C is 85 (hypotenuse),
- Side from B to C is labeled 68,
- Side from A to B is unlabeled,
- Question: BC = ?

But BC is already labeled 68 — so this is confusing.

Possibility: Maybe the 68 is not BC but AB?

But the diagram shows:
- From A to B: 68?
Wait, the diagram says:
> At vertex B, angle is right.
> From A to C: 85
> From B to C: ?
> From A to B: 68?

Wait — no, looking carefully: the side from A to B is not labeled, but the side from B to C is labeled 68, and from A to C is 85.

So:
- AB = ?
- BC = 68
- AC = 85 (hypotenuse)

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

But the question says "BC = ?", but BC is already given as 68.

This must be a mislabeling. Likely, the question should be AB = ?, but it says BC = ?.

Alternatively, maybe the 68 is on AB, and BC is unknown?

Wait — let's check the image description:
"12. BC = ?" with triangle ABC, right angle at B, AC = 85, and one leg is 68.

If 68 is adjacent to B, then it could be either AB or BC.

But if BC = ?, then 68 must be AB.

So:
- AB = 68
- AC = 85 (hypotenuse)
- Find BC

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

✔ So BC = 51

So the 68 is AB, and BC = ?

Thus, answer is $\boxed{51}$

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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:

| Problem | Answer |
|--------|--------|
| 1 | 13 |
| 2 | 5 |
| 3 | 10 |
| 4 | 25 |
| 5 | 15 |
| 6 | 17 |
| 7 | 16 |
| 8 | 12 |
| 9 | 24 |
| 10 | 32 |
| 11 | 60 |
| 12 | 51 |
| 13 | $\sqrt{130}$ ≈ 11.4 |
| 14 | $\sqrt{217}$ ≈ 14.73 |
| 15 | $\sqrt{346}$ ≈ 18.6 |
| 16 | $\sqrt{119}$ ≈ 10.9 |
| 17 | $\sqrt{168} = 2\sqrt{42}$ ≈ 12.96 |
| 18 | $\sqrt{715}$ ≈ 26.74 |

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🔍 Summary:

All problems use the Pythagorean Theorem:
$$
a^2 + b^2 = c^2
$$
- If you know both legs → find hypotenuse.
- If you know hypotenuse and one leg → find other leg.

Always identify the right angle and the hypotenuse first!

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