Geometry practice problems using the Pythagorean theorem.

You’ve uploaded a worksheet with several right triangle problems, mostly applying the Pythagorean Theorem:

> Pythagorean Theorem: In a right triangle,
> a² + b² = c²,
> where *c* is the hypotenuse (side opposite the right angle), and *a* and *b* are the legs.

Let’s solve each problem step by step.

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Problems 4–9: Find the missing side

#### Problem 4
Triangle: legs = 12 yd, ?; hypotenuse = 13 yd
We’re missing one leg. Let’s call it *x*.

> x² + 12² = 13²
> x² + 144 = 169
> x² = 169 - 144 = 25
> x = √25 = 5 yd

✔ Answer: 5 yd

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#### Problem 5
Legs: 25 cm, 38 cm → find hypotenuse *c*

> c² = 25² + 38² = 625 + 1444 = 2069
> c = √2069 ≈ 45.49 cm (rounded to two decimals)

✔ Answer: ≈45.49 cm

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#### Problem 6
Legs: 0.8 mi, 0.5 mi → find hypotenuse *c*

> c² = 0.8² + 0.5² = 0.64 + 0.25 = 0.89
> c = √0.89 ≈ 0.94 mi

✔ Answer: ≈0.94 mi

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#### Problem 7
Leg: 15 ft, hypotenuse: 25 ft → find other leg *x*

> x² + 15² = 25²
> x² + 225 = 625
> x² = 400
> x = √400 = 20 ft

✔ Answer: 20 ft

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#### Problem 8
Isosceles triangle with two sides = 40 in. — but no right angle shown? Wait — this looks like an isosceles triangle, not necessarily right-angled. But if we assume the height splits it into two right triangles (which is standard for finding height), then:

Each half of the base is unknown. But wait — the diagram doesn’t show the base or height labeled. Actually, since both sides are 40 in., and if we assume it's isosceles with base unknown, we can't solve without more info — unless we assume it’s equilateral? No, it doesn’t say that.

Wait — looking again: Problem 8 just says “40 in.” on two sides, and nothing else. But in context, maybe it’s implying to find the height? But no base is given.

Actually, this might be a trick — perhaps it’s meant to be solved as an isosceles triangle split into two right triangles, but we need the base. Since it’s not given, and no right angle is marked, this problem may be incomplete or mislabeled.

But let’s check the image description — you mentioned "each side of an equilateral triangle" in #13, so #8 is likely not equilateral. Without more data, we cannot solve #8 definitively.

However, if we assume the triangle is isosceles with two equal sides of 40 in. and base also 40 in. (i.e., equilateral), then we can proceed — but that’s assumption.

Alternatively, maybe the diagram shows a height? Since you said “I uploaded an image”, and I don’t have it, I’ll note:

> ❗ Problem 8 is unsolvable as stated unless more info (like base or angle) is provided.

But since it’s grouped with right triangles, maybe the height is drawn? If so, and if the base is implied to be split equally, still — we need base length.

Let me skip for now and come back.

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#### Problem 9
Legs: 70 m, ?; hypotenuse: 100 m → find other leg *x*

> x² + 70² = 100²
> x² + 4900 = 10000
> x² = 5100
> x = √5100 ≈ 71.41 m

✔ Answer: ≈71.41 m

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Now the word problems:

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#### Problem 10: TV Screen Diagonal

Screen: 48 in wide, 27 in high → diagonal = ?

This forms a right triangle! Use Pythagorean Theorem.

> d² = 48² + 27² = 2304 + 729 = 3033
> d = √3033 ≈ 55.07 in

✔ Answer: ≈55.07 inches

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#### Problem 11: Ladder against wall

Ladder = 18 ft (hypotenuse), base = 7 ft from wall → how high up the wall?

> h² + 7² = 18²
> h² + 49 = 324
> h² = 275
> h = √275 ≈ 16.58 ft

✔ Answer: ≈16.58 feet

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#### Problem 12: Hulk’s Walk

West 8 blocks, North 6 blocks → distance from home?

Each block = 500 ft → so:

- West: 8 × 500 = 4000 ft
- North: 6 × 500 = 3000 ft

These form legs of a right triangle. Distance from home = hypotenuse.

> d² = 4000² + 3000² = 16,000,000 + 9,000,000 = 25,000,000
> d = √25,000,000 = 5000 ft

✔ Answer: 5000 feet

*(Note: This is a classic 3-4-5 triangle scaled by 1000: 3000-4000-5000)*

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#### Problem 13: Height of Equilateral Triangle

Each side = 9 cm. Find height *h*.

In an equilateral triangle, the height splits the base into two equal parts (each 4.5 cm), forming two congruent right triangles.

So:

> h² + (4.5)² = 9²
> h² + 20.25 = 81
> h² = 60.75
> h = √60.75 = √(243/4) = (√243)/2 = (9√3)/2 ≈ 7.794 cm

Exact value: (9√3)/2 cm
Approximate: ≈7.79 cm

✔ Answer: (9√3)/2 cm or ≈7.79 cm

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Back to Problem 8

Since it’s labeled as “40 in.” on two sides and no base or angle given, and no right angle marked, it’s ambiguous.

However, if we assume it’s an isosceles triangle with two equal sides of 40 in. and the base is also 40 in. (equilateral), then it’s identical to Problem 13.

Then height = (√3 / 2) × side = (√3 / 2) × 40 = 20√3 ≈ 34.64 in

But since Problem 13 already asks for equilateral, and Problem 8 doesn’t specify, it’s safer to say:

> ❗ Problem 8: Insufficient information. Possibly intended to be solved as equilateral? If so, height = 20√3 in ≈34.64 in.

But without confirmation, we should flag it.

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## ✔ Final Answers:

4. 5 yd
5. ≈45.49 cm
6. ≈0.94 mi
7. 20 ft
8. ❗ Insufficient info — possibly 20√3 in if equilateral
9. ≈71.41 m
10. ≈55.07 in
11. ≈16.58 ft
12. 5000 ft
13. (9√3)/2 cm ≈7.79 cm

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Let me know if you want diagrams or step-by-step visuals for any of these!