Geometry worksheet featuring triangle and rectangle problems for calculating side lengths and angles.

Let’s solve each problem one by one. We’ll use the Pythagorean Theorem when needed:
a² + b² = c², where c is the hypotenuse (the longest side, opposite the right angle).

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Problem 1: Rectangle ABCD — Find diagonal AC

We’re told AB = 3 and BC = 4. In a rectangle, adjacent sides are perpendicular → triangle ABC is a right triangle with legs 3 and 4.

So, diagonal AC is the hypotenuse:

AC² = AB² + BC²
AC² = 3² + 4² = 9 + 16 = 25
AC = √25 = 5

✔ Answer for #1: 5

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Problem 2: Right triangle with legs 7 and 24 — Find hypotenuse x

Use Pythagorean Theorem:

x² = 7² + 24² = 49 + 576 = 625
x = √625 = 25

✔ Answer for #2: 25

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Problem 3: Square with side 8 — Find diagonal d

In a square, diagonal splits it into two right triangles with legs = side length.

d² = 8² + 8² = 64 + 64 = 128
d = √128 = √(64 × 2) = 8√2 ≈ but we can leave as exact value unless asked to approximate.

But let’s check if they want simplified radical or decimal? Since others are integers, maybe simplify:

√128 = √(64×2) = 8√2 — that’s simplest form.

Wait — actually, in many school problems, they accept 8√2. But let me double-check: Is there a mistake?

No — correct. Diagonal of square with side s is s√2 → so 8√2.

But looking at other answers being whole numbers… Hmm. Maybe I misread? No — problem says “calculate the length of the diagonal” — no specification. So 8√2 is correct.

But wait — let me recalculate:

8² + 8² = 64 + 64 = 128 → √128 = 8√2 — yes.

Alternatively, perhaps they expect decimal? But since others are exact, better to give exact.

Actually, let me see problem 4 — rhombus — might be similar.

Hold on — maybe I should just write 8√2.

But let me check online or standard practice — usually in such worksheets, if it simplifies nicely, they do. 8√2 is fine.

Wait — actually, √128 can also be written as 8√2 — that’s standard.

✔ Answer for #3: 8√2

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Problem 4: Rhombus with diagonals 10 and 24 — Find side s

Important: In a rhombus, diagonals bisect each other at right angles.

So, half of 10 is 5, half of 24 is 12. These form legs of a right triangle, and the side of the rhombus is the hypotenuse.

s² = 5² + 12² = 25 + 144 = 169
s = √169 = 13

✔ Answer for #4: 13

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Problem 5: Right triangle with leg 9, hypotenuse 15 — Find missing leg x

Use Pythagorean Theorem:

x² + 9² = 15²
x² + 81 = 225
x² = 225 - 81 = 144
x = √144 = 12

✔ Answer for #5: 12

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Problem 6: Right triangle with legs 5 and 12 — Find hypotenuse x

x² = 5² + 12² = 25 + 144 = 169
x = √169 = 13

✔ Answer for #6: 13

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Now, compiling all answers:

1. 5
2. 25
3. 8√2
4. 13
5. 12
6. 13

Wait — problem 3: Is 8√2 acceptable? Let me think again.

Looking back at the worksheet — it says “Calculate the length...” — and in problem 1, 2, 4, 5, 6 — all answers are integers. Only problem 3 gives irrational.

But mathematically, it’s correct. Perhaps they want decimal? But no instruction.

Alternatively — did I misread the figure? It says “square” with side 8 — yes.

Maybe they expect us to write √128? But 8√2 is simpler.

I think 8√2 is correct. Some teachers prefer simplified radicals.

Alternatively, maybe calculate numerical value? √128 ≈ 11.31 — but that seems unlikely without calculator.

Given context, I’ll stick with 8√2.

But let me verify with another approach: diagonal of square = side × √2 → 8 × √2 — same thing.

Yes.

Final Answers:

1. 5
2. 25
3. 8√2
4. 13
5. 12
6. 13

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Final Answer:
1. 5
2. 25
3. 8√2
4. 13
5. 12
6. 13