Math problems demonstrating the Pythagorean theorem with areas and perimeters of squares.

Let’s solve both problems step by step.

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Problem 9: Which of the following could be the areas of the three squares below?

We are given a diagram with three squares built on the sides of a right triangle. The key idea here is the Pythagorean theorem:
> In a right triangle, the square of the hypotenuse (the longest side) equals the sum of the squares of the other two sides.

Since the squares are built on each side, their areas are equal to the squares of the side lengths.

So if the side lengths of the triangle are `a`, `b`, and `c` (with `c` being the hypotenuse), then:

> Area of square on side a = `a²`
> Area of square on side b = `b²`
> Area of square on side c = `c²`

And from Pythagoras:
> `a² + b² = c²`

That means: The area of the largest square should equal the sum of the areas of the two smaller squares.

Let’s test each option:

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Option A: 12 ft², 16 ft², and 20 ft²

Check: Does 12 + 16 = 20? → 28 ≠ 20 → ✘ No

Wait — maybe we need to check which one is the largest? Let’s assume 20 is the largest.

12 + 16 = 28 ≠ 20 → Still no.

What if 16 is the largest? 12 + 20 = 32 ≠ 16 → No.

What if 12 is the largest? 16 + 20 = 36 ≠ 12 → No.

→ So Option A is wrong.

But wait — let me double-check my math. Maybe I misread.

Actually, looking again at the student’s work in red ink: they wrote “a² + b² = c²” multiple times, and for problem 10 they used 3-4-5 triangle.

For problem 9, perhaps we should look for numbers where two add up to the third.

Try Option D: 8, 16, 24

Is 8 + 16 = 24? → YES! ✔

So that works.

Check others just to be sure.

Option B: 10, 18, 30

10 + 18 = 28 ≠ 30 → ✘

Option C: 4, 5, 12

4 + 5 = 9 ≠ 12 → ✘

Only Option D satisfies: small + medium = large → 8 + 16 = 24.

✔ So Problem 9 answer is D

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Problem 10: Which of the following could be the perimeters of the three squares below?

Same diagram — three squares on the sides of a right triangle.

Now we’re asked about perimeters, not areas.

Perimeter of a square = 4 × side length.

So if the triangle has sides `a`, `b`, `c` (right triangle, so `a² + b² = c²`), then:

Perimeter of square on side `a` = `4a`
Perimeter of square on side `b` = `4b`
Perimeter of square on side `c` = `4c`

We need to find which set of perimeters corresponds to side lengths that satisfy the Pythagorean theorem.

In other words, if perimeters are P₁, P₂, P₃, then side lengths are P₁/4, P₂/4, P₃/4.

Then check: does (P₁/4)² + (P₂/4)² = (P₃/4)² ?

Multiply both sides by 16:
→ P₁² + P₂² = P₃² ? Wait — NO!

Wait — careful:

If sides are `a`, `b`, `c`, then:

`a² + b² = c²`

But perimeters are `4a`, `4b`, `4c`

So let’s denote:

Let p = 4a → a = p/4
q = 4b → b = q/4
r = 4c → c = r/4

Then:

(p/4)² + (q/4)² = (r/4)²
→ p²/16 + q²/16 = r²/16
→ Multiply both sides by 16:
→ p² + q² = r²

Wait — that would mean the perimeters themselves must satisfy the Pythagorean theorem? That can’t be right unless the scaling factor cancels out.

Actually, let’s plug in actual numbers.

Look at the student’s red writing: they drew a 3-4-5 triangle.

Side lengths: 3, 4, 5 → since 3² + 4² = 9 + 16 = 25 = 5² → valid.

Then perimeters of squares:

Square on side 3: perimeter = 4×3 = 12
Square on side 4: perimeter = 4×4 = 16
Square on side 5: perimeter = 4×5 = 20

So perimeters: 12, 16, 20

Check options:

Option A: 12 ft, 16 ft, and 20 ft → matches exactly! ✔

Check others quickly:

Option B: 20, 16, 24

Divide by 4: sides = 5, 4, 6 → check 4² + 5² = 16 + 25 = 41 ≠ 36 → ✘

Option C: 40, 80, 120

Sides: 10, 20, 30 → 10² + 20² = 100 + 400 = 500 ≠ 900 → ✘

Option D: 16, 24, 28

Sides: 4, 6, 7 → 4² + 6² = 16 + 36 = 52 ≠ 49 → ✘

Only Option A works.

Also, note the student circled A and wrote “12:4” meaning 12÷4=3, etc., confirming 3-4-5 triangle.

✔ So Problem 10 answer is A

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Final Answer:
Problem 9: D
Problem 10: A