Mathematics › 🔍 Grade 10 › Glass Box: Finding the Space Diagonal with 3D Pythagoras

Glass Box: Finding the Space Diagonal with 3D Pythagoras

Apply the Pythagorean theorem twice — once across the floor, once up through the box — and you can measure the diagonal of any rectangular room.

Grade 10Geometry / Algebra 2

Glass Box: Finding the Space Diagonal with 3D Pythagoras — illustration

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The big idea: The longest diagonal through a rectangular box — corner to opposite corner, cutting straight through the interior — can be found by applying the Pythagorean theorem twice. First find the diagonal across the base, then treat that diagonal and the height as the two legs of a second right triangle whose hypotenuse is the space diagonal.

🎯 By the end, you'll be able to

Find the diagonal of a rectangle using the 2D Pythagorean theorem
Explain why applying Pythagoras twice gives the space diagonal of a box
Use the formula d = √(l² + w² + h²) to find a box's space diagonal
Apply the formula to real box and room dimensions

📎 You should already know

Pythagorean theorem (2D)
Basic 3D solid vocabulary (length, width, height)

From flat diagonals to a space diagonal

You already know how to find the diagonal across a flat rectangle: it's the hypotenuse of a right triangle formed by the length and width. Now imagine a rectangular box — a room, a shipping crate, a fish tank — and ask for the diagonal that runs from one corner all the way through the interior to the opposite corner. That's the space diagonal, and it takes the same idea, used twice.

\[ d_{\text{base}} = \sqrt{l^2 + w^2} \]

First, the ordinary 2D diagonal of the rectangular base, from the length and width.

A second right triangle, standing up

That base diagonal doesn't just sit on the floor — it becomes one leg of a brand-new right triangle. Stand the height h up from one end of the base diagonal to the top corner directly above the opposite end. The base diagonal and the height are the two legs of this second triangle, and its hypotenuse is exactly the space diagonal you're after.

🔑 Pythagoras, twice

Find the base diagonal with l and w, then use that diagonal together with the height h in a second right triangle. Combining both steps into one formula gives the space diagonal directly.

\[ d = \sqrt{l^2 + w^2 + h^2} \]

The space diagonal of a rectangular box, in terms of its length, width, and height.

🎮 Glass Box LIVE

Find the space diagonal of a box by applying Pythagoras twice.

📝 Worked example: A box has length 3, width 4, and height 12. Find its space diagonal.

Base diagonal: √(3² + 4²) = √(9 + 16) = √25 = 5.
Now use that 5 as one leg, with the height 12 as the other leg of the second triangle.
Space diagonal = √(5² + 12²) = √(25 + 144) = √169.

✓ The space diagonal is <strong>13</strong> — a 3–4–5 base diagonal feeding straight into a 5–12–13 triangle.

📝 Worked example: A box has length 2, width 3, and height 6. Find its space diagonal.

Base diagonal: √(2² + 3²) = √(4 + 9) = √13.
Space diagonal = √(13 + 6²) = √(13 + 36) = √49.

✓ The space diagonal is <strong>7</strong>.

⚠️ Square, then add, then root — don't shortcut it

It is tempting to just add l + w + h directly, but that is not the diagonal. You must square each dimension, add the squares, and only then take the square root: d = √(l² + w² + h²), never √l + √w + √h or l + w + h.

Check your understanding

1. To find the diagonal of the base rectangle of a box with length 3 and width 4, you compute:

The base diagonal is the hypotenuse of a right triangle with legs 3 and 4: √(9+16) = √25 = 5.

2. A box has length 3, width 4, height 12. What is the length of its space diagonal?

The base diagonal is 5 (3-4-5 triangle), then 5 and 12 form a second right triangle with hypotenuse √(25+144) = √169 = 13.

3. The formula for the space diagonal of a rectangular box with dimensions l, w, h is:

The space diagonal combines both applications of Pythagoras into one formula: the square root of the sum of the squares of all three dimensions.

4. A box has length 2, width 3, height 6. What is its space diagonal?

√(2² + 3² + 6²) = √(4 + 9 + 36) = √49 = 7.

5. Why does finding a box's space diagonal require applying the Pythagorean theorem twice?

The base diagonal becomes a leg of a second right triangle with the height as the other leg — that second application produces the space diagonal.

✅ Key takeaways

The space diagonal of a box runs from one corner to the opposite corner through the interior.
First find the base diagonal with Pythagoras: d_base = √(l² + w²).
Then use that base diagonal and the height as legs of a second right triangle.
Combined, the space diagonal is d = √(l² + w² + h²).
Always square each dimension, sum the squares, then take one square root at the end.