Can I use the d squared equals l squared plus w squared plus h squared formula for any 3D shape?

No. That shortcut only works for cuboids (rectangular boxes). For cylinders, cones, pyramids, and other shapes, apply Pythagoras step by step on the right triangles inside the solid.

Can I use the d squared equals l squared plus w squared plus h squared formula for any 3D shape?

No. That shortcut only works for cuboids (rectangular boxes). For cylinders, cones, pyramids, and other shapes, apply Pythagoras step by step on the right triangles inside the solid.

Year 11 General Shape And Measurement

Pythagoras Theorem In Three Dimensions

Q: How does Pythagoras' theorem work in 3D?

Apply Pythagoras twice. First, find a diagonal inside one face of the solid using the 2D theorem. Then use that diagonal as a side of a new right triangle to find the space diagonal.

Q: What is the formula for the space diagonal of a cuboid?

d squared equals length squared plus width squared plus height squared. Take the square root to get d. This is the shortcut that combines two applications of 2D Pythagoras.

Q: How do you find the longest object that fits inside a cylinder?

The longest object sits diagonally. Use Pythagoras with the diameter of the base and the height of the cylinder: L squared equals diameter squared plus height squared.

Q: What is the slant height of a cone?

The slant height is the distance from the apex down the side to the edge of the base. It connects the radius and the perpendicular height via Pythagoras: s squared equals r squared plus h squared.

Q: Why does the space diagonal formula add three squares?

Because two applications of Pythagoras combine: base diagonal squared equals l squared plus w squared, then space diagonal squared equals base diagonal squared plus h squared. Substituting gives l squared plus w squared plus h squared.

10 practice questions 1 video lesson Theory + worked examples

Theory

Pythagoras' theorem extends to 3D by applying it twice: first to a 2D triangle inside the solid, then to another triangle that uses the first result. For a cuboid with length $ℓ$ , width $w$ , and height $h$ , the space diagonal follows the shortcut $d^{2} = ℓ^{2} + w^{2} + h^{2}$ .

Pythagoras' theorem also works in three dimensions. The key idea is to apply it twice: first to a 2D triangle inside the solid, then to a second triangle that uses your first result as one of its sides.

The classic case is the space diagonal of a cuboid — the line from one corner to the opposite corner. For a box with length $ℓ$ , width $w$ , and height $h$ , the space diagonal $d$ satisfies $d^{2} = ℓ^{2} + w^{2} + h^{2}$ — a shortcut that combines two Pythagoras steps into one.

The same two-step idea handles other 3D shapes. A cylinder: the longest pencil that fits inside has length $\sqrt{(2 r)^{2} + h^{2}}$ — diameter and height form the right triangle. A cone: the slant height $s$ connects radius and height via $s^{2} = r^{2} + h^{2}$ .

The shortcut only works for cuboids. For any other shape, do the two steps explicitly: sketch the right triangle, apply Pythagoras to it, then sketch the next one.

d² = ℓ² + w² + h² for a cuboid

Find b first, then use b with h to find d

Two-step Pythagoras on a cuboid:

b^{2} = ℓ^{2} + w^{2} then d^{2} = b^{2} + h^{2}

The cuboid space-diagonal shortcut (combining the two steps):

d^{2} = ℓ^{2} + w^{2} + h^{2}

d^{2} = ℓ^{2} + w^{2} + h^{2}

Longest object inside a cylinder (diameter $2 r$ , height $h$ ):

L = \sqrt{(2 r)^{2} + h^{2}}

L = \sqrt{{(2 r)}^{2} + h^{2}}

Slant height of a cone (radius $r$ , height $h$ ):

s^{2} = r^{2} + h^{2}

s^{2} = r^{2} + h^{2}

The cuboid shortcut only works for cuboids. For other 3D shapes, identify each right triangle and apply 2D Pythagoras one step at a time.

How to apply Pythagoras in 3D

Sketch the 3D shape and identify the line whose length you want.
Find a 2D right triangle inside the solid — typically using one face or the base.
Apply Pythagoras to that triangle to get a useful intermediate length.
Find a second right triangle that uses your intermediate length and another known dimension.
Apply Pythagoras again to get the final answer. For a cuboid, the shortcut $d^{2} = ℓ^{2} + w^{2} + h^{2}$ does both steps at once.

Example 1 — Space diagonal of a cuboid

A box is

8

cm long,

6

cm wide, and

2

cm tall. Find the space diagonal (2 dp).

Solution

Use the cuboid shortcut.

$d^{2}$	$=$	$8^{2} + 6^{2} + 2^{2}$
$d^{2}$	$=$	$64 + 36 + 4 = 104$
$d$	$=$	$\sqrt{104}$
$d$	$\approx$	$10.20$ cm

d \approx 10.20

The space diagonal is about $10.20$ cm.

Example 2 — Two-step on a cuboid

A box is

12

m long,

5

m wide, and

84

m tall. Find the space diagonal in two steps.

Solution

Step 1 — base diagonal:

$b^{2}$	$=$	$12^{2} + 5^{2}$
$b^{2}$	$=$	$144 + 25 = 169$
$b$	$=$	$13$ m

Step 2 — space diagonal:

$d^{2}$	$=$	$13^{2} + 84^{2}$
$d^{2}$	$=$	$169 + 7,056 = 7,225$
$d$	$=$	$\sqrt{7,225} = 85$ m

d = 85

The space diagonal is exactly $85$ m.

Example 3 — Pencil in a cylinder

A cylindrical tin has radius

3

cm and height

8

cm. Find the longest pencil that fits inside (2 dp).

Solution

The pencil's length, the diameter, and the height form a right triangle.

$diameter$	$=$	$2 \times 3 = 6$
$L^{2}$	$=$	$6^{2} + 8^{2}$
$L^{2}$	$=$	$36 + 64 = 100$
$L$	$=$	$10$ cm

L = 10

The longest pencil is $10$ cm — a 6-8-10 triple.

Example 4 — Slant height of a cone

A cone has base radius

5

cm and vertical height

12

cm. Find its slant height.

Solution

The slant is the hypotenuse of the triangle formed by $r$ and $h$ .

$s^{2}$	$=$	$r^{2} + h^{2}$
$s^{2}$	$=$	$5^{2} + 12^{2} = 169$
$s$	$=$	$13$ cm

s = 13

The slant height is $13$ cm (a 5-12-13 triple).

Common pitfalls

Using the shortcut on a non-cuboid. The formula

d^{2} = ℓ^{2} + w^{2} + h^{2}

only works for rectangular boxes. For cylinders, cones, and pyramids, identify the right triangles inside the solid and apply Pythagoras step by step.

Skipping the sketch. 3D problems require drawing the right triangle you're solving each time. Without it, it's easy to use the wrong sides or pick the wrong hypotenuse.

Confusing slant height with vertical height. In a cone or pyramid, the perpendicular height runs straight down from the apex to the centre of the base. The slant runs from apex to the edge. They are different — use the one the question gives you.

Using radius where diameter belongs. The longest object in a cylinder uses the diameter across the base, not the radius. Double the radius first.

Frequently asked questions

How does Pythagoras' theorem work in 3D?

Apply it twice. First, find a diagonal inside one face using 2D Pythagoras. Then use that diagonal as a side of a new right triangle to find the line you want.

What is the formula for the space diagonal of a cuboid?

$d^{2} = ℓ^{2} + w^{2} + h^{2}$ . Take the square root for $d$ . It's the shortcut for two combined steps of 2D Pythagoras.

How do you find the longest object that fits inside a cylinder?

Use the diameter across the base and the height: $L^{2} = (2 r)^{2} + h^{2}$ .

What is the slant height of a cone?

The distance from the apex down the side to the edge of the base. It connects radius and vertical height via $s^{2} = r^{2} + h^{2}$ .

Can I use the d² = l² + w² + h² formula for any 3D shape?

No. That shortcut only works for cuboids. For other shapes, apply Pythagoras one step at a time on the right triangles inside the solid.

Why does the space diagonal formula add three squares?

The two steps combine: base diagonal squared = $ℓ^{2} + w^{2}$ ; space diagonal squared = base diagonal squared + $h^{2}$ . Substituting gives $ℓ^{2} + w^{2} + h^{2}$ .