Cambridge IGCSE — Extended Maths
3D Pythagoras & Trigonometry
The Big Idea: any messy 3D shape can be broken down into flat, ordinary 2D right-angled triangles — find the ones hiding inside it, draw them flat on paper, then just use Pythagoras' theorem and SOHCAHTOA like normal.
Summary — What This Chapter Covers
- Right-angled triangles are hidden inside 3D shapes (cuboids, prisms, cones, pyramids) — your job is to spot them.
- Pythagoras' theorem (a² = b² + c²) finds missing lengths once you've isolated a 2D triangle.
- There's a 3D version of Pythagoras — d² = x² + y² + z² — for the straight-line distance between two corners, but you never actually need to memorise it.
- SOHCAHTOA finds missing angles or sides once a right-angled triangle is isolated, exactly like in 2D.
- The "angle between a line and a plane" is a specific, common exam question — it needs a very particular triangle to be drawn correctly.
- Complicated problems are solved by chaining together multiple 2D triangles, one feeding into the next.
1. Pythagoras' Theorem in 3D Shapes
Here's the mindset shift for this whole chapter: 3D geometry is just 2D geometry that's been folded up. Every cone, cuboid, pyramid, and prism you'll ever be asked about is secretly hiding a bunch of ordinary flat triangles inside it. Your entire job is to find one, redraw it flat on your page (ignoring the 3D shape around it), and then solve it exactly the way you've solved 2D triangles for years.
Why this works
A right-angled triangle doesn't care whether it's sitting flat on a page or tilted inside a 3D cone — the relationship a² = b² + c² is exactly the same either way. The 3D-ness of the problem is just packaging. Once you strip a triangle out of the shape, it's a completely normal 2D problem.
Worked Example — The Cone
Problem
A cone has base radius 5 cm and perpendicular height 12 cm. Find the slant height,
l.
Cone (side view, but we only care about ONE triangle inside it):
/|
/ |
l / | 12 cm (perpendicular height)
/ |
/_____|
5 cm (radius)
This right-angled triangle is formed by:
• the radius (base)
• the perpendicular height (vertical)
• the slant height (hypotenuse, l)
Step 1 — Isolate the triangle: radius = 5, height = 12, slant = l (hypotenuse).
Step 2 — Apply Pythagoras: l² = 5² + 12² = 25 + 144 = 169
Step 3 — Square root: l = √169 = 13 cm
Practice Question 1.1
A cone has base radius 9 cm and slant height 15 cm. Find the perpendicular height of the cone.
2. The 3D Version of Pythagoras' Theorem
Sometimes you're not given a neat 2D triangle to start with — you're just asked for the distance between two corners of a cuboid, like from one bottom corner to the diagonally opposite top corner. This is where the 3D Pythagoras formula comes in.
The "Two 2D Problems" Trick
without
twice in a row
3. SOHCAHTOA in 3D
angle
SOHCAHTOA (reminder)