3D Pythagoras (AQA GCSE Maths): Revision Notes

Step-by-Step Breakdown:

Step 1: First, you create a right-angled triangle using two sides of the base ($a$ and $b$) to find the diagonal across the base ($e$): $e^2 = a^2 + b^2$

Step 2: Then you form another right-angled triangle using this base diagonal ($e$) and the height ($c$) to find the space diagonal ($d$): $d^2 = e^2 + c^2$

When you substitute the first equation into the second, you get: $d^2 = (a^2 + b^2) + c^2 = a^2 + b^2 + c^2$

Worked Example: Finding the Space Diagonal

Find the exact length of diagonal BH in a cube with 4cm sides.

Step 1: Write down the formula: $a^2 + b^2 + c^2 = d^2$

Step 2: Substitute the values: $4^2 + 4^2 + 4^2 = BH^2$

Step 3: Calculate: $48 = BH^2$

Step 4: Take the square root: $BH = \sqrt{48} = 4\sqrt{3}$ cm

Worked Example: Finding the Vertical Height of a Square-Based Pyramid

For a square-based pyramid where M is the midpoint of the base:

Step 1: Label the key point as the midpoint of the base edge

Step 2: Think of the problem as three sides of a cuboid, where the pyramid's slant edge becomes the space diagonal

Step 3: Sketch the full cuboid to visualise the problem clearly

Step 4: Apply the 3D Pythagoras formula: $a^2 + b^2 + c^2 = d^2$

Step 5: Substitute using appropriate labels and solve for the unknown height

In this case: $EN^2 + NM^2 + AM^2 = AE^2$

• $3.5^2 + 3.5^2 + AM^2 = 9^2$
• $AM = \sqrt{81 - 2 \times 12.25} = 7.52$ cm