Pythagoras theorem calculator: find the missing side

Square, add, square root. For shorter sides 6 and 8: $c^2 = 6^2 + 8^2 = 36 + 64 = 100$, so $c = \sqrt{100} = 10$.

Square, subtract, square root. For a hypotenuse of 13 and one side of 5: $a^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $a = 12$.

A quick check: the hypotenuse must be the longest side. If your answer for a shorter side is bigger than the hypotenuse, you added instead of subtracting.

Some whole-number sides work exactly. Recognising them saves time:

• Distance between two points: the horizontal and vertical differences are the shorter sides.
• Ladders, ramps and diagonals of rectangles are classic exam contexts.
• Mistakes: squaring the hypotenuse but adding it, forgetting the final square root, or using Pythagoras on a triangle without a right angle (use the sine or cosine rule instead).

Pythagoras' theorem is on the Digital SAT reference sheet; for angles, see right-triangle trigonometry.

To find the distance from $(1, 2)$ to $(7, 10)$, draw a right-angled triangle: the horizontal side is $7 - 1 = 6$ and the vertical side is $10 - 2 = 8$. The distance is the hypotenuse: $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$. This is the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, which is just Pythagoras in disguise.

For the diagonal of a cuboid with sides $a$, $b$ and $c$, use Pythagoras twice, or in one go: $d = \sqrt{a^2 + b^2 + c^2}$. A box 3 cm by 4 cm by 12 cm has a space diagonal of $\sqrt{9 + 16 + 144} = \sqrt{169} = 13$ cm. 3D Pythagoras is a Higher-tier GCSE topic.