Pythagoras — Shorter Side & Two-Step

Section 1 of 3

Finding a Shorter Side

Rearrange

If you know the hypotenuse and one short side, subtract: $O^2 = H^2 - A^2$.

Worked example — hypotenuse and one side

The hypotenuse is $6$ and one short side is $3$. Find the other side $x$.

$x^2 + 3^2 = 6^2$

$x^2 + 9 = 36 \Rightarrow x^2 = 27$

$x = \sqrt{27} = 3\sqrt{3}$

$x = 3\sqrt{3}$

You try

Hypotenuse $9$, one side $5$. Find the other side $x$.

$x^2 + 5^2 = 9^2$.

$x^2 = 81 - 25 = 56 \Rightarrow x = \sqrt{56}$

$x = \sqrt{56}$

You try

Hypotenuse $8$, one side $3$. Find the other side $x$.

$x^2 + 3^2 = 8^2$.

$x^2 = 64 - 9 = 55 \Rightarrow x = \sqrt{55}$

$x = \sqrt{55}$

You try

Hypotenuse $8$, one side $4$. Find the other side $x$.

$x^2 + 4^2 = 8^2$.

$x^2 = 64 - 16 = 48 \Rightarrow x = \sqrt{48} = 4\sqrt{3}$

$x = 4\sqrt{3}$

Section 2 of 3

Squares & Equal Sides

Worked example — a diagonal of a square

A square has side $6$. Find the diagonal $x$.

$x^2 = 6^2 + 6^2 = 72$

$x = \sqrt{72} = 6\sqrt{2}$

$x = 6\sqrt{2}$

You try

A right-angled triangle has hypotenuse $10$ and two equal short sides $x$. Find $x$.

$x^2 + x^2 = 10^2$.

$2x^2 = 100 \Rightarrow x^2 = 50$

$x = \sqrt{50}$

You try

A square has diagonal $12$. Find the side $x$.

$x^2 + x^2 = 12^2$.

$2x^2 = 144 \Rightarrow x^2 = 72$

$x = \sqrt{72}$

Section 3 of 3

Two-Step Problems

One triangle at a time

For a shape made of two right-angled triangles, find the shared side first, then use it in the second triangle.

Worked example — diagonal then far side

A shape splits into two right-angled triangles. The first has short sides $7$ and $3$; its hypotenuse $y$ is a side of the second triangle, whose other side is $5$. Find $y$ and then $x$ (the second hypotenuse).

$y^2 = 7^2 + 3^2 = 58 \Rightarrow y = \sqrt{58}$

$x^2 = (\sqrt{58})^2 + 5^2 = 58 + 25 = 83$

$x = \sqrt{83}$

$y = \sqrt{58},\ x = \sqrt{83}$

You try

A triangle of height $x$ has hypotenuse $12$ and base $6$; then a second right-angled triangle has that same height and base $8$, with hypotenuse $y$. Find $x$ and $y$.

$x^2 + 6^2 = 12^2$, then $y^2 = 8^2 + x^2$.

$x^2 = 144 - 36 = 108 \Rightarrow x = \sqrt{108} = 6\sqrt{3}$

$y^2 = 8^2 + (\sqrt{108})^2 = 64 + 108 = 172 \Rightarrow y = \sqrt{172}$

$x = 6\sqrt{3},\ y = \sqrt{172}$

That’s Class 6.

Finding a shorter side, diagonals of squares, and two-step problems. Class 7: circles — parts, tangents and chords.