G6, G20 - Pythagoras theorem in two dimensions

Pythagoras' theorem links the three side lengths of a right-angled triangle. You will understand its area relationship, find missing lengths, test whether an angle is right, and choose the useful right triangles inside larger two-dimensional problems.

The hypotenuse and the square relationship

A $5\,\mathrm{m}$ ladder stands with its foot $1.4\,\mathrm{m}$ from a vertical wall. How high does it reach? Subtracting $1.4$ from $5$ will not work: these lengths run in different directions.

You already know

A right angle is $90^\circ$ . Squaring multiplies a number by itself; taking a positive square root recovers a positive length. A triangle's area is $\frac12bh$ .

The two sides meeting at the right angle are perpendicular. The remaining side crosses opposite the right angle, even when the triangle is tilted on the page.

Hypotenuse

The hypotenuse is the side opposite the right angle in a right-angled triangle. It is always that triangle's longest side. A non-right-angled triangle has no hypotenuse.

For perpendicular lengths $3$ and $4$ , a hypotenuse of $5$ gives $3^2+4^2=9+16=25=5^2$ . Squares built on the two perpendicular sides together have the same area as the square built on the hypotenuse.

A loop marked into $12$ equal sections can therefore set out a right angle: stretch it into sides of $3$ , $4$ and $5$ sections. The angle between the $3$ - and $4$ -section sides is exactly right.

Pythagoras' theorem

a^2+b^2=c^2

The triangle must be right-angled; $a,b$ meet at the right angle and $c$ is the hypotenuse.

On the formula sheet (2026 version; the 2027 sheet will have the same content).

Here is why it works for every right triangle, rather than just the $3,4,5$ example. Arrange four congruent right triangles inside a square of side $a+b$ . Their hypotenuses enclose a central square of side $c$ . Its angles are right because the two acute triangle angles together make $90^\circ$ , leaving $90^\circ$ at each meeting point.

[DIAGRAM: asset_name: Four-triangle area derivation of Pythagoras theorem; asset_slug: 1ma1_m58_area_derivation; caption: The large square contains four equal right triangles and a central hypotenuse square; description: A square of side a plus b contains four congruent blue right triangles with legs a and b. Their hypotenuses, all labelled c, enclose a sage central square. The chosen plotted leg lengths are 3 and 4, but the labels and area equation show the general argument.]
The large square contains four equal right triangles and a central hypotenuse square

Calculate the same outer area in two ways:

\begin{aligned} (a+b)^2&=4\left(\frac12ab\right)+c^2 &&\text{four triangles and square}\\ a^2+2ab+b^2&=2ab+c^2 &&\text{expand and simplify}\\ a^2+b^2&=c^2 &&\text{subtract }2ab \end{aligned}

The picture therefore explains a relationship between areas, although its final use is often to find a length.

Worked example 1: identifying the square relationship

Question. Triangle $ABC$ is right-angled at $B$ , with $AB=3$ , $BC=4$ , $AC=5$ units. Identify the hypotenuse and check Pythagoras' relationship.

Plan. Locate the side opposite $B$ , then compare squares.

\begin{aligned} c&=AC=5 &&\text{opposite the right angle}\\ AB^2+BC^2&=3^2+4^2=25 &&\text{square the perpendicular sides}\\ AC^2&=5^2=25 &&\text{square the hypotenuse} \end{aligned}

Why this step? Side roles come from the right angle, not from which side looks horizontal.

Answer. $AC$ is the hypotenuse, and $3^2+4^2=5^2$ .

Check. $5$ is longer than both $3$ and $4$ .

Change the two perpendicular lengths to $5$ and $12$ units. The area construction still works, including when the lengths are unequal.

Worked example 2: checking the area construction numerically

Question. Four right triangles with perpendicular sides $5$ , $12$ units form the construction above. Find the central square's area and side.

Plan. Subtract the four triangle areas from the outer square area.

\begin{aligned} A_{\text{outer}}&=(5+12)^2=289 &&\text{outer side is }17\\ A_{\text{triangles}}&=4\left(\frac12\times5\times12\right)=120 &&\text{four congruent pieces}\\ c^2&=289-120=169 &&\text{central area}\\ c&=13 &&\text{positive square root} \end{aligned}

Why this step? Subtracting the triangle regions leaves exactly the central square, without gaps or overlap.

Answer. Its area is $169$ square units and its side is $13$ units.

Check. $5^2+12^2=25+144=169$ .

An accurate relationship must square all three lengths.

Watch out

Wrong: $3+4=5$ . That adds lengths, whereas Pythagoras adds square areas. The correct line is $3^2+4^2=5^2$ , or $9+16=25$ .

The next check uses the same side roles with letters.

Finding a hypotenuse or shorter side

Identify the unknown side before deciding whether to add or subtract squares.

If the hypotenuse is missing, add the two perpendicular-side squares. If a perpendicular side is missing, subtract the known perpendicular-side square from the hypotenuse square. Both come from rearranging the same printed formula.

Worked example 3: finding the hypotenuse

Question. A right triangle has perpendicular sides $6\,\mathrm{cm}$ and $8\,\mathrm{cm}$ . Find its hypotenuse $c$ .

Plan. Add the squares, then take the positive square root.

\begin{aligned} c^2&=6^2+8^2 &&\text{unknown is the hypotenuse}\\ &=36+64=100 &&\text{square then add}\\ c&=\sqrt{100}=10\,\mathrm{cm} &&\text{return to a length} \end{aligned}

Why this step? Pythagoras gives $c^2$ , so the last square root is essential.

Answer. $10\,\mathrm{cm}$ .

Check. The hypotenuse is longer than both given sides.

Keep the same triangle but hide the $8\,\mathrm{cm}$ perpendicular side instead. Changing the unknown changes the rearrangement.

Worked example 4: finding a shorter side

Question. A right triangle has hypotenuse $10\,\mathrm{cm}$ , one perpendicular side $6\,\mathrm{cm}$ and other side $b$ . Find $b$ .

Plan. Subtract the known leg square, then take a positive root.

\begin{aligned} 6^2+b^2&=10^2 &&\text{place the hypotenuse correctly}\\ b^2&=100-36=64 &&\text{subtract }6^2\\ b&=\sqrt{64}=8\,\mathrm{cm} &&\text{positive length} \end{aligned}

Why this step? Adding here would calculate a side longer than the already-known hypotenuse.

Answer. $8\,\mathrm{cm}$ .

Check. $6^2+8^2=100=10^2$ .

The same subtraction gives $\sqrt{13^2-5^2}=12\,\mathrm{cm}$ for a hypotenuse of $13\,\mathrm{cm}$ and leg of $5\,\mathrm{cm}$ . Results need not be integers: legs $5$ , $7\,\mathrm{cm}$ give $\sqrt{74}\,\mathrm{cm}$ , about $8.60\,\mathrm{cm}$ to three significant figures. Keep the root or full calculator value until final rounding.

Watch out

Wrong: $b=10^2-6^2=64\,\mathrm{cm}$ . The subtraction gives $b^2$ , not $b$ . Take $\sqrt{64}$ to get $8\,\mathrm{cm}$ . Square units become length units after the root.

A negative value under the square root is a signal to check the side roles or the data: the hypotenuse must be longest.

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G6, G20 - Pythagoras theorem in two dimensions

The hypotenuse and the square relationship

Finding a hypotenuse or shorter side

Continue this lesson.

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