Sec 2 Pythagoras’ Theorem: Sides, Converse & Examples

Secondary 2 G3 Mathematics · Study guide

Understand the relationship between the three sides of a right-angled triangle, use its converse carefully, and choose the right triangle in a longer problem.

Concept → worked examples → practice14 min read · practice linked below

Three sides, one right angle

Pythagoras’ theorem connects lengths in a right-angled triangle. If the two shorter sides are $a$ and $b$ , and the hypotenuse is $c$ , then $a^2+b^2=c^2$ . The squares are essential: the theorem does not say that the two shorter lengths add to the longest length.

Before you begin: be comfortable with squares, square roots, rearranging an equation and converting length units. Identify the right angle first. The side opposite it is always the hypotenuse, even when the triangle has been rotated.

The equation says that the area of the square on the hypotenuse equals the sum of the areas of the squares on the other two sides. This explains why we add squared lengths, then take a square root to recover a length.

Find the hypotenuse: add the squares

Worked example 1 — the diagonal of a rectangle

A rectangular noticeboard is 8 cm wide and 6 cm high. Find its diagonal.

The diagonal divides the rectangle into two right-angled triangles. In either triangle, the diagonal $d$ is opposite the right angle, so it is the hypotenuse.

$d^2=8^2+6^2=64+36=100.$

Taking the positive square root gives $d=10\text{ cm}$ . We reject $-10$ because a length is positive.

The 8 cm and 6 cm sides meet at the right angle. The diagonal is the longest side.

Check: 10 cm is longer than either 8 cm or 6 cm, and $10^2=8^2+6^2$ . Adding the sides would give 14 cm, which is the distance around two edges, not the diagonal.

Find a shorter side: subtract before taking the root

Worked example 2 — a cable and a vertical pole

A 13 m straight cable connects the top of a vertical pole to a point on level ground 5 m from its base. Find the height of the pole.

The pole and ground are perpendicular. The cable is opposite that right angle, so its length, 13 m, belongs alone on the hypotenuse side of the equation. Let the height be $h$ metres.

$h^2+5^2=13^2\quad\Rightarrow\quad h^2=169-25=144.$

Thus $h=\sqrt{144}=12\text{ m}$ . Subtracting the lengths, $13-5$ , would not preserve the equation.

Check: $12^2+5^2=144+25=169=13^2$ , and the pole is shorter than the cable. Keep all dimensions in the same unit before squaring.

If the required length is not a whole number, retain the square root or the full calculator value during working. Round only the final answer to the precision requested.

The converse: can you prove a right angle?

Pythagoras’ theorem starts with a known right angle and finds a length. Its converse starts with three side lengths and tests whether the angle opposite the longest side is a right angle. Do not assume a right angle just because a sketch looks square.

Worked example 3 — compare both sides of the test

Triangle A has side lengths 7.5 cm, 10 cm and 12.5 cm. Triangle B has side lengths 7.5 cm, 10 cm and 12 cm. Determine which is right-angled.

For triangle A, the longest side is 12.5 cm. Calculate the two quantities separately:

$7.5^2+10^2=56.25+100=156.25,\qquad12.5^2=156.25.$

They are equal, so triangle A is right-angled opposite the 12.5 cm side, by the converse of Pythagoras’ theorem.

For triangle B, the longest side is 12 cm. Its square is $12^2=144$ , whereas the sum of the other two squares is still 156.25. These are unequal, so triangle B is not right-angled.

Check: both sets can form triangles because the two shorter sides sum to more than the longest side. A triple such as 2, 3, 6 cannot form a triangle at all. Use exact given lengths in the equality test; rounded measurements need allowance for their stated accuracy.

A longer problem: choose one triangle at a time

Worked example 4 — a diagonal and the area of a kite

A kite-shaped quadrilateral $ABCD$ consists of two isosceles triangles $ABC$ and $ADC$ on opposite sides of $AC$ . Their perpendicular heights meet at the midpoint $O$ of $AC$ . Given $AC=16\text{ cm}$ , $AB=BC=10\text{ cm}$ and $AD=CD=17\text{ cm}$ , find $BD$ and the area of the kite.

Split the kite along the shared base. Each height belongs to a different right-angled triangle.

Since $O$ is the midpoint, $AO=OC=16\div2=8\text{ cm}$ .
In right-angled triangle $AOB$ , $BO^2=10^2-8^2=36$ , so $BO=6\text{ cm}$ .
In right-angled triangle $AOD$ , $DO^2=17^2-8^2=225$ , so $DO=15\text{ cm}$ .
$B$ and $D$ lie on opposite sides of $AC$ , so $BD=6+15=21\text{ cm}$ . Add the two triangular areas: $\frac12(16)(6)+\frac12(16)(15)=168\text{ cm}^2$ .

Check: $8^2+6^2=10^2$ and $8^2+15^2=17^2$ . The area also equals $\frac12\times AC\times BD=\frac12(16)(21)=168\text{ cm}^2$ , since the diagonals are perpendicular.

Explore how the sides are connected

Build a right-angled triangle

Change the two perpendicular sides. Predict whether doubling both sides will double the hypotenuse, then compare the squared lengths.

The triangle is drawn to scale. The square marks the right angle; c is always opposite it.

8² + 6² = 64 + 36 = 100 cm². Therefore c = √100 = 10 cm. The hypotenuse is longer than either perpendicular side.

Common mistakes and reliable checks

Using the theorem in any triangle: establish a right angle first. If only three sides are given, use the converse as a test.
Adding when finding a shorter side: write the complete equation with the hypotenuse first, then rearrange.
Forgetting the square root: $x^2=50$ gives $x=\sqrt{50}$ , not 50. A squared length has square units; a final length has ordinary units.
Using a whole base instead of half: mark the midpoint and label the specific right triangle before substituting.

Choose the next step

Try the Sec 2 Pythagoras and trigonometry practice bank, starting with converse questions and then connected triangles. When a problem gives an angle and one side, move to trigonometric ratios.