Pythagoras, congruent and similar triangles · Angles, triangles & polygons

Pythagoras’ theorem

In a right-angled triangle, the longest side, opposite the right angle, is the hypotenuse $c$ . The square on it equals the squares on the other two sides added together:

c^2 = a^2 + b^2

To find a shorter side, take away instead: $a^2 = c^2 - b^2$ . Some whole-number sets come up again and again, the Pythagorean triples: $3, 4, 5$ ; $5, 12, 13$ ; $8, 15, 17$ ; $7, 24, 25$ , and any multiple of them, such as $6, 8, 10$ .

Pythagorasc is opposite the right angle: c² = a² + b²

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q4 (a)

In the diagram, $PQRS$ is a quadrilateral, $\angle PQR = \angle PRS = 90^\circ$ , $|PQ| = 3\text{ cm}$ , $|QR| = 4\text{ cm}$ and $|PS| = 13\text{ cm}$ . Find the area of the quadrilateral.

Split the shape
Draw the diagonal $PR$ . It splits $PQRS$ into triangle $PQR$ (right angle at $Q$ ) and triangle $PRS$ (right angle at $R$ ).
Think first. Which diagonal makes two right-angled triangles?
The shared side PR
$|PR| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ cm.
Think first. In triangle PQR, which side is the hypotenuse?
The side RS
$|RS| = \sqrt{13^2 - 5^2} = \sqrt{144} = 12$ cm.
Think first. In triangle PRS, PS = 13 is the hypotenuse. Add or subtract?
Add the areas
Area $= \frac12 \times 3 \times 4 + \frac12 \times 5 \times 12 = 6 + 30 = 36\text{ cm}^2$ .
Think first. Each triangle's base and height are the two sides next to its right angle.

Congruent triangles

Two triangles are congruent when they are exactly the same shape and size: every side and angle of one matches one of the other. You don’t have to check all six parts. Any of these four is enough:

SSS: all three sides match.
SAS: two sides and the angle between them match.
ASA (or AAS): two angles and a side match.
RHS: both have a right angle, the hypotenuses match, and one other side matches.

Two sides and an angle that is not between them (SSA) is not enough: the same parts can make two different triangles.

SSSThree sides

SASTwo sides and the angle between them

ASATwo angles and a side

RHSRight angle, hypotenuse and a side

Similar triangles

Two triangles are similar when they have the same angles. Then one is an enlargement of the other: every side is multiplied by the same scale factor $k$ . If two angles match, the third must too (they all add to $180^\circ$ ), so two equal angles are enough.

To use similar triangles, match the corners by their equal angles first. Then the sides opposite matching angles go together:

\frac{\text{side of big}}{\text{matching side of small}} = k \quad\text{for every pair}

Lengths scale by $k$ , but areas scale by $k^2$ .

Similar trianglesSame angles; every side × k; area × k²

Try it

Similar trianglesSlide the cut

0.4PH ÷ PQ = HK ÷ QR = PK ÷ PR0.4² = 0.16area of PHK ÷ area of PQR0.84trapezium HQRK ÷ area of PQR

k = PH ÷ PQ = 0.4

HK is parallel to QR, so triangle PHK has the same angles as PQR: the triangles are similar. Every side of PHK is 0.4 of the matching side: PH = 4 cm, HK = 5.6 cm, PK = 4.8 cm. The area goes down by 0.4² = 0.16, not by 0.4: lengths scale by k, areas by k².

A line parallel to one side of a triangle cuts off a smaller triangle with the same angles (corresponding angles), so the two are similar. Slide the cut: all three ratios stay equal, while the area ratio is the square. Then try “Pythagoras” and look for whole-number answers.

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q3 (a)

In the diagram, $QSP$ and $QTR$ are straight lines, $|PS| = 6\text{ cm}$ , $|QS| = 4\text{ cm}$ , $|QT| = 5\text{ cm}$ and $\angle QTS = \angle RPQ$ . Calculate $|TR|$ .

Show the triangles are similar
Triangles $QTS$ and $QPR$ share the angle at $Q$ , and $\angle QTS = \angle QPR$ (given). Two equal angles, so they are similar.
Think first. Triangles QTS and QPR share one angle. Which? And which other angles are equal?
Match the corners
$Q \leftrightarrow Q$ , $T \leftrightarrow P$ (the equal given angles), so $S \leftrightarrow R$ .
Think first. Q matches Q. Which corner of QPR matches T? Which matches S?
Write the ratio
$\dfrac{|QT|}{|QP|} = \dfrac{|QS|}{|QR|}$ . With $|QP| = 4 + 6 = 10$ : $\dfrac{5}{10} = \dfrac{4}{|QR|}$ , so $|QR| = 8$ cm.
Think first. QT matches QP, and QS matches which side?
Answer the question
$|TR| = |QR| - |QT| = 8 - 5 = 3$ cm.
Think first. The question asks for TR, not QR.

Your turn

WAEC 2021 · Paper 2 · Q3 (a)

(a)
A vertical pole $6\text{ m}$ high casts a shadow $9\text{ m}$ long at the same time that a tree casts a shadow $30\text{ m}$ long. Find the height of the tree.

Worked solution (try it first)

(a)

At the same time of day, heights and shadows are in the same ratio (similar triangles): $\frac{h}{30} = \frac{6}{9}$ , so $h = 30 \times \frac23 = 20$ m.

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