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Year 10 Maths Core Geometry

Similar Triangles

20 practice questions 1 video lesson Theory + worked examples

Similar Triangles in NSW Year 10 Core maths applies similarity to triangles, which often appear inside other shapes (NESA outcome MA5‑GEO‑C‑01). When a line is drawn parallel to one side of a triangle, or two right-angled triangles share an angle, the triangles are similar and their sides are in proportion.

You will match sides and angles using a similarity statement, decide whether two triangles are similar, and find unknown lengths in overlapping and nested triangles. You will also find how high a ladder reaches, the height of a gum tree using a mirror on the ground, the height of a tower from a line of sight over a pole, and the area of a larger similar right-angled triangle. It is part of Year 10 Core Geometry.

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Theory

Similar triangles have equal matching angles and matching sides in the same ratio. This Year 10 maths page shows how to spot similar triangles, match their sides, and find unknown lengths, including parallel lines inside a triangle, triangles that cross at a point, and real problems with shadows, mirrors and rivers.

Two triangles are similar if one is an enlargement of the other. Their matching angles are equal and their matching sides are in the same ratio, the scale factor. We write \(\triangle PQR \,|||\, \triangle XYZ\) (sometimes \(\sim\)), with the vertices in matching order.

Two common ways to see that triangles are similar: all three pairs of sides are in the same ratio, or two pairs of angles are equal (then the third pair is equal too, by the angle sum).

Equal angles often come from a common angle shared by both triangles, a pair of right angles, vertically opposite angles, or corresponding and alternate angles on parallel lines.

Matching sides are opposite matching angles. The shortest side matches the shortest side and the longest matches the longest.

Similar triangles let you measure things you cannot reach, such as the height of a tree or the width of a river, from lengths you can measure.

A triangle cut by a line parallel to its baseTriangle ABC with DE parallel to BC. The small triangle ADE shares the angle at A, and the angles at D and B are corresponding angles, so triangle ADE is similar to triangle ABC. A B C D E
\(DE \parallel BC\): \(\triangle ADE \,|||\, \triangle ABC\). The common angle at \(A\) and the corresponding angles at \(D\) and \(B\) are equal.
Two triangles crossing at a point between parallel linesPQ is parallel to RS. The lines PS and QR cross at X. The angles at X are vertically opposite, and the angles at P and S are alternate angles, so triangle PQX is similar to triangle SRX. P Q X R S
\(PQ \parallel RS\): \(\triangle PQX \,|||\, \triangle SRX\). Vertically opposite angles at \(X\), alternate angles at \(P\) and \(S\).

If \(\triangle PQR \,|||\, \triangle XYZ\):

\[\dfrac{XY}{PQ} = \dfrac{YZ}{QR} = \dfrac{XZ}{PR} = k\]
XYPQ=YZQR=XZPR=k

To find an unknown \(x\), write it over its matching side and set it equal to another known pair:

\[\dfrac{x}{\text{matching side}} = \dfrac{\text{known side}}{\text{its matching side}}\]
xmatching side=known sideits matching side
Parallel line inside a triangle. If \(PQ \parallel ST\) inside a triangle with apex \(R\), \(RP = x\) and \(PS = d\), the large triangle's side is \(x + d\). An equation such as \(\dfrac{x}{x+d} = \dfrac{PQ}{ST}\) is solved by cross-multiplying.
Height to the hypotenuse. In a right-angled triangle, the perpendicular from the right angle to the hypotenuse makes two triangles similar to each other. If it cuts the hypotenuse into \(p\) and \(q\), its length \(h\) satisfies \(\dfrac{h}{p} = \dfrac{q}{h}\), so \(h^2 = pq\).
Right-angled similar triangles. If only two sides are given, Pythagoras' theorem finds the third first. Then the scale factor gives the other triangle's sides, and from them its area.

How to find an unknown side in similar triangles

  1. Show the triangles are similar: give two pairs of equal angles with reasons, or three sides in the same ratio.
  2. Separate overlapping triangles. Sketch them side by side, the same way round, using whole sides.
  3. Match the sides: each side is opposite a matching angle.
  4. Find the scale factor from a known pair, or write the ratio equation with the unknown on top.
  5. Solve and check that the bigger triangle has the bigger sides.

Real problems

  1. Shadows: height matches height, shadow matches shadow (same sun angle).
  2. Mirror on the ground: the angles at the mirror are equal, so eye height \(\div\) distance to the mirror matches object height \(\div\) its distance to the mirror.
  3. Lining up two tops: a short pole and a tall tower seen from one point share the angle at that point. Use the whole distance to the tower.
Example 1 β€” Are they similar?
Triangle \(KLM\) has sides \(6\), \(8\) and \(9\) cm. Triangle \(XYZ\) has sides \(9\), \(12\) and \(13.5\) cm. Are the triangles similar? If so, which angle of \(\triangle XYZ\) matches \(\angle K\)?
Solution
Two triangles with sides 6, 8, 9 and 9, 12, 13.5Triangle KLM has sides KL 9, LM 6 and MK 8. Triangle XYZ has sides XY 13.5, YZ 9 and ZX 12. Not to scale. 9 6 8 13.5 9 12 K L M X Y Z

Match shortest with shortest, middle with middle, longest with longest, and divide.

\(\dfrac{9}{6}\)\(=\)\(1.5\)
\(\dfrac{12}{8}\)\(=\)\(1.5\)
\(\dfrac{13.5}{9}\)\(=\)\(1.5\)
96=128=13.59=1.5

All three ratios are equal, so the triangles are similar with scale factor \(1.5\).

\(\angle K\) is opposite the \(6\) cm side \(LM\). The matching side is \(YZ = 9\) cm, which is opposite \(\angle X\). So \(\angle K\) matches \(\boldsymbol{\angle X}\).

Example 2 β€” A parallel line inside a triangle
In \(\triangle ABC\), \(DE \parallel BC\). \(AD = 9\) cm, \(DB = 6\) cm and \(DE = 12\) cm.
(i) Find \(BC\).
(ii) \(AC = 17.5\) cm. Find \(AE\).
Solution
Triangle ABC with DE parallel to BCTriangle ABC with D on AB and E on AC, DE parallel to BC. AD is 9, DB is 6 and DE is 12. BC is unknown. 9 6 12 ? A B C D E

\(\triangle ADE \,|||\, \triangle ABC\) (common angle at \(A\), corresponding angles at \(D\) and \(B\)). The large triangle uses the whole side \(AB = 9 + 6 = 15\) cm.

\(k\)\(=\)\(\dfrac{AB}{AD} = \dfrac{15}{9} = \dfrac{5}{3}\)
\(BC\)\(=\)\(12 \times \dfrac{5}{3} = \mathbf{20}\text{ cm}\)
\(AE\)\(=\)\(17.5 \div \dfrac{5}{3} = \mathbf{10.5}\text{ cm}\)
BC=12×159=20

Keep \(k\) as the fraction \(\dfrac{5}{3}\) rather than \(1.67\), so the answers stay exact.

Example 3 β€” Shadows
At the same time of day, a \(1.8\) m tall student casts a \(2.4\) m shadow and a tree casts a \(14\) m shadow. Find the height of the tree.
Solution
A person and a tree with their shadowsA person 1.8 m tall casts a 2.4 m shadow. At the same time a tree of height h casts a 14 m shadow. The sun's rays make the same angle with the ground in both. 1.8 m 2.4 m h 14 m

Both triangles have a right angle, and the sun's rays meet the ground at the same angle. So the triangles are similar: height matches height and shadow matches shadow.

\(\dfrac{h}{1.8}\)\(=\)\(\dfrac{14}{2.4}\)
\(h\)\(=\)\(\dfrac{1.8 \times 14}{2.4}\)
\(\)\(=\)\(\mathbf{10.5}\text{ m}\)
h1.8=142.4,h=10.5

Writing the ratios as unknown over its match puts \(h\) on top, so one multiplication finishes it.

Example 4 β€” Width of a river
A tree \(T\) is directly opposite \(B\) across a river. \(BC = 36\) m and \(CD = 15\) m along the bank. \(DE = 20\) m is at right angles to the bank, and \(T\), \(C\) and \(E\) are in a straight line. Find the width \(w = BT\).
Solution
Measuring the width of a river with similar trianglesA tree T is on the far bank directly opposite B. C is 36 m along the near bank from B, and D is 15 m further. E is 20 m from D at right angles to the bank, placed so that T, C and E are in a straight line. The width w is BT. river 36 m 15 m 20 m w T B C D E

\(\angle TCB = \angle ECD\) (vertically opposite) and \(\angle TBC = \angle EDC = 90^\circ\), so \(\triangle TBC \,|||\, \triangle EDC\). Match the sides: \(BT \leftrightarrow DE\) and \(BC \leftrightarrow DC\).

\(\dfrac{w}{20}\)\(=\)\(\dfrac{36}{15}\)
\(w\)\(=\)\(20 \times \dfrac{36}{15}\)
\(\)\(=\)\(\mathbf{48}\text{ m}\)
w=20×3615=48

The side next to the shared corner \(C\) on the bank matches the side next to \(C\) on the other triangle.

Common pitfalls

Use the whole side. With a parallel line inside a triangle, the large triangle's side is \(AD + DB\), not \(DB\). Using the piece \(DB\) is the most common error.
Match by angles, not by position. Turned or flipped triangles put matching sides in different places. Match each side by the angle opposite it.
Keep the ratios the same way up. If you write small over large on one side of the equation, write small over large on the other side too.
Equal angles, not equal sides. Similar triangles need not be the same size. If they are also the same size, they are congruent.

Frequently asked questions

How do you know if two triangles are similar?

Two triangles are similar if two pairs of angles are equal, or if all three pairs of sides are in the same ratio. Equal angles often come from a shared angle, right angles or parallel lines.

How do you find the missing side of a similar triangle?

Match the sides, find the scale factor from a known pair, then multiply or divide. Or write the unknown over its matching side equal to another known ratio and solve.

Which sides match in similar triangles?

Sides opposite equal angles match. The shortest side matches the shortest, and the longest matches the longest. In a similarity statement, the letters in the same positions match.

How do you use shadows to find the height of a tree?

At the same moment the sun makes the same angle for every object, so the triangles are similar. Tree height \(\div\) tree shadow \(=\) your height \(\div\) your shadow.

What is the difference between similar and congruent triangles?

Similar triangles have the same shape but can be different sizes. Congruent triangles have the same shape and the same size, so the scale factor is \(1\).