Chapter 6 · Triangles

6.1 Similar Figures

Two figures are similar when one is an exact enlargement or reduction of the other. Make that idea precise with two conditions — equal angles and proportional sides — and learn to write and read a similarity statement.

Take a photograph and print it twice, once as a small $6\ \text{cm}\times4\ \text{cm}$ picture and once as a large $15\ \text{cm}\times10\ \text{cm}$ poster. Nobody would say the poster has a different shape; it is the same picture made bigger. Now stretch the photograph so that it fills a $15\ \text{cm}\times 15\ \text{cm}$ square. The shape has changed — faces look wide and fat. What exactly is the difference between "bigger" and "stretched"?

In geometry, figures that have the same shape (but not necessarily the same size) are called similar. Our job in this section is to turn that picture into a precise test. We will find that a shape is preserved when two things hold together: all the angles stay the same and all the lengths are multiplied by the same number. We then focus on triangles, learn how to write a similarity statement correctly, and use it to find unknown lengths — which is the skill needed for the rest of the chapter.

What does "same shape" mean?

📘 Definition 6.1 — Similar figures
Two polygons with the same number of sides are similar if both of these conditions hold:
  1. their corresponding angles are equal, and
  2. their corresponding sides are in the same ratio (they are proportional).

"Corresponding" means "in the same place": the first vertex of one polygon matches the first vertex of the other, the second matches the second, and so on, and a side corresponds to the side joining the corresponding vertices. We write $\sim$ for "is similar to". For example, the poster rectangle and the photograph rectangle above are similar: all four angles are $90^\circ$, and $\dfrac{15}{6}=\dfrac{10}{4}=2.5$.

📘 Definition 6.2 — Scale factor
If the polygons are similar, the common ratio of corresponding sides, $\;k=\dfrac{\text{a side of the second figure}}{\text{the corresponding side of the first figure}}$, is called the scale factor (or ratio of similitude) from the first figure to the second.

If $k\gt1$ the second figure is an enlargement; if $0\lt k\lt1$ it is a reduction; and if $k=1$ the two figures are the same size and shape — they are congruent. So congruent figures are the special case of similar figures with scale factor $1$. A scale factor multiplies lengths: every side, and also the perimeter, becomes $k$ times as large. It never changes an angle.

Some similar figures need no checking at all, because their shape is completely fixed: any two circles are similar (the scale factor is the ratio of the radii), any two squares are similar, and any two equilateral triangles are similar (all angles are $60^\circ$, and the three side ratios are all equal to the ratio of the side lengths). Any two regular polygons with the same number of sides are likewise similar.

⚠️ Both conditions are needed — one is not enough

Equal angles alone do not make polygons similar. A $6\times2$ rectangle and a $6\times3$ rectangle both have four right angles, but the sides $\frac66=1$ and $\frac32=1.5$ are not in the same ratio — the second is a stretched version of the first.

Proportional sides alone do not make polygons similar. A square of side $4$ and a rhombus of side $8$ have sides in the same ratio $2$, but the square has right angles and the rhombus (say, with angles $60^\circ$ and $120^\circ$) does not: it is a "pushed-over" square.

We shall see in Section 6.3 that for triangles a smaller amount of information is enough — but for polygons with four or more sides, always check both.

✏️ Example 1 — Are the two rectangles similar?
Decide whether (a) a $12\ \text{cm}\times8\ \text{cm}$ rectangle and a $9\ \text{cm}\times6\ \text{cm}$ rectangle, (b) a $12\ \text{cm}\times8\ \text{cm}$ rectangle and a $15\ \text{cm}\times9\ \text{cm}$ rectangle are similar. If they are, give the scale factor from the first to the second.
Angles Every rectangle has four angles of $90^\circ$, so Condition 1 holds in both (a) and (b).
(a) Sides Match long side with long side and short side with short side: $\dfrac{9}{12}=\dfrac34$ and $\dfrac{6}{8}=\dfrac34$. The ratios are equal, so the rectangles are similar, with scale factor $k=\tfrac34$ (a reduction).
(b) Sides $\dfrac{15}{12}=1.25$ but $\dfrac{9}{8}=1.125$. The ratios differ, so the rectangles are not similar.
Check In (b), if we tried to enlarge the $12\times8$ rectangle until its length is $15$, its width would be $8\times1.25=10$, not $9$. The $15\times9$ rectangle is a differently shaped one.
Interactive 1 — The Scale Machine: Enlarge Without Changing the Shape

The blue triangle $ABC$ has sides $AB=6$, $BC=5$, $CA=4$. The red triangle $PQR$ is the same triangle multiplied by the scale factor $k$. Drag $k$ (or press Play): every side is multiplied by $k$, so the three ratios $\frac{PQ}{AB}$, $\frac{QR}{BC}$, $\frac{RP}{CA}$ stay equal — and the green angle arcs show that no angle ever changes.

1.5
sides $PQ$, $QR$, $RP$—
ratios $\tfrac{PQ}{AB},\tfrac{QR}{BC},\tfrac{RP}{CA}$—
angles of $PQR$ ($P$, $Q$, $R$)—
angles of $ABC$ ($A$, $B$, $C$)—

Similar triangles and the similarity statement

Triangles are the building blocks of all polygons, so they are the figures we study most. For two triangles Definition 6.1 becomes a precise statement with three equal angles and three equal ratios.

📘 Definition 6.3 — Similar triangles
$\triangle ABC$ is similar to $\triangle PQR$, written $\triangle ABC\sim\triangle PQR$, if $$\angle A=\angle P,\quad\angle B=\angle Q,\quad\angle C=\angle R\qquad\text{and}\qquad\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}.$$

The order of the letters in a similarity statement carries the information: the first letters ($A$, $P$) are corresponding vertices, so are the second ($B$, $Q$) and the third ($C$, $R$). Each side is named by the two vertices at its ends, so $AB$ corresponds to $PQ$, $BC$ to $QR$ and $CA$ to $RP$. A handy check: corresponding sides are opposite to equal angles — $BC$ is opposite $\angle A$, and $QR$ is opposite $\angle P=\angle A$.

🧭 Problem-solving strategy — using a similarity statement
  1. Match the vertices in the order written: $\triangle ABC\sim\triangle PQR$ means $A\leftrightarrow P$, $B\leftrightarrow Q$, $C\leftrightarrow R$.
  2. List the three equal angle pairs and the three corresponding side pairs. Writing them in two columns avoids mix-ups.
  3. Find the scale factor from a pair of sides that are both known.
  4. Multiply (or divide) the other known sides by the scale factor. Use equal angles to find missing angles; the angle sum is $180^\circ$ for the rest.
  5. Check that the longest side of one triangle corresponds to the longest side of the other, and the shortest to the shortest.
✏️ Example 2 — Reading a similarity statement
$\triangle ABC\sim\triangle PQR$ with $AB=6$ cm, $BC=9$ cm, $CA=12$ cm and $PQ=4$ cm. Also $\angle A=50^\circ$ and $\angle B=60^\circ$. Find $QR$, $RP$ and the angles $P$, $Q$, $R$.
Match $AB\leftrightarrow PQ$, $BC\leftrightarrow QR$, $CA\leftrightarrow RP$.
Scale factor From $ABC$ to $PQR$: $k=\dfrac{PQ}{AB}=\dfrac46=\dfrac23$.
Sides $QR=\dfrac23\times9=6$ cm and $RP=\dfrac23\times12=8$ cm.
Angles $\angle P=\angle A=50^\circ$ and $\angle Q=\angle B=60^\circ$. In $\triangle ABC$: $\angle C=180^\circ-50^\circ-60^\circ=70^\circ$, so $\angle R=\angle C=70^\circ$.
Check $\dfrac{AB}{PQ}=\dfrac64=1.5$, $\dfrac{BC}{QR}=\dfrac96=1.5$, $\dfrac{CA}{RP}=\dfrac{12}{8}=1.5$ ✓. The angles $50^\circ+60^\circ+70^\circ=180^\circ$ ✓.
✏️ Example 3 — Writing the correct statement
In $\triangle LMN$ and $\triangle RST$ we know $\angle L=\angle S$, $\angle M=\angle T$ and $\angle N=\angle R$, and the triangles are similar. Write the similarity statement, and write the three equal ratios of sides.
Match vertices $L\leftrightarrow S$, $M\leftrightarrow T$, $N\leftrightarrow R$. Write $L,M,N$ first, then the partners in the same order: $S,T,R$.
Statement $\triangle LMN\sim\triangle STR$. (It is not $\triangle RST$: that would pair $L$ with $R$.)
Ratios $\dfrac{LM}{ST}=\dfrac{MN}{TR}=\dfrac{NL}{RS}$. Each side is named by the matched vertices in the same order.

Finding lengths with similar figures

Once two figures are known to be similar, the scale factor gives every missing length. Because the sides are in proportion, we can also work by setting up a proportion and solving it: if $\dfrac{a}{b}=\dfrac{c}{d}$, then $ad=bc$ (multiply both sides by $bd$ to see why). Two natural uses are scale drawings and shadows.

Shadows. The Sun is so far away that its rays are almost parallel. An upright stick and an upright pole at the same moment make angles of the same size with the ground, and both make a right angle with the ground; so the two right triangles formed by object, shadow and ray have equal angles. They are similar, and heights are in the same ratio as shadow lengths. (Section 6.3 explains why equal angles are enough for triangles.)

✏️ Example 4 — Height of a flagpole
Aarav is $1.6$ m tall and his shadow is $2.4$ m long. At the same moment the shadow of the school flagpole is $9$ m long. How tall is the flagpole?
Model Let the pole be $h$ m tall. The triangle (Aarav, shadow, ray) is similar to the triangle (pole, shadow, ray), so $\dfrac{\text{height}}{\text{shadow}}$ is the same for both.
Proportion $\dfrac{h}{9}=\dfrac{1.6}{2.4}=\dfrac23$.
Solve $h=9\times\dfrac23=6$.
Answer The flagpole is $6$ m tall. Check: $1.6\times\dfrac{9}{2.4}=1.6\times3.75=6$ ✓.
✏️ Example 5 — An enlarged poster
A triangular sail-shaped logo has sides $5$ cm, $7$ cm and $8$ cm. It is enlarged on a poster by the scale factor $2.5$. Find the sides and the perimeter of the enlarged logo, and compare the two perimeters.
Sides Multiply every side by $2.5$: $5\times2.5=12.5$ cm, $7\times2.5=17.5$ cm, $8\times2.5=20$ cm.
Perimeters Original: $5+7+8=20$ cm. Enlarged: $12.5+17.5+20=50$ cm.
Compare $\dfrac{50}{20}=2.5$. The perimeter is multiplied by the same scale factor, because it is the sum of lengths that are each multiplied by $k$: $k(a+b+c)=ka+kb+kc$.
✏️ Example 6 — An unknown inside the ratios
$\triangle ABC\sim\triangle PQR$ with $AB=(x+2)$ cm, $BC=(3x-2)$ cm, $PQ=9$ cm and $QR=15$ cm. Find $x$ and the lengths $AB$ and $BC$.
Match $AB\leftrightarrow PQ$ and $BC\leftrightarrow QR$, so $\dfrac{AB}{PQ}=\dfrac{BC}{QR}$, i.e. $\dfrac{x+2}{9}=\dfrac{3x-2}{15}$.
Clear fractions $15(x+2)=9(3x-2)$, so $15x+30=27x-18$.
Solve $48=12x$, so $x=4$.
Lengths $AB=4+2=6$ cm and $BC=3(4)-2=10$ cm.
Check $\dfrac{6}{9}=\dfrac23$ and $\dfrac{10}{15}=\dfrac23$ ✓.
Interactive 2 — Same Angles or Same Ratios? You Need Both

The blue figure is fixed. Change the red figure with the two sliders and see when it is similar to the blue one. In "Rectangles" the angles are always $90^\circ$ — only the ratios can fail. In "Square and rhombus" the sides are always in proportion — only the angles can fail. Press Play to sweep the second slider and watch the verdict.

95
angle condition—
side ratios—
similar?—

Exercises

I
Are the two figures always, sometimes or never similar? If they are always similar, say why; if not, give a counter-example.
  • 1 Any two squares
  • 2 Any two rectangles
  • 3 Any two rhombuses
  • 4 Any two equilateral triangles
  • 5 Any two circles
  • 6 Any two isosceles triangles
  • 7 A $12\ \text{cm}\times8\ \text{cm}$ rectangle and a $9\ \text{cm}\times6\ \text{cm}$ rectangle
  • 8 A $10\ \text{cm}\times6\ \text{cm}$ rectangle and a $15\ \text{cm}\times8\ \text{cm}$ rectangle
  • 9 Two regular hexagons, with sides $2$ cm and $5$ cm
  • 10 A square of side $5$ cm and a rhombus of side $10$ cm with one angle $70^\circ$
1 Always similar. All angles are $90^\circ$, and for sides $s_1$ and $s_2$ all four ratios equal $\dfrac{s_2}{s_1}$.
2 Not always. Angles are equal, but $6\times2$ and $6\times3$ have side ratios $1$ and $1.5$.
3 Not always. Sides are proportional, but a rhombus with angles $60^\circ,120^\circ$ and one with $90^\circ$ (a square) have different angles.
4 Always similar. All angles are $60^\circ$, and the three side ratios all equal $\dfrac{s_2}{s_1}$.
5 Always similar. A circle has no corners; one is an enlargement of the other with scale factor equal to the ratio of the radii.
6 Not always. An isosceles triangle with apex angle $40^\circ$ and one with apex angle $100^\circ$ have different angles.
7 Similar. $\dfrac{9}{12}=\dfrac{6}{8}=\dfrac34$ and all angles are $90^\circ$.
8 Not similar. $\dfrac{15}{10}=1.5$ but $\dfrac{8}{6}=1.33\ldots$
9 Similar. Every angle of a regular hexagon is $120^\circ$, and all sides are in the ratio $\dfrac52=2.5$.
10 Not similar. Sides are in the ratio $\dfrac{10}{5}=2$ for all four, but the angles are $90^\circ$ in the square and $70^\circ/110^\circ$ in the rhombus.
II
Similarity statements and scale factors.
  • 1 $\triangle ABC\sim\triangle DEF$ with $AB=6$ cm, $BC=8$ cm, $CA=10$ cm and $DE=9$ cm. Find $EF$ and $FD$.
  • 2 $\triangle ABC\sim\triangle PQR$ with $\angle A=40^\circ$ and $\angle B=75^\circ$. Find $\angle P$, $\angle Q$ and $\angle R$.
  • 3 In $\triangle LMN$ and $\triangle RST$: $\angle L=\angle S$, $\angle M=\angle T$, $\angle N=\angle R$, and the triangles are similar. Write the similarity statement and the equal ratios of sides.
  • 4 $\triangle ABC\sim\triangle QRP$ with $AB=8$ cm, $BC=12$ cm, $CA=10$ cm and $QR=6$ cm. Find $RP$ and $PQ$.
  • 5 $\triangle ABC\sim\triangle DEF$, the perimeter of $\triangle ABC$ is $24$ cm and $\dfrac{DE}{AB}=2.5$. Find the perimeter of $\triangle DEF$.
  • 6 Two similar triangles have corresponding sides in the ratio $3:5$. The perimeter of the larger one is $40$ cm. Find the perimeter of the smaller one.
  • 7 $\triangle ABC\sim\triangle DEF$. Which side of $\triangle DEF$ corresponds to $AC$? If $AC=14$ cm and that side is $21$ cm, what is the scale factor from $ABC$ to $DEF$?
1 $k=\dfrac{DE}{AB}=\dfrac96=1.5$. $EF=1.5\times8=12$ cm and $FD=1.5\times10=15$ cm.
2 $\angle P=\angle A=40^\circ$, $\angle Q=\angle B=75^\circ$, and $\angle R=\angle C=180^\circ-40^\circ-75^\circ=65^\circ$.
3 $\triangle LMN\sim\triangle STR$ ($L\leftrightarrow S$, $M\leftrightarrow T$, $N\leftrightarrow R$). Ratios: $\dfrac{LM}{ST}=\dfrac{MN}{TR}=\dfrac{NL}{RS}$.
4 $A\leftrightarrow Q$, $B\leftrightarrow R$, $C\leftrightarrow P$. So $AB\leftrightarrow QR$, $BC\leftrightarrow RP$, $CA\leftrightarrow PQ$. $\dfrac{AB}{QR}=\dfrac86=\dfrac43$. Then $RP=12\times\dfrac34=9$ cm and $PQ=10\times\dfrac34=7.5$ cm.
5 Perimeters scale by the same factor: $24\times2.5=60$ cm.
6 The smaller perimeter is $\dfrac35$ of the larger: $\dfrac35\times40=24$ cm.
7 $A\leftrightarrow D$ and $C\leftrightarrow F$, so $AC\leftrightarrow DF$. The scale factor is $\dfrac{DF}{AC}=\dfrac{21}{14}=1.5$.
III
Applications.
  • 1 A boy $1.5$ m tall casts a shadow $2$ m long. At the same time a lamp-post casts a shadow $12$ m long. How tall is the lamp-post?
  • 2 On a map with scale $1:50\,000$ (that is, $1$ cm on the map is $50\,000$ cm on the ground), two towns are $6.4$ cm apart. How far apart are they, in km?
  • 3 A triangular field has sides $2$ cm, $3$ cm and $4$ cm on the same map (scale $1:50\,000$). Find its real side lengths in km and its real perimeter.
  • 4 A $15\ \text{cm}\times10\ \text{cm}$ photograph is enlarged, without distortion, so that its longer side is $24$ cm. Find the new shorter side. Will the enlargement fit in a frame $24\ \text{cm}\times14\ \text{cm}$?
  • 5 A model aeroplane is made to the scale $1:40$. The real wingspan is $32$ m and the model is $45$ cm long. Find the model's wingspan and the real length of the aeroplane.
  • 6 A pole $4$ m tall casts a shadow $6$ m long. At the same moment a building casts a shadow $45$ m long. Find the height of the building.
  • 7 A slide $4$ cm wide and $3$ cm high is projected on a screen so that the picture is $1.6$ m wide. How high is the picture on the screen, if the picture is similar to the slide?
1 $\dfrac{h}{12}=\dfrac{1.5}{2}$, so $h=12\times0.75=9$ m.
2 $6.4\times50\,000=320\,000$ cm $=3200$ m $=3.2$ km.
3 Each side: $2\times50\,000=100\,000$ cm $=1$ km; $3\to1.5$ km; $4\to2$ km. Perimeter $=1+1.5+2=4.5$ km.
4 $k=\dfrac{24}{15}=1.6$, so the shorter side is $10\times1.6=16$ cm. The enlargement is $24\ \text{cm}\times16$ cm, which is taller than the $14$ cm frame, so it does not fit (it would have to be shrunk to $\tfrac{14}{16}$ of its size).
5 Model wingspan: $\dfrac{32\ \text{m}}{40}=0.8$ m $=80$ cm. Real length: $45\times40=1800$ cm $=18$ m.
6 $\dfrac{H}{45}=\dfrac46$, so $H=45\times\dfrac23=30$ m.
7 $k=\dfrac{160}{4}=40$ (using cm), so the height is $3\times40=120$ cm $=1.2$ m.
IV
Algebra and reasoning.
  • 1 $\triangle ABC\sim\triangle PQR$ with $AB=(2x-1)$ cm, $BC=(x+5)$ cm, $PQ=10$ cm, $QR=15$ cm. Find $x$, and hence $AB$ and $BC$.
  • 2 A $12\times8$ rectangle is similar to a rectangle with sides $(x+3)$ and $x$ (long side first). Find $x$.
  • 3 Use Definition 6.3 to show that any two equilateral triangles are similar.
  • 4 A triangle has sides $4$ cm, $6$ cm and $8$ cm. A similar triangle has perimeter $54$ cm. Find its sides.
  • 5 A lawn is $20$ m by $12$ m. A path of uniform width $x$ m ($x\gt0$) runs round it. Show that the outer rectangle is not similar to the lawn.
  • 6 If $\triangle ABC\sim\triangle DEF$ and $\triangle DEF\sim\triangle PQR$, show that $\triangle ABC\sim\triangle PQR$.
1 $\dfrac{2x-1}{10}=\dfrac{x+5}{15}$, so $15(2x-1)=10(x+5)$, $30x-15=10x+50$, $20x=65$, $x=3.25$. Then $AB=5.5$ cm and $BC=8.25$ cm. Check: $\tfrac{5.5}{10}=0.55$, $\tfrac{8.25}{15}=0.55$ ✓.
2 $\dfrac{x+3}{12}=\dfrac{x}{8}$, so $8x+24=12x$, $x=6$. The rectangle is $9\times6$, scale factor $\tfrac34$ ✓.
3 Let the triangles have sides $s$ and $t$. All six angles are $60^\circ$, so the three angle conditions hold. The three ratios of corresponding sides are $\dfrac ts,\dfrac ts,\dfrac ts$ — equal. So both conditions of Definition 6.3 hold.
4 The original perimeter is $18$ cm, so $k=\dfrac{54}{18}=3$. The sides are $12$ cm, $18$ cm and $24$ cm.
5 Outer sides $20+2x$ and $12+2x$. Similarity needs $\dfrac{20+2x}{20}=\dfrac{12+2x}{12}$, so $12(20+2x)=20(12+2x)$, i.e. $240+24x=240+40x$, so $16x=0$ and $x=0$. A path of positive width ($x\gt0$) can never give similar rectangles. (Frames and borders change the shape of a picture!)
6 The angles of $ABC$ equal those of $DEF$, which equal those of $PQR$. If $\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{CA}{FD}=k_1$ and $\dfrac{DE}{PQ}=\dfrac{EF}{QR}=\dfrac{FD}{RP}=k_2$, then $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{CA}{RP}=k_1k_2$. So the conditions hold: $\triangle ABC\sim\triangle PQR$.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case study

How this is examined: board papers ask for the definition-level facts as 1-mark MCQs or Assertion–Reason items (which pairs are always similar, which angle matches which), and for short numerical problems — find the missing sides of similar triangles, or an unknown $x$ from equal ratios (2–3 marks). Always write the similarity statement with the vertices in the correct order, because the order earns the first mark.

E1. Board · 1 mark Which pair of figures is always similar?  (A) two rectangles  (B) two rhombuses  (C) two equilateral triangles  (D) two isosceles triangles

Test both conditions Two equilateral triangles have all angles $60^\circ$ and all side ratios equal to the ratio of their sides.
Trap Rectangles (A) have equal angles but sides can be in different ratios; rhombuses (B) have proportional sides but different angles; isosceles triangles (D) can have different apex angles.
Answer (C).

E2. Board · 1 mark Assertion (A): A square of side $3$ cm and a rhombus of side $6$ cm with one angle $60^\circ$ are similar. Reason (R): Two polygons are similar if their corresponding angles are equal and their corresponding sides are proportional.
(A) Both A and R are true and R is the correct explanation of A.  (B) Both A and R are true but R is not the correct explanation of A.  (C) A is true but R is false.  (D) A is false but R is true.

Check R This is exactly Definition 6.1 — equal angles and proportional sides. R is true.
Check A The sides are in the ratio $\tfrac63=2$, but the square has angles $90^\circ$ while the rhombus has angles $60^\circ$ and $120^\circ$. The angle condition fails, so A is false.
Answer (D). R is true, A is false. Option (B) is tempting because "proportional sides" sounds like enough, but the rhombus is a pushed-over square.

E3. Board · 2 marks $\triangle ABC\sim\triangle DEF$, $AB=7$ cm, $BC=6$ cm, $CA=7$ cm and $EF=9$ cm. Find the perimeter of $\triangle DEF$.

Scale factor $BC\leftrightarrow EF$, so $k=\dfrac{EF}{BC}=\dfrac96=1.5$. (1 mark)
Perimeter Perimeter of $ABC=7+6+7=20$ cm, so perimeter of $DEF=1.5\times20=30$ cm. (Sides: $DE=10.5$, $EF=9$, $FD=10.5$; sum $30$ ✓.) (1 mark)

E4. Board · 3 marks $\triangle ABC\sim\triangle PQR$ with $AB=x$ cm, $BC=(x+4)$ cm, $CA=10$ cm, $PQ=6$ cm and $QR=12$ cm. Find $x$ and the length $RP$.

Equal ratios $\dfrac{AB}{PQ}=\dfrac{BC}{QR}$, so $\dfrac{x}{6}=\dfrac{x+4}{12}$. (1 mark)
Solve $12x=6(x+4)$, so $6x=24$ and $x=4$. Then $AB=4$ cm and $BC=8$ cm. (1 mark)
RP $\dfrac{CA}{RP}=\dfrac{BC}{QR}=\dfrac{8}{12}=\dfrac23$, so $RP=10\times\dfrac32=15$ cm. (1 mark) Check: $\tfrac{4}{6}=\tfrac{8}{12}=\tfrac{10}{15}=\tfrac23$ ✓.

E5. Board · Case study A town planner draws a plan of a rectangular park to the scale $1:500$ (so $1$ cm on the plan is $500$ cm $=5$ m on the ground). On the plan the park measures $12$ cm by $8$ cm.
(i) Find the real length and breadth of the park. (2 marks)
(ii) A rectangular pond in the park measures $15$ m by $10$ m. Is the pond similar to the park? (1 mark)
(iii) The perimeter of the park on the plan is $40$ cm. Find the real perimeter and show that it is $500$ times the plan perimeter. (1 mark)

(i) Length: $12\times5=60$ m. Breadth: $8\times5=40$ m. The park is $60\ \text{m}\times40\ \text{m}$.
(ii) Angles are all $90^\circ$. Ratios: $\dfrac{15}{60}=\dfrac14$ and $\dfrac{10}{40}=\dfrac14$. Equal, so the pond is similar to the park, with scale factor $\tfrac14$.
(iii) Real perimeter $=2(60+40)=200$ m $=20\,000$ cm. And $40\times500=20\,000$ cm ✓, so lengths and perimeters both scale by $500$.
🔬 Research–Teaching NexusMini project · optional, about 45 minutes

Research question. Can you measure the height of a tree, a lamp-post or your school building using only a stick, a tape measure and the Sun — and how accurate is the answer?

Why it matters. Surveyors, foresters and archaeologists measure things they cannot climb or reach. The oldest recorded use of this method is by the Greek thinker Thales, who is said to have found the height of a pyramid from its shadow. The same idea of scaling from a small, measurable triangle to a huge one is used today in map-making, in building scale models of bridges and aircraft for wind-tunnel tests, and in comparing medical scans taken at different sizes.

What to do.

  1. Choose a tall object (a tree, flagpole, or building) with level ground beside it. On a sunny day, push a straight stick or a metre rule upright into the ground, or hold it vertically.
  2. At the same moment, measure the stick's height $s$, the length $t$ of the stick's shadow, and the length $T$ of the object's shadow (from its base to the shadow tip). Use a tape measure; if the base is hidden, measure to a fixed mark and add the distance.
  3. Calculate the object's height $H=\dfrac{s}{t}\times T$. Explain in a sentence why the two triangles are similar.
  4. Repeat three times at intervals of $15$ minutes. The shadows change, but the ratio $\dfrac{T}{t}$ should stay nearly the same.
  5. If you can, measure the real height another way (the height of a floor is about $3$ m — count floors, or ask the caretaker) and find the percentage error.

Tools: A straight stick or metre rule, a tape measure, a notebook, a calculator or spreadsheet

Hand in: A neat sketch of the two triangles, a table of your three sets of measurements with the calculated heights, your best estimate with the percentage difference from your check, and two sentences on what causes errors (stick not vertical, ground not level, fuzzy shadow edge…).

Going further. What if the ground slopes? Try a cloudy day with a lamp instead of the Sun: the lamp's rays are no longer parallel — does the method still work? (Hint: draw the triangles.)

Teacher/parent note: the method rests on the Sun's rays being effectively parallel and both objects being vertical, so the two shadow triangles have equal angles. Typical student errors are measuring the stick's shadow at a different time from the object's, and measuring to the shadow of the top of a tree crown that is not directly above its trunk. Errors of $5$–$10\%$ are normal.

🧪 Self-check quiz

0/5 correct

Q1. Two polygons with the same number of sides are similar when

Q2. If $\triangle ABC\sim\triangle PQR$, then $\dfrac{AB}{PQ}$ equals

Q3. $\triangle ABC\sim\triangle DEF$, $\angle A=50^\circ$ and $\angle E=60^\circ$. Then $\angle C$ is

Q4. A rectangle $8\times6$ is similar to a rectangle with sides $12$ and $x$, where $12$ corresponds to $8$. Then $x=$

Q5. A triangle has sides $4$, $5$ and $7$ cm. A similar triangle has perimeter $48$ cm. Its longest side is

← Chapter 6 — Triangles6.2 The Basic Proportionality Theorem →