1. Introduction to Triangles and Similarity
A triangle is a three-sided polygon. Every triangle has three sides, three angles and an important fixed property: the sum of its three interior angles is always \(180^{\circ}\). If you shrink or stretch a triangle uniformly, its angles stay exactly the same while its sides grow or shorten in one common proportion — that is the birth of similarity.
Similar figures surround us: a photo and its enlargement, the map of a city and the city itself, a toy car and the real car, a shadow and the object that casts it. In each case the shape is identical but the size differs by a fixed scale factor. This chapter formalises that idea for triangles and uses it to measure heights, widths and scaled drawings without ever walking to the object you are measuring.
Similar figures
Same shape, different (or equal) sizes.
Scale factor
The fixed multiplier \(k\) applied to every side.
Similar triangles
Corresponding angles equal, corresponding sides proportional.
BPT
A line parallel to a side divides the other two proportionally.
What you will learn in this chapter:
- Tell similar figures apart from congruent ones.
- Write and interpret the similarity symbol \(\triangle ABC \sim \triangle DEF\) and always use the correct vertex order.
- Name corresponding sides and angles confidently.
- State, prove and apply the Basic Proportionality Theorem and its converse.
- Decide similarity using the AA, SSS and SAS criteria.
- Connect perimeters (same ratio) and areas (squared ratio) of similar triangles.
- Solve shadow, map, model and surveying problems with a single proportion.
- Answer competency and case-based questions with reasoning, not just calculation.
2. Similar Figures
Two figures are similar when they have exactly the same shape. Size does not matter: a small equilateral triangle and a large equilateral triangle are similar because they differ only by a uniform scale. The circles below and the pair of triangles show the idea — each pair has the same shape but a different size.
Two figures having the same shape (not necessarily the same size) are called similar figures. For polygons this means two things together: their corresponding angles are equal and their corresponding sides are in the same ratio.
Which figures are always similar?
| Figure | Always similar? | Why |
|---|---|---|
| Any two line segments | Yes | Same shape; which is "larger" is just a scale |
| Any two circles | Yes | Every circle has the same shape |
| Any two squares | Yes | All angles \(90^{\circ}\) and all sides equal ratio |
| Any two equilateral triangles | Yes | All three angles \(60^{\circ}\) in each |
| Any two rectangles | No | Only if length : breadth ratios match |
| Any two isosceles triangles | No | Only if the vertex angles, hence the ratios, match |
Same shape is not the same as same size. Two figures that are identical in both shape and size are called congruent; congruent figures are a special case of similarity where the scale factor is exactly \(1\).
3. Similar Figures vs Congruent Figures
Congruence compares size as well as shape; similarity compares only shape. A scaled-up photograph is similar to the original but not congruent to it, while a photocopy at 100% is both.
| Property | Congruent figures | Similar figures |
|---|---|---|
| Shape | Same | Same |
| Size | Same | Same or different |
| Corresponding angles | Equal | Equal |
| Corresponding sides | Equal in length | In the same ratio |
| Scale factor | Exactly 1 | Any positive number \(k\) |
| Symbol | \(\cong\) | \(\sim\) |
Congruent = same size and shape; similar = same shape. Every pair of congruent figures is similar (with scale factor 1), but most similar figures are not congruent.
4. Similar Triangles — Notation and Meaning
We write \(\triangle ABC \sim \triangle DEF\), read as "triangle ABC is similar to triangle DEF". The order of the letters is not optional: it tells you exactly which parts correspond.
In \(\triangle ABC \sim \triangle DEF\):
- Vertex \(A\) corresponds to vertex \(D\), so \(\angle A = \angle D\).
- Vertex \(B\) corresponds to vertex \(E\), so \(\angle B = \angle E\).
- Vertex \(C\) corresponds to vertex \(F\), so \(\angle C = \angle F\).
- Side \(AB\) corresponds to side \(DE\) and \(\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{CA}{FD}\).
| \(\triangle ABC\) | Corresponds to | \(\triangle DEF\) |
|---|---|---|
| Vertex \(A\) | ↔ | Vertex \(D\) |
| Vertex \(B\) | ↔ | Vertex \(E\) |
| Vertex \(C\) | ↔ | Vertex \(F\) |
| Side \(AB\) | ↔ | Side \(DE\) |
| Side \(BC\) | ↔ | Side \(EF\) |
| Side \(CA\) | ↔ | Side \(FD\) |
| \(\angle A\) | ↔ | \(\angle D\) |
Writing \(\triangle ABC \sim \triangle DEF\) while using the side pairing \(AB\leftrightarrow EF\) is wrong. The written order decides the correspondence: \(A\leftrightarrow D\), \(B\leftrightarrow E\), \(C\leftrightarrow F\), and the ratios must be written in that order.
5. Corresponding Sides and Angles
Corresponding parts of similar triangles behave predictably. The two triangles below are similar: \(\triangle ABC \sim \triangle DEF\). Angles that sit at corresponding positions are equal, and the three pairs of corresponding sides share one common ratio.
\(\triangle ABC \sim \triangle DEF\) with \(AB = 6\text{ cm}\), \(BC = 9\text{ cm}\), \(CA = 12\text{ cm}\) and \(DE = 4\text{ cm}\). Find \(EF\) and \(FD\).
The common ratio is \(\dfrac{AB}{DE} = \dfrac{6}{4} = \dfrac{3}{2}\). Hence
\[EF = BC \times \frac{2}{3} = 9 \times \frac{2}{3} = 6\text{ cm},\qquad FD = CA \times \frac{2}{3} = 12 \times \frac{2}{3} = 8\text{ cm}\]
So \(EF = 6\) cm and \(FD = 8\) cm ✓ (check: \(6,9,12\) is every corresponding side of \(4,6,8\) multiplied by \(\frac{3}{2}\)).
To find the common ratio, use the pair of sides where both lengths are given, then apply that single ratio to every other corresponding pair.
6. Basic Proportionality Theorem (BPT)
The Basic Proportionality Theorem, also called Thales' Theorem, is the backbone of this chapter. It says that a line drawn inside a triangle, parallel to one side, cuts the other two sides in the same ratio.
If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio.
In \(\triangle ABC\), if \(DE \parallel BC\) with \(D\) on \(AB\) and \(E\) on \(AC\), then:
\[ \boxed{\frac{AD}{DB} = \frac{AE}{EC}} \]
Proof of the BPT
In \(\triangle ABC\), \(DE \parallel BC\), \(AD = 3\text{ cm}\), \(DB = 5\text{ cm}\) and \(AE = 4.5\text{ cm}\). Find \(EC\).
By BPT, \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\), so \(\dfrac{3}{5} = \dfrac{4.5}{EC}\).
\[EC = \frac{5 \times 4.5}{3} = 7.5\text{ cm}\]
So \(EC = 7.5\) cm.
In \(\triangle ABC\), \(DE \parallel BC\), \(AD = 2\text{ cm}\), \(DB = 3\text{ cm}\), \(AC = 10\text{ cm}\). Find \(AC\)'s split into \(AE\) and \(EC\) — that is, find \(AE\) and \(EC\).
Using \(\dfrac{AD}{DB} = \dfrac{AE}{EC} = \dfrac{2}{3}\), the whole of \(AC\) is divided in the ratio \(2 : 3\):
\[AE = \frac{2}{2+3} \times 10 = 4\text{ cm},\qquad EC = \frac{3}{2+3} \times 10 = 6\text{ cm}\]
Check: \(4 + 6 = 10\) cm ✓, and \(\frac{AD}{DB} = \frac{2}{3} = \frac{AE}{EC} = \frac{4}{6}\) ✓.
BPT needs the line to be parallel to a side. A line that merely cuts two sides without being parallel to the third side does not give the equal-ratio property.
7. Converse of BPT
The reverse statement is equally powerful: if a line splits two sides in the same ratio, the line must be parallel to the third side.
If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
In \(\triangle ABC\), if \(D\) lies on \(AB\), \(E\) lies on \(AC\), and \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\), then \(DE \parallel BC\).
In \(\triangle ABC\), \(AD = 2.4\text{ cm}\), \(DB = 3.6\text{ cm}\), \(AE = 2.8\text{ cm}\), \(EC = 4.2\text{ cm}\). Is \(DE \parallel BC\)?
Compute both ratios:
\[\frac{AD}{DB} = \frac{2.4}{3.6} = \frac{2}{3},\qquad \frac{AE}{EC} = \frac{2.8}{4.2} = \frac{2}{3}\]
The ratios are equal, so by the converse of BPT, \(DE \parallel BC\).
Simplify each ratio before comparing. Writing \(2.4 : 3.6 = 2 : 3\) and \(2.8 : 4.2 = 2 : 3\) makes the equality obvious and carries marks in board solutions.
8. Criteria for Similarity: AA, SSS and SAS
Checking all six conditions (three angles and three sides) is tedious. The three criteria below let you prove similarity from much less information. They are the similarity counterparts of the congruence rules, with "equal sides" relaxed to "proportional sides".
Criterion 1 — AA similarity
If two angles of one triangle are equal to two corresponding angles of another triangle, the triangles are similar. (The third angles are then automatically equal, since each sum is \(180^{\circ}\).)
Criterion 2 — SSS similarity
If the three sides of one triangle are proportional to the three sides of another triangle, the triangles are similar.
Criterion 3 — SAS similarity
If one angle of a triangle equals one angle of another triangle and the two sides including that angle are proportional, the triangles are similar. The equal angle must be the included angle — the angle between the two sides being compared.
Two sides and a non-included angle (an "SSA" match) does not guarantee similarity. In SAS, the given angle must sit between the two sides, exactly as \(\angle A\) lies between \(AB\) and \(AC\).
In \(\triangle ABC\), \(\angle A = 40^{\circ}\) and \(\angle B = 65^{\circ}\). In \(\triangle DEF\), \(\angle D = 40^{\circ}\) and \(\angle E = 65^{\circ}\). Are the triangles similar?
\(\angle A = \angle D = 40^{\circ}\) and \(\angle B = \angle E = 65^{\circ}\), so by the AA criterion, \(\triangle ABC \sim \triangle DEF\). (The third angles agree too: both are \(180^{\circ} - 105^{\circ} = 75^{\circ}\).)
\(\triangle ABC\) has sides \(3, 4, 5.5\) and \(\triangle DEF\) has sides \(6, 8, 11\). Are they similar?
Check the ratios in matching order:
\[\frac{6}{3} = 2,\qquad \frac{8}{4} = 2,\qquad \frac{11}{5.5} = 2\]
All three ratios are equal, so by the SSS criterion, \(\triangle ABC \sim \triangle DEF\) with scale factor \(\frac{1}{2}\).
In \(\triangle ABC\), \(AB = 5\), \(AC = 7\) and \(\angle A = 60^{\circ}\). In \(\triangle DEF\), \(DE = 10\), \(DF = 14\) and \(\angle D = 60^{\circ}\). Are they similar?
The equal angle \(\angle A = \angle D = 60^{\circ}\) is the included angle in both pairs, and
\[\frac{AB}{DE} = \frac{5}{10} = \frac{1}{2},\qquad \frac{AC}{DF} = \frac{7}{14} = \frac{1}{2}\]
So by the SAS criterion, \(\triangle ABC \sim \triangle DEF\) with scale factor \(\frac{1}{2}\).
Under AA, only two angles are needed, because the third follows for free. Under SSS, all three side-ratios must match. Under SAS, two sides and the included angle match.
9. Important Results
From the definition and the criteria, a cluster of results follows almost immediately. Learn them as a set; board questions reuse them constantly.
| Result | Statement |
|---|---|
| 1. Transitivity | If \(\triangle ABC \sim \triangle PQR\) and \(\triangle PQR \sim \triangle XYZ\), then \(\triangle ABC \sim \triangle XYZ\). |
| 2. Similarity is not equality of sides | Similar triangles can always be laid one over the other after scaling; equal sides make them congruent, a sub-case. |
| 3. Perimeters in the same ratio | If \(\triangle ABC \sim \triangle DEF\) with scale factor \(k\), then their perimeters are also in the ratio \(k : 1\). |
| 4. Areas in the squared ratio | The areas of similar triangles are in the square of the ratio of corresponding sides. |
| 5. Equal angles, proportional sides | Corresponding angles of similar triangles are equal and corresponding sides are proportional — the two defining properties always occur together. |
| 6. Scale factor below 1 | If the scale factor \(k < 1\), the image triangle is smaller; if \(k > 1\) it is larger; if \(k = 1\) the triangles are congruent. |
\(\triangle ABC \sim \triangle DEF\) with \(AB = 8\), \(BC = 10\), \(CA = 6\) and \(DE = 12\). Find the perimeter of \(\triangle DEF\).
Scale factor = \(\dfrac{DE}{AB} = \dfrac{12}{8} = \dfrac{3}{2}\). Perimeter of \(\triangle ABC = 8 + 10 + 6 = 24\).
\[P(\triangle DEF) = 24 \times \frac{3}{2} = 36\]
So the perimeter of \(\triangle DEF\) is 36. Direct check: \(12 + 15 + 9 = 36\) ✓.
10. Ratio of Areas of Similar Triangles
This is one of the most frequently examined results of the chapter. Whenever triangles are similar, their sides stay in one ratio \(k\), but their areas grow by \(k^{2}\) because area depends on two linear dimensions (base and height).
If \(\triangle ABC \sim \triangle DEF\), with corresponding sides in the ratio \(k\), then
\[ \boxed{\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle DEF)} = k^{2} = \left(\frac{AB}{DE}\right)^{2} = \left(\frac{BC}{EF}\right)^{2} = \left(\frac{CA}{FD}\right)^{2}} \]
Derivation
\(\triangle ABC \sim \triangle DEF\) and the ratio of corresponding sides is \(2 : 5\) (i.e. \(k = \frac{2}{5}\), \(\triangle ABC\) smaller). If \(\operatorname{ar}(\triangle ABC) = 20\text{ cm}^{2}\), find \(\operatorname{ar}(\triangle DEF)\).
Areas are in the ratio \(k^{2} = 4 : 25\). With \(\triangle ABC = 20\text{ cm}^{2}\),
\[\operatorname{ar}(\triangle DEF) = 20 \times \frac{25}{4} = 125\text{ cm}^{2}\]
So \(\operatorname{ar}(\triangle DEF) = \mathbf{125\text{ cm}^{2}}\).
Two similar triangles have areas \(49\text{ cm}^{2}\) and \(81\text{ cm}^{2}\). If a side of the smaller triangle is \(7\text{ cm}\), find the corresponding side of the larger triangle.
\[\frac{\text{smaller area}}{\text{larger area}} = \frac{49}{81} \implies k^{2} = \frac{49}{81} \implies k = \frac{7}{9}\]
Corresponding side of larger \(= 7 \times \frac{9}{7} = 9\text{ cm}\).
So the corresponding side is 9 cm.
For similar figures the perimeters use the ratio \(k\), but the areas use \(k^{2}\). A scale factor \(3\) multiplies sides by \(3\) and areas by \(9\). Mixing these two is the single most common error in this chapter.
11. Important Proofs
Board papers repeatedly ask for whole proofs. Practise boiling each one down to Given, To prove, Construction and Proof steps.
Write "By BPT" / "By converse of BPT" at the exact line you apply them. Even a correct answer loses marks if the theorem name and the ratio it produces are not written explicitly.
12. Solved Examples
Twelve fully worked problems covering every idea in the chapter. Try each one on paper before reading the solution.
13. Application-Based Problems
Similarity's power is measurement at a distance. The pattern is always the same: identify two similar triangles, write the proportion, and solve for the unknown length. The diagrams below show the three classic setups: a shadow, a survey across a river, and a scaled drawing.
Shadows: heights from shadows
A \(1.8\text{ m}\) girl casts a shadow \(2.4\text{ m}\) long. The shadow of a tree at the same moment is \(12\text{ m}\). Height of tree \(= \dfrac{1.8}{2.4} \times 12 = 9\text{ m}\). So the tree is 9 m tall.
Surveying: measuring a distance you cannot walk
To find the width of a river, a surveyor forms two similar triangles sharing vertex \(A\). A small triangle with base \(2\text{ m}\) gives a scale of \(1 : 12\) against the full triangle, whose base is the river width \(W\). So \(W = 2 \times 12 = 24\text{ m}\). The river is 24 m wide.
Scale drawings and maps
On a tourist map, \(1\text{ cm}\) represents \(5\text{ km}\). Two towns are \(4.6\text{ cm}\) apart on the map. Actual distance \(= 4.6 \times 5 = 23\text{ km}\). The towns are 23 km apart.
A model car is \(12\text{ cm}\) long and represents a real car of length \(4.8\text{ m}\). Scale = \(12\text{ cm} : 480\text{ cm} = 1 : 40\). The model is built at one fortieth scale.
A photo is \(8\text{ cm}\) tall and \(12\text{ cm}\) wide. It is enlarged so the height becomes \(20\text{ cm}\). Scale factor \(= \dfrac{20}{8} = 2.5\), so the new width \(= 12 \times 2.5 = 30\text{ cm}\).
A triangular bracket \(ABC\) has a strut \(DE \parallel BC\) with \(AD = 6\), \(DB = 12\), \(AE = 8\). By BPT, \(\dfrac{AE}{EC} = \dfrac{AD}{DB} = \dfrac{1}{2}\), so \(EC = 16\). The whole bracket arm \(AC = 8 + 16 = 24\).
A small triangular logo has area \(12\text{ cm}^{2}\). The same logo is printed at 3 times the scale. Area scales by \(3^{2} = 9\), so the enlarged logo occupies \(12 \times 9 = 108\text{ cm}^{2}\).
A vertical pole \(2\text{ m}\) tall casts a shadow \(2.5\text{ m}\). The flagpole's shadow at the same time is \(30\text{ m}\). Height of flagpole \(= \dfrac{2}{2.5} \times 30 = 24\text{ m}\). So the flagpole is 24 m tall.
14. Competency and Reasoning Questions
Competency items test why the tools work and whether you can justify a statement, not just compute. These questions each carry the necessary reasoning.
Two triangles have two equal pairs of angles. Why does that force similarity, without checking the third angle or any side?
Reasoning. Each angle sum is \(180^{\circ}\), so the third angles are automatically equal (\(180^{\circ} -\) the two known ones). Two equal angles fix the whole shape; sides then must be proportional, giving similarity.
Every pair of congruent triangles is also similar. Explain the scale factor involved.
Reasoning. Congruent triangles share all angles (equal) and all sides (equal). For similar triangles, equal sides mean a common ratio of \(1:1\), i.e. scale factor \(k = 1\).
If the ratio of corresponding sides of two similar triangles is \(5 : 6\), what is the ratio of their perimeters? Justify.
Reasoning. A perimeter is the sum of the three sides, and each side is scaled by the same \(k\). Summing the three scaled sides gives \(k\) times the original perimeter, so the perimeter ratio is also \(5 : 6\).
Two rectangles have the same area. Must they be similar?
Reasoning. No. Similarity needs proportional sides (same length : breadth ratio), not equal areas. A \(2 \times 6\) and a \(3 \times 4\) rectangle both have area 12 but their side ratios \(1:3\) and \(3:4\) differ, so they are not similar.
Why must the correspondence be written in order in \(\triangle ABC \sim \triangle DEF\)?
Reasoning. The symbol lists the matching vertices: \(A\leftrightarrow D\), \(B\leftrightarrow E\), \(C\leftrightarrow F\). A scrambled order like \(\triangle ABC \sim \triangle EFD\) would assert \(\angle A = \angle E\), which is generally false and would give the wrong side ratios.
A student claims, "If all angles of two triangles are equal, the triangles are congruent." Is that right?
Reasoning. No. Equal angles give similarity, not congruence. Congruence additionally requires equal (not merely proportional) sides; a large and a small equilateral triangle have equal angles but different sides.
A triangle is scaled by \(k = 0.5\). Does its area shrink by 0.5?
Reasoning. No. Area multiplies by \(k^{2} = 0.25\), so the image has a quarter of the area. A scale of a half shrinks sides to half but area to one quarter.
In \(\triangle ABC\), a line cuts \(AB\) at \(P\) and \(AC\) at \(Q\) with \(AP = 2\), \(PB = 3\), \(AQ = 2.5\), \(QC = 4\). Is \(PQ \parallel BC\)?
Reasoning. Check \(\dfrac{AP}{PB} = \dfrac{2}{3}\) but \(\dfrac{AQ}{QC} = \dfrac{2.5}{4} = \dfrac{5}{8}\). The ratios are not equal, so by the converse of BPT, \(PQ\) is not parallel to \(BC\).
Explain why a poster scaled up by factor 3 needs 9 times the paint.
Reasoning. Paint covers area, and for a fixed shape the area scales with the square of the linear scale factor. Factor 3 in lengths means factor \(3^{2} = 9\) in area, hence 9 times the paint.
You know all three sides of a triangle and only two sides of another. Which similarity criterion can you possibly use?
Reasoning. With one side and two sides given, SSS is unusable (needs three pairs). If the included angle between the known sides is also known, SAS may apply; otherwise you cannot decide from the given data.
15. Case-Based Examples
Case studies present a situation and 2-3 related sub-questions. Read the whole case first, then answer using only its data, showing all working.
Case Study 1 — Shadows on a Playground
At 4:00 pm, a student of height \(1.5\text{ m}\) casts a shadow \(2\text{ m}\) long. Two playground structures, a slide and a swing frame, cast shadows of \(8\text{ m}\) and \(10\text{ m}\) respectively at the same moment.
Question 1
State the similarity criterion that lets you compare the student and the slide.
Question 2
Find the height of the slide.
Question 3
Find the height of the swing frame.
Solution.
(1) The sun rays are parallel, so the two shadow triangles share equal angles; by the AA criterion the triangles are similar.
(2) \(\dfrac{\text{height}}{\text{shadow}} = \dfrac{1.5}{2} = \dfrac{h_s}{8} \implies h_s = \dfrac{1.5 \times 8}{2} = 6\text{ m}\). The slide is 6 m tall.
(3) \(h_f = \dfrac{1.5 \times 10}{2} = 7.5\text{ m}\). The swing frame is 7.5 m tall.
Case Study 2 — The Town Map
A town map is drawn to scale \(1 : 5000\). On the map, the park is a rectangle \(6\text{ cm}\) by \(4\text{ cm}\), and the distance between the park and the museum is \(8\text{ cm}\).
Question 1
Interpret the scale \(1 : 5000\) in words.
Question 2
Find the actual dimensions of the park.
Question 3
Find the actual distance between the park and the museum.
Solution.
(1) Every \(1\text{ cm}\) on the map equals \(5000\text{ cm} = 50\text{ m}\) on the ground.
(2) Length \(= 6 \times 50 = 300\text{ m}\), breadth \(= 4 \times 50 = 200\text{ m}\). The park is 300 m by 200 m.
(3) Distance \(= 8 \times 50 = 400\text{ m}\). The museum is 400 m from the park.
Case Study 3 — Surveying the River
A surveying team wants the width of a river. They mark point \(A\) on one bank and create the similar pair of triangles shown in Section 13, with parallel bases \(BC \parallel DE\). On the small triangle \(ABC\), the side \(BC = 2\text{ m}\). The full triangle \(ADE\) has base \(DE\) crossing the river at a measured value of \(9\text{ m}\) in the drawing scale where the small base was measured.
Question 1
State the similarity criterion used between \(\triangle ABC\) and \(\triangle ADE\).
Question 2
Find the scale factor from the small triangle to the full triangle.
Question 3
If the river width is the base \(DE\) and the survey recorded \(DE = 24\text{ m}\), verify that \(BC\) stays consistent with the two given measurements.
Solution.
(1) Since \(BC \parallel DE\), \(\angle ABC = \angle ADE\) and \(\angle ACB = \angle AED\) (corresponding angles). By the AA criterion, \(\triangle ABC \sim \triangle ADE\).
(2) \(\dfrac{DE}{BC} = \dfrac{24}{2} = 12\), so the scale factor is 12.
(3) Using the recorded \(DE = 24\text{ m}\) and \(BC = 2\text{ m}\), the ratio \(\dfrac{DE}{BC} = 12\) matches the scale factor found in part (2), so the measurements are consistent.
Case Study 4 — School Banner
A triangular school banner is \(4\text{ m}\) wide at the base and \(1.5\text{ m}\) high. A scaled-down version for a desk flag has a base of \(16\text{ cm}\), using the same shape.
Question 1
Find the scale factor from the banner to the desk flag.
Question 2
Find the height of the desk flag.
Question 3
What is the ratio of the areas of the banner and the desk flag?
Solution.
(1) Banner base \(4\text{ m} = 400\text{ cm}\); desk flag base \(16\text{ cm}\). Scale factor \(= \dfrac{16}{400} = \dfrac{1}{25}\).
(2) Desk flag height \(= 150\text{ cm} \times \dfrac{1}{25} = 6\text{ cm}\).
(3) Areas are in the square of the scale factor: \(\left(\dfrac{1}{25}\right)^{2} = \dfrac{1}{625}\). The banner has 625 times the desk flag's area.
16. CBSE Board-Style Questions
The following are original items written in the CBSE board pattern and difficulty for this chapter, with fog-free workings.
Q-B1. (1 mark)
\(\triangle ABC \sim \triangle DEF\), \(AB = 3\), \(DE = 6\), \(BC = 4\). Find \(EF\).
Q-B2. (1 mark)
The ratio of the areas of two similar triangles is \(100 : 121\). Find the ratio of their corresponding sides.
Q-B3. (2 marks)
In \(\triangle ABC\), \(DE \parallel BC\), \(AD = 5\), \(DB = 15\), \(AE = 3\). Find \(EC\).
Q-B4. (2 marks)
Justify whether two triangles with angles \(50^{\circ}, 70^{\circ}, 60^{\circ}\) and \(70^{\circ}, 60^{\circ}, 50^{\circ}\) are similar.
Q-B5. (3 marks)
A rod \(2\text{ m}\) long casts a shadow \(1.6\text{ m}\). A building beside it casts a shadow \(14.4\text{ m}\) at the same moment. Find the building's height.
Q-B6. (3 marks)
In \(\triangle ABC\), \(DE \parallel BC\) with \(AD : DB = 1 : 3\). If the area of \(\triangle ADE\) is \(8\text{ cm}^{2}\), find the area of \(\triangle ABC\).
Q-B7. (5 marks)
Prove the Basic Proportionality Theorem: if \(DE \parallel BC\) in \(\triangle ABC\), then \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\).
Q-B8. (5 marks)
State and prove that the areas of two similar triangles are in the square of the ratio of their corresponding sides. Also use it: two similar triangles have area ratio \(16 : 49\) and their total area is \(650\text{ cm}^{2}\). Find each area.
Answers. (B1) \(\dfrac{EF}{BC} = \dfrac{6}{3} = 2\), so \(EF = 8\). (B2) \(\sqrt{100 : 121} = 10 : 11\). (B3) \(EC = \dfrac{15 \times 3}{5} = 9\). (B4) The two triangles have the same three angles, so by AA they are similar. (B5) \(h = \dfrac{2}{1.6} \times 14.4 = 18\text{ m}\). (B6) \(\dfrac{AD}{AB} = \dfrac{1}{4}\), so the area ratio is \(1 : 16\) and \(\operatorname{ar}(\triangle ABC) = 8 \times 16 = 128\text{ cm}^{2}\). (B7) See Section 6 for the full proof; the key steps are the area ratios from common altitudes and the equal areas of \(\triangle DBE\) and \(\triangle DEC\). (B8) Proof in Section 10. Areas: let them be \(16x\) and \(49x\); \(65x = 650\), so \(x = 10\), giving 160 cm² and 490 cm².
17. Common Examination Mistakes
| Common mistake | Correct method |
|---|---|
| Writing \(\triangle ABC \sim \triangle DEF\) in the wrong order | Keep the order so that \(A\leftrightarrow D\), \(B\leftrightarrow E\), \(C\leftrightarrow F\); it fixes every other correspondence |
| Applying the area ratio \(k\) instead of \(k^{2}\) | Areas of similar triangles use the square of the side ratio |
| Using SAS with a non-included angle | The given angle must lie between the two given sides |
| Using BPT when the line is not parallel to a side | BPT and its converse both require the parallelism/split to be stated correctly |
| Saying equal angles alone give congruence | Equal angles give similarity; equal sides give congruence |
| Forgetting to square when going from areas back to sides | Side ratio is the square root of the area ratio |
| Mixing units in shadow or map problems | Convert to one unit, e.g. metres, before writing the proportion |
| Quoting a criterion without its name | Write "by AA criterion", "by SSS criterion" or "by SAS criterion" in the solution |
| Not simplifying ratios before comparing | Reduce \(2.4 : 3.6\) to \(2 : 3\); the equality then shows instantly |
| Ignoring a given "line is parallel" twice over | Reuse the same parallel in several sub-parts; it gives one proportion that answers them all |
18. CBSE Exam Tips
- Start every similarity answer by declaring the criterion: "By AA criterion, \(\triangle ABC \sim \triangle DEF\)". The name gains a mark even before you compute.
- For BPT questions, write the proportion \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\) first, then substitute; signs and units then check themselves.
- When the area ratio is given and you need the side ratio, take the square root; when the side ratio is given and you need the area ratio, square it.
- In shadow problems state that the sun's rays are parallel, so the shadow triangles are similar by AA — a reason instead of a guess.
- Cross-check proportions numerically: e.g. a scale \(1:12\) must inflate every measured length by 12, and areas by 144.
- For "is \(PQ \parallel BC\)?" questions, write both ratios, simplify, and compare before concluding.
- Present proofs in the four blocks (Given / To prove / Construction / Proof); keep each block short and label the theorem used.
- Convert all lengths to one unit before forming a ratio, and cancel units completely before writing the final answer.
19. Quick Revision
| Concept | Key point |
|---|---|
| Similar figures | Same shape, not necessarily same size |
| Congruent figures | Same shape and size; a similar case with scale factor 1 |
| Similar triangles | Corresponding angles equal and corresponding sides proportional |
| Notation | \(\triangle ABC \sim \triangle DEF\) means \(A\leftrightarrow D\), \(B\leftrightarrow E\), \(C\leftrightarrow F\) |
| BPT | \(DE \parallel BC \implies \dfrac{AD}{DB} = \dfrac{AE}{EC}\) |
| Converse of BPT | \(\dfrac{AD}{DB} = \dfrac{AE}{EC} \implies DE \parallel BC\) |
| AA criterion | Two equal corresponding angles suffice |
| SSS criterion | All three side ratios equal |
| SAS criterion | Two proportional sides and the equal included angle |
| Perimeter ratio | Same as the side ratio \(k\) |
| Area ratio | Square of the side ratio \(k^{2}\) |
| Real-life uses | Shadows, maps, models, photos, surveying, banners |
20. Formula and Theorem Box
\[DE \parallel BC \implies \frac{AD}{DB} = \frac{AE}{EC}\]
\[\frac{AD}{DB} = \frac{AE}{EC} \implies DE \parallel BC\]
\[\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} \implies \triangle ABC \sim \triangle DEF\]
\[\angle A = \angle D \text{ and } \frac{AB}{DE} = \frac{AC}{DF} \implies \triangle ABC \sim \triangle DEF\]
\[\angle A = \angle D \text{ and } \angle B = \angle E \implies \triangle ABC \sim \triangle DEF\]
\[P(\triangle ABC) : P(\triangle DEF) = AB : DE \qquad (k : 1)\]
\[ \boxed{\frac{\operatorname{ar}(\triangle ABC)}{\operatorname{ar}(\triangle DEF)} = k^{2}} \]
Keep the correspondence order straight in every formula: ratio of corresponding sides only, included angle only for SAS, and parallel-base only for BPT.
21. Self-Practice Questions
Forty original questions written in the CBSE pattern, in four graded levels. Answers are in Section 22.
Level 1 — Conceptual
Q1.
Define similar figures with an example.
Q2.
State the two conditions for \(\triangle ABC \sim \triangle DEF\).
Q3.
If \(\triangle ABC \sim \triangle DEF\), write the three pairs of equal angles.
Q4.
What is the scale factor between two congruent triangles?
Q5.
Are all equilateral triangles similar? Give a reason.
Q6.
Are all isosceles triangles similar? Give a reason.
Q7.
State the Basic Proportionality Theorem.
Q8.
State the converse of the Basic Proportionality Theorem.
Q9.
If the ratio of the sides of two similar triangles is \(2 : 3\), what is the ratio of their perimeters?
Q10.
If the ratio of the sides of two similar triangles is \(2 : 3\), what is the ratio of their areas?
Level 2 — Standard
Q11.
\(\triangle ABC \sim \triangle DEF\), \(AB = 5\), \(DE = 7.5\), \(BC = 6\). Find \(EF\).
Q12.
\(\triangle ABC \sim \triangle DEF\) with area ratio \(4 : 9\). If \(AB = 6\), find \(DE\).
Q13.
In \(\triangle ABC\), \(DE \parallel BC\), \(AD = 3\), \(DB = 12\), \(AE = 2.5\). Find \(EC\).
Q14.
In \(\triangle ABC\), \(AD = 2\), \(DB = 3\), \(AE = 2.8\), \(EC = 4.2\). Is \(DE \parallel BC\)?
Q15.
In \(\triangle ABC\), \(DE \parallel BC\), \(AD = 4\), \(DB = 8\), \(AC = 18\). Find \(AE\) and \(EC\).
Q16.
A triangle has sides \(5, 7, 9\). A similar triangle has its first side \(12.5\). Find its third side.
Q17.
\(\triangle ABC \sim \triangle PQR\) with perimeters \(40\) and \(60\). If \(\operatorname{ar}(\triangle ABC) = 16\), find \(\operatorname{ar}(\triangle PQR)\).
Q18.
Similar triangles have side ratio \(3 : 5\) and the smaller area is \(54\text{ cm}^{2}\). Find the larger area.
Q19.
A pole \(1.6\text{ m}\) tall casts a shadow \(2\text{ m}\). A building casts a \(25\text{ m}\) shadow. Find its height.
Q20.
A map has scale \(1 : 50000\). Two towns are \(3.4\text{ cm}\) apart on the map. Find the actual distance in km.
Level 3 — Application
Q21.
A tree \(5\text{ m}\) tall casts a \(6.25\text{ m}\) shadow. A pole casts a \(10\text{ m}\) shadow at the same time. Find the pole's height.
Q22.
A \(10\text{ cm} \times 15\text{ cm}\) photo is enlarged so its width is \(24\text{ cm}\). Find the new height.
Q23.
A model is built at scale \(1 : 25\). The model is \(12\text{ cm}\) tall. Find the real height in metres.
Q24.
A rectangular field \(120\text{ m} \times 80\text{ m}\) is drawn at scale \(1 : 2000\). Find its dimensions on the drawing in centimetres.
Q25.
In \(\triangle ABC\), \(DE \parallel BC\) with \(AD = \frac{2}{5}AB\). If \(AE = 6\), find \(EC\) and then \(AC\).
Q26.
A triangular logo has area \(20\text{ cm}^{2}\). It is enlarged at scale 3. Find the new area.
Q27.
A banner \(4\text{ m} \times 2\text{ m}\) is scaled up by factor \(\frac{5}{2}\). Find its new dimensions and area.
Q28.
In a river survey, a \(2\text{ m}\) base triangle is similar to a \(17\text{ m}\) base triangle (scale \(1 : 8.5\)). The river width corresponds to \(8\text{ cm}\) on a \(1:425\) drawing. Find the actual width in metres.
Q29.
Are triangles with sides \(2.5, 6, 6.5\) and \(10, 24, 26\) similar? If yes, find the ratio of their areas.
Q30.
An equilateral triangle of side \(5\text{ cm}\) is scaled by factor 3. Find the new side and the new perimeter.
Level 4 — Competency / HOTS
Q31.
Explain why checking two angles is enough in the AA criterion.
Q32.
For similar triangles with scale factor \(k\), write the ratio of perimeters and the ratio of areas. Justify briefly.
Q33.
Why are two circles always similar, but two rectangles not always?
Q34.
In \(\triangle ABC\), \(D\) and \(E\) are the midpoints of \(AB\) and \(AC\). Show that \(\triangle ADE \sim \triangle ABC\) and find the ratio of their areas.
Q35.
A student says triangles with sides \(3, 4, 6\) and \(6, 8, 12\) are similar. Verify by SSS.
Q36.
A rectangle is scaled by factor 3. By what factor does its area increase? Explain.
Q37.
The ratio of the areas of two similar triangles is \(4 : 49\). Find the ratio of their corresponding sides.
Q38.
In \(\triangle ABC\), \(AD = 1.8\), \(DB = 2.7\), \(AE = 2.4\), \(EC = 3.6\). Show that \(DE \parallel BC\).
Q39.
In \(\triangle ABC\), \(M\) is the midpoint of \(AB\). A line through \(M\) parallel to \(BC\) meets \(AC\) at \(N\). Prove that \(AN = NC\).
Q40.
In the figure with \(BC \parallel DE\), \(\triangle ABC \sim \triangle ADE\) with \(\dfrac{AD}{AB} = \dfrac{2}{5}\). Find the ratio of the areas \(\operatorname{ar}(\triangle ADE) : \operatorname{ar}(\triangle ABC)\).
22. Answers to Self-Practice Questions
A1.
Figures with the same shape but not necessarily the same size, e.g. any two squares.
A2.
Corresponding angles are equal and corresponding sides are in the same ratio.
A3.
\(\angle A = \angle D\), \(\angle B = \angle E\), \(\angle C = \angle F\).
A4.
\(1\) (congruent triangles are similar with \(k = 1\)).
A5.
Yes; every equilateral triangle has all three angles \(60^{\circ}\), so any two are similar by AA.
A6.
No; two isosceles triangles can have different vertex angles, so their sides need not be proportional.
A7.
If \(DE \parallel BC\), then \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\).
A8.
If \(\dfrac{AD}{DB} = \dfrac{AE}{EC}\), then \(DE \parallel BC\).
A9.
\(2 : 3\) (perimeters follow the side ratio).
A10.
\(4 : 9\) (areas follow the square of the side ratio).
A11.
\(\dfrac{7.5}{5} = 1.5\), so \(EF = 6 \times 1.5 = 9\).
A12.
Side ratio \(\sqrt{4 : 9} = 2 : 3\), so \(DE = 6 \times \dfrac{3}{2} = 9\).
A13.
\(EC = \dfrac{12 \times 2.5}{3} = 10\).
A14.
\(\dfrac{2}{3} = \dfrac{2.8}{4.2}\), so yes, \(DE \parallel BC\).
A15.
\(\dfrac{AD}{DB} = \dfrac{1}{2}\), so \(AE = \frac{1}{3} \times 18 = 6\), \(EC = 12\).
A16.
Scale \(\dfrac{12.5}{5} = 2.5\); third side \(= 9 \times 2.5 = 22.5\).
A17.
Side ratio \(40 : 60 = 2 : 3\); areas \(4 : 9\); \(\operatorname{ar}(\triangle PQR) = 16 \times \frac{9}{4} = 36\).
A18.
\(\dfrac{54}{\text{larger}} = \dfrac{9}{25}\), so larger \(= 54 \times \frac{25}{9} = 150\text{ cm}^{2}\).
A19.
\(h = \dfrac{1.6}{2} \times 25 = 20\text{ m}\).
A20.
\(3.4 \times 50000 = 170000\text{ cm} = 1.7\text{ km}\).
A21.
\(\dfrac{5}{6.25} \times 10 = 8\text{ m}\).
A22.
Scale \(2.4\); new height \(= 15 \times 2.4 = 36\text{ cm}\).
A23.
\(12 \times 25 = 300\text{ cm} = 3\text{ m}\).
A24.
\(\dfrac{120}{2000} = 0.06\text{ m} = 6\text{ cm}\); \(\dfrac{80}{2000} = 4\text{ cm}\).
A25.
\(\dfrac{AD}{DB} = \dfrac{2}{3} \implies \dfrac{AE}{EC} = \dfrac{2}{3}\); \(EC = 9\), \(AC = 15\).
A26.
Areas \(\times 3^{2} = 9\); new area \(= 20 \times 9 = 180\text{ cm}^{2}\).
A27.
New dimensions \(10\text{ m} \times 5\text{ m}\); area ratio \(= \left(\frac{5}{2}\right)^{2} = \frac{25}{4}\), new area \(= 8 \times \frac{25}{4} = 50\text{ m}^{2}\).
A28.
Actual width \(= 8 \text{ cm} \times 425 = 3400\text{ cm} = 34\text{ m}\).
A29.
Ratios all \(4 : 1\), so similar (SSS). Area ratio \(= 16 : 1\).
A30.
New side \(= 15\text{ cm}\); perimeter \(= 45\text{ cm}\).
A31.
Two equal angles force the third to be equal too (each angle sum is \(180^{\circ}\)), and equal angles fix the shape, giving proportional sides.
A32.
Perimeters \(k : 1\) (sum of \(k\)-scaled sides); areas \(k^{2} : 1\) (base and height both scale by \(k\)).
A33.
Every circle has the same shape; rectangles differ in length : breadth ratio, so only those with matching ratios are similar.
A34.
D and E are midpoints, so \(AD = DB\), \(AE = EC\), hence \(\dfrac{AD}{AB} = \dfrac{AE}{AC} = \frac{1}{2}\) and by SAS, \(\triangle ADE \sim \triangle ABC\); area ratio \(= \left(\frac{1}{2}\right)^{2} = 1 : 4\).
A35.
\(\dfrac{6}{3} = 2\), \(\dfrac{8}{4} = 2\), \(\dfrac{12}{6} = 2\); equal ratios, so similar by SSS.
A36.
Factor \(9\); area scales with \(k^{2} = 3^{2}\).
A37.
\(\sqrt{4 : 49} = 2 : 7\).
A38.
\(\dfrac{AD}{DB} = \dfrac{1.8}{2.7} = \frac{2}{3}\); \(\dfrac{AE}{EC} = \dfrac{2.4}{3.6} = \frac{2}{3}\); equal, so by the converse of BPT, \(DE \parallel BC\).
A39.
By BPT, \(\dfrac{AM}{MB} = \dfrac{AN}{NC}\). Since \(AM = MB\), the left ratio is 1, so \(\dfrac{AN}{NC} = 1\) and \(AN = NC\).
A40.
Area ratio equals \(\left(\frac{AD}{AB}\right)^{2} = \left(\frac{2}{5}\right)^{2} = 4 : 25\).