Similarity | ICSE Class 10 Maths Notes

This note covers similar figures, size transformations, correspondence, comparison with congruency, the angle-angle (AA), side-side-side (SSS) and side-angle-side (SAS) conditions for similar triangles, the Basic Proportionality Theorem, areas of similar triangles, and applications to indirect measurement, maps and models.

What does similarity mean, and how does it compare with congruency?

Definition: Similar figures have the same shape but not necessarily the same size. Congruent figures have both the same shape and the same size.

A size transformation enlarges or reduces a figure while preserving its shape. Corresponding parts are the parts that match between the original figure and its image. Every corresponding length changes by the same multiplier, called the scale factor.

An enlargement increases lengths; a reduction decreases them. Angles between corresponding sides remain equal. Similarity therefore describes a relationship between whole shapes, not merely a resemblance between one pair of sides.

Which familiar figures are similar?

All circles are similar, all squares are similar, and all equilateral triangles are similar. An equilateral triangle has three equal sides. Different sizes do not prevent these figures from having the same shape.

All congruent figures are similar, but similar figures need not be congruent. The phrase “not necessarily the same size” matters: similar figures may have equal sizes. Similarity does not require the figures to have different sizes.

A polygon is a closed plane figure made from straight line segments. Two polygons with the same number of sides are similar when their corresponding angles are equal and their corresponding sides are proportional. Proportional means that the ratios of matching lengths are equal. A ratio compares two quantities by division.

For polygons in general, one of those conditions alone is insufficient. Equal angles do not by themselves establish similarity between a square and a rectangle. Proportional sides do not by themselves establish similarity between a square and a rhombus, a quadrilateral with four equal sides.

How do correspondence and scale factors control a calculation?

Let A, B and C be the vertices, or corners, of one triangle, and D, E and F those of another. The notation △ABC ∼ △DEF means that triangle ABC is similar to triangle DEF. Here △ means “triangle” and ∼ means “is similar to”.

The order gives the correspondence: A matches D, B matches E, and C matches F. AB means the length of the segment joining A and B; the same convention applies to other pairs of letters. The symbol ∠ means “angle”.

Thus ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. A three-letter angle name identifies its vertex by the middle letter: ∠ABC is the angle at B between BA and BC. The symbol = means “is equal to”.

The matching sides satisfy AB/DE = BC/EF = CA/FD. The slash denotes division. A ratio can also use a colon: AB:DE compares AB with DE in that order. Reversing every ratio is valid; reversing just one generally destroys the equality.

Which direction does the scale factor describe?

Let k be the scale factor from the original figure to its image. Then k = image length/original length. Multiplying any original length by k gives its corresponding image length. The direction must be stated before using the number.

Worked example 1. A photograph is enlarged from a size of 35 mm to the corresponding size of 45 mm. Here mm means millimetres. Find the enlargement multiplier and its reverse reduction multiplier.

Answer: Enlargement multiplier = 45/35 = 9/7. Reduction multiplier = 35/45 = 7/9. Each compares corresponding lengths, with the destination length in the numerator, the number above the fraction bar.

The denominator is the number below the fraction bar. A scale factor is a ratio of lengths expressed in the same unit, so those units cancel. It is not itself a length and should not carry a length unit.

Writing the vertices in the correct order is part of the mathematics. For the correspondence above, △BAC ∼ △EDF is valid because both lists were reordered consistently. △ABC ∼ △EDF would describe a different matching.

How does the AA condition establish similarity?

Theorem: AA similarity

AA means angle-angle. If two angles of one triangle respectively equal two angles of another triangle, the triangles are similar. “Respectively” means that the equalities follow the stated correspondence, rather than pairing the angles at random.

The angle sum property states that the three interior angles of a triangle add to 180°. The symbol ° denotes degrees, a unit for measuring angles. Once two pairs of angles are equal, the third pair is equal too.

This also explains AAA, meaning angle-angle-angle: equality of all three corresponding angles establishes similarity. AA uses only two pairs because the third pair follows from the angle sum. The corresponding sides are then proportional, even if their lengths were not given.

Worked example 2. In triangles ABC and DEF, ∠B = ∠E = 60°, ∠C = ∠F = 40°, BC = 3 cm and EF = 5 cm. Here cm means centimetres. Find the remaining angles and the ratio of corresponding sides.

Answer: ∠A = ∠D = 180° − 60° − 40° = 80°. The symbol − means subtraction. By AA, △ABC ∼ △DEF, so AB/DE = BC/EF = CA/FD = 3/5 = 0.6.

How are the matching vertices found?

Start with the given angle equalities: B matches E and C matches F. The remaining vertices A and D must then match. Write the similarity statement in that order before using side ratios. The appearance or orientation of a sketch does not replace this matching.

In geometric problems, equal angles may arise from parallel lines, a shared angle, or vertically opposite angles. Parallel lines are lines in the same plane that do not meet. Vertically opposite angles are the opposite angles formed by two intersecting lines; these are equal.

A common angle belongs to both triangles being compared. If the triangles share an angle, this supplies one equality, not the whole AA condition. Identify a second matching pair before concluding similarity and writing the required proportion.

How does the SSS condition establish similarity?

Theorem: SSS similarity

SSS means side-side-side. If the three sides of one triangle are respectively proportional to the three sides of another triangle, the triangles are similar. Their corresponding angles are consequently equal; these angles need not be supplied separately.

The word proportional distinguishes the similarity condition from SSS congruency. For congruency, corresponding sides are equal. For similarity, all three ratios must have a common value, which need not be one. A single matching ratio cannot establish the condition.

Worked example 3. Triangles ABC and DEF have AB = 3 cm, BC = 6 cm, CA = 8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm. Determine whether they are similar.

Answer: AB/DE = 3/4.5 = 2/3; BC/EF = 6/9 = 2/3; CA/FD = 8/12 = 2/3. All three corresponding side ratios agree, so △ABC ∼ △DEF by SSS.

These triangles have different side lengths, yet their matching sides change by the same factor. From ABC to DEF, the multiplier is 3/2; in the opposite direction, it is 2/3. Both descriptions express the same similarity with opposite directions of comparison.

What should be checked before concluding SSS?

1. Identify the proposed matching sides from their endpoints and lengths.
2. Put lengths from the same triangle in every numerator.
3. Reduce all three ratios and check that they are equal.
4. State SSS and write the triangles with corresponding vertices in the same positions.

The conclusion gives ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. It does not give numerical values for these angles. Similarity establishes equal corresponding angles without requiring their individual measures to be calculated.

How does the SAS condition establish similarity?

Theorem: SAS similarity

SAS means side-angle-side. Two triangles are similar if two sides of one triangle are proportional to two sides of another triangle and their included angles are equal. The included angle is the angle between the two sides being compared.

For sides AB and AC, the included angle is ∠BAC, or ∠A. For sides DE and DF, it is ∠EDF, or ∠D. Equal angles elsewhere in the triangles do not supply the stated SAS condition.

Worked example 4. In triangles ABC and DEF, AB = 2 cm, AC = 4 cm, DE = 3 cm, DF = 6 cm and ∠A = ∠D = 50°. Establish similarity and state the third side ratio.

Answer: AB/DE = 2/3 and AC/DF = 4/6 = 2/3. The included angles ∠A and ∠D are equal, so △ABC ∼ △DEF by SAS. Hence BC/EF = 2/3.

The conclusion about BC and EF follows after similarity has been established. Their actual lengths cannot be found from the ratio alone. The ratio tells how they compare, while a numerical length would require enough additional information.

How can the three conditions be distinguished?

Select the condition supported by the data. When two angle equalities are available, AA establishes similarity without checking side lengths. When all three pairs of side lengths are available, SSS can establish similarity without calculating angles. SAS connects side information with one specified angle.

Note: A correct ratio is not enough by itself. Name the condition and show every part of that condition before using the remaining corresponding sides or angles.

How is the Basic Proportionality Theorem used?

Theorem: Basic Proportionality Theorem

The Basic Proportionality Theorem, abbreviated BPT, states that a line parallel to one side of a triangle, intersecting the other two sides at distinct points, divides those sides in the same ratio. Distinct points are points that are different from one another.

In triangle ABC, let D lie on AB and E lie on AC, with DE parallel to BC. The symbol ∥ means “is parallel to”, so the condition is DE ∥ BC. The resulting proportion is AD/DB = AE/EC.

AD and DB are the two parts of AB; AE and EC are the two parts of AC. Compare the segment from A with the remaining segment on each side. Do not use a whole side in just one denominator while retaining the divided-side interpretation.

Triangle ABC has D on AB and E on AC, with DE parallel to BC. The drawing also joins B to E and C to D. Dotted perpendicular segments DM and EN meet AC and AB respectively.

See Fig. 6.10 in your NCERT textbook

A perpendicular meets a line at a right angle, an angle of 90°. In the diagram, M and N name the points where those perpendiculars meet the sides. These additional lines support an area argument; the application proportion itself uses AD, DB, AE and EC.

Worked example 5. In triangle ABC, D is on AB, E is on AC and DE ∥ BC. Given AD = 1.5 cm, DB = 3 cm and AE = 1 cm, find EC.

Answer: By BPT, AD/DB = AE/EC. Therefore 1.5/3 = 1/EC. Cross-multiplication, which equates the products across a proportion, gives 1.5 × EC = 3 × 1. Here × means multiplication. Thus EC = 2 cm.

Worked example 6. In triangle ABC, D is on AB, E is on AC and DE ∥ BC. Given DB = 7.2 cm, AE = 1.8 cm and EC = 5.4 cm, find AD.

Answer: AD/7.2 = 1.8/5.4 = 1/3 by BPT. Hence AD = 7.2/3 = 2.4 cm.

How can whole sides be used instead?

Since AB = AD + DB and AC = AE + EC, the same configuration also gives AD/AB = AE/AC. The symbol + means addition. This compares each segment next to A with its complete side, using the same type of comparison on both sides.

Use the form that matches the available data. If AB is given with AD, first find DB by subtraction when the divided-side form is needed. If AE and AC are given, the whole-side form may avoid that extra step.

When can proportional segments establish parallel lines?

Theorem: Converse of the Basic Proportionality Theorem

A converse exchanges the hypothesis and conclusion of a statement. The converse of BPT says that if a line divides two sides of a triangle in the same ratio, it is parallel to the third side.

Let E and F lie on sides PQ and PR of triangle PQR. If PE/EQ = PF/FR, then EF ∥ QR. This time proportional division is the given or calculated fact, and parallelism is the conclusion.

Worked example 7. E and F lie on PQ and PR of triangle PQR. Given PE = 4 cm, EQ = 4.5 cm, PF = 8 cm and FR = 9 cm, decide whether EF ∥ QR.

Answer: PE/EQ = 4/4.5 = 8/9, while PF/FR = 8/9. Since the side-division ratios are equal, EF ∥ QR by the converse of BPT.

Keep the logical direction clear. BPT starts from a parallel line and supplies a ratio. Its converse starts from the ratio and establishes that the line is parallel. Stating which direction is being used explains why the conclusion follows.

What if the two ratios differ?

For PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm, the ratios are 3.9/3 = 1.3 and 3.6/2.4 = 1.5. They differ, so EF is not parallel to QR: parallelism would require equal ratios by BPT.

A midpoint divides a segment into two equal parts. When E and F are midpoints of PQ and PR, both divided-side ratios equal one. The converse therefore gives EF ∥ QR. Conversely, a line through one midpoint parallel to another side bisects the remaining side.

To bisect a segment means to divide it into equal parts. The midpoint application is thus a direct use of proportional division, with the two parts on each side equal.

How are the areas of similar triangles related?

Theorem: Areas of similar triangles

The areas of two similar triangles are proportional to the squares of their corresponding sides. Area measures the region enclosed by a plane figure. Squaring a quantity means multiplying it by itself; the superscript ² denotes that operation.

Write ar(ABC) for the area of triangle ABC. If △ABC ∼ △DEF, then ar(ABC)/ar(DEF) = (AB/DE)² = (BC/EF)² = (CA/FD)². Parentheses group the entire side ratio that must be squared.

This result requires similar triangles. A single pair of side lengths from arbitrary triangles does not determine their area ratio. If similarity is not given, first establish it from AA, SSS or SAS before applying the area formula.

Worked example 8. Triangles ABC and DEF have AB = 3 cm, BC = 6 cm, CA = 8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm. Find ar(ABC):ar(DEF).

Answer: AB/DE = BC/EF = CA/FD = 2/3, so △ABC ∼ △DEF by SSS. The area ratio is (2/3)² = 4/9. Therefore ar(ABC):ar(DEF) = 4:9.

Why is the scale factor squared?

A triangle's area equals half its base multiplied by its perpendicular height. A base is the side chosen for this calculation. The corresponding height, or altitude, is the perpendicular distance from the opposite vertex to the line containing that base.

For similar triangles, corresponding bases and heights change by the same length factor. Multiplying base by height therefore applies the factor twice. The factor of one-half occurs in both areas and cancels when their ratio is taken.

In the example, the side ratio is 2:3 and the area ratio is 4:9. These are different kinds of comparison. An area ratio cannot be substituted unchanged as a side ratio. To reverse the calculation, take the positive square root of the area ratio.

A positive square root is the positive number whose square equals the given number. Lengths are positive, so the positive root is appropriate. Reversing the triangle order reverses both ratios: DEF to ABC has side ratio 3:2 and area ratio 9:4.

Note: Similarity controls area through a squared length ratio. Keep the triangle order identical in the area ratio and the side ratio before squaring.

How is similarity applied to maps and models?

A map scale compares a distance on a map with the corresponding ground distance. A model scale compares a length on a model with the corresponding length of the actual object. These are applications of a common length ratio.

The representative fraction expresses the represented length divided by the corresponding actual length, with both expressed in the same unit. “Represented length” means the length shown on the map or model. The comparison must use matching lengths and a clearly stated direction.

How can a scale calculation be organised?

Let m denote a map length, g its corresponding ground length, and s the map-to-ground scale factor. These letters name quantities; here m is a variable, not a unit abbreviation. In common length units, s = m/g.

Consequently, m = s × g and g = m/s. The first equation finds the map length from the ground length; the second finds the ground length from the map length. For a reduction, the represented length is smaller than the corresponding actual length.

1. Identify the two corresponding lengths and which one is required.
2. Express both lengths in the same unit before forming the scale ratio.
3. Write represented length divided by actual length as the stated scale factor.
4. Rearrange the proportion, calculate, and attach the correct length unit to the answer.

The same sequence applies to a model, using model length in place of map length and actual length in place of ground length. Different features must follow the common scale if the model represents a similar shape.

How do triangular areas change on a representation?

If a triangular region is represented by a similar triangle at length scale s, its represented area divided by its actual area equals s². Use common area units for this ratio. The area theorem connects the linear scale with the scale of triangular areas.

Do not multiply an angle by the length scale. Corresponding angles remain equal in a similarity transformation. Lengths use the scale factor, while corresponding triangular areas use its square. Keeping those quantities separate prevents a correct scale from being applied to the wrong measurement.

How can similar triangles measure heights and shadows?

Indirect measurement finds a length through geometric relationships instead of measuring that length directly. A vertical object and its shadow can form a right triangle. A right triangle contains a right angle, and vertical here means perpendicular to the level ground.

When a pole and a tower cast shadows at the same time on level ground, take the sunlight rays as parallel. Their height-and-shadow triangles then have equal right angles and equal angles with the ground, so AA establishes similarity.

Worked example 9. A vertical pole 6 m high casts a shadow 4 m long. At the same time, a vertical tower casts a shadow 28 m long on level ground. Here m after a number means metres. Taking sunlight rays as parallel, find the tower's height.

Answer: Let h be the tower's height in metres. By AA, the height-and-shadow triangles are similar, so h/6 = 28/4. Hence h = 6 × 28/4 = 42 m.

How is a lamp-post shadow different to set up?

For a lamp-post, the larger triangle's base includes both the distance from the post to the person and the person's shadow. Comparing only the post-to-person distance with the shadow would omit part of the larger base.

AB is the vertical lamp-post, CD is the girl, and B, D and E lie along the ground. The sloping line from lamp A passes through C to shadow tip E. DE is the girl's shadow.

See Fig. 6.32 in your NCERT textbook

Worked example 10. A girl 90 cm tall starts at the base of a lamp-post and walks away on level ground at 1.2 m/s. The unit m/s means metres per second. The lamp is 3.6 m above the ground. Find her shadow length after 4 seconds, taking the girl and post as vertical.

Answer: Her height is 0.9 m and her distance from the post is 1.2 × 4 = 4.8 m. Let x be her shadow length in metres. AA gives (4.8 + x)/x = 3.6/0.9 = 4. Thus 4.8 + x = 4x, so 3x = 4.8 and x = 1.6 m.

Here △ABE ∼ △CDE. The angles at B and D are right angles, and the triangles share the angle at E. The correspondence pairs the larger ground length BE with DE and the lamp height AB with the girl's height CD.

What makes the equation complete?

The equation uses BE = BD + DE, because D lies between B and E. The starting position, walking speed and elapsed time determine BD. The two heights determine the vertical ratio. Both sets of information are needed before solving for the shadow.

Use a common unit before substituting heights or distances. The 90 cm = 0.9 m conversion lets every length in the lamp-post proportion be compared in metres. Retaining 90 beside 3.6 without conversion would change the ratio and give an incorrect result.

Glossary

• Similarity — A relationship between figures having the same shape but not necessarily the same size.
• Congruency — A relationship between figures having both the same shape and the same size.
• Corresponding parts — Matching vertices, angles or sides identified by the order of a similarity statement.
• Scale factor — The common multiplier taking each original length to its corresponding image length.
• Proportional sides — Corresponding sides whose lengths give equal ratios when compared in a consistent order.
• Included angle — The angle formed between the two sides being compared in the SAS condition.
• AA similarity — The condition establishing similarity from two pairs of equal corresponding angles in triangles.
• SSS similarity — The condition establishing similarity from three pairs of proportional corresponding sides in triangles.
• SAS similarity — The condition requiring two proportional side pairs and equality of their included angles.
• Basic Proportionality Theorem — A parallel to one triangle side divides the other two sides in equal ratios.
• Converse — A statement formed by exchanging the hypothesis and conclusion of an original statement.
• Altitude — The perpendicular distance from a triangle vertex to the line containing its opposite side.
• Representative fraction — Represented length divided by actual corresponding length, with both expressed in the same unit.
• Indirect measurement — Finding an unknown length through geometric relationships involving other lengths that can be measured.

Common errors and misconceptions

• Misconception: Similar figures must have different sizes. Correct: They have the same shape but not necessarily different sizes. Congruent figures are also similar.
• Misconception: Vertex order is optional. Correct: The order fixes corresponding vertices, angles and sides. Reordering one triangle alone can make the claimed correspondence incorrect.
• Misconception: SSS similarity requires equal side lengths. Correct: It requires three equal ratios of corresponding sides; equal lengths are the stronger congruency condition.
• Misconception: Any equal angle works with two proportional side pairs. Correct: SAS requires the included angle between those particular sides in each triangle.
• Misconception: BPT permits mixing a divided segment with a whole side. Correct: Use AD/DB = AE/EC or AD/AB = AE/AC consistently.
• Misconception: Areas have the same ratio as corresponding sides. Correct: For similar triangles, square the corresponding side ratio to obtain the area ratio.
• Misconception: A lamp-post triangle uses only the distance to the girl as its base. Correct: Its complete base also includes the girl's shadow length.

Exam-style questions with model answers

Q1. State the relationship between similarity and congruency. Must similar figures have different sizes? [2 marks]

1. Congruent figures have the same shape and size, so every pair of congruent figures is also similar.
2. Similar figures have the same shape but not necessarily the same size; their sizes may be equal or different.

Q2. In triangles ABC and DEF, ∠B = ∠E = 60°, ∠C = ∠F = 40°, BC = 3 cm and EF = 5 cm. Find the remaining angles, establish similarity and give the corresponding side ratio. [3 marks]

1. The remaining angles are ∠A = ∠D = 180° − 60° − 40° = 80°, using the angle sum property in each triangle.
2. The pairs ∠B = ∠E and ∠C = ∠F establish △ABC ∼ △DEF by AA, with A corresponding to D.
3. Corresponding sides are proportional, so AB/DE = BC/EF = CA/FD = 3/5. Each ratio compares ABC with DEF in the same order.

Q3. Triangle ABC has AB = 3 cm, BC = 6 cm and CA = 8 cm. Triangle DEF has DE = 4.5 cm, EF = 9 cm and FD = 12 cm. Establish similarity and find their area ratio, ABC to DEF. [4 marks]

1. The first corresponding side ratio is AB/DE = 3/4.5 = 2/3, comparing a side of ABC with its proposed match in DEF.
2. The other ratios are BC/EF = 6/9 = 2/3 and CA/FD = 8/12 = 2/3. All three ratios are equal.
3. Therefore △ABC ∼ △DEF by SSS, with A, B and C matching D, E and F respectively.
4. The area ratio is the square of the corresponding side ratio: ar(ABC)/ar(DEF) = (2/3)² = 4/9, or 4:9.

Q4. In triangles ABC and DEF, AB = 2 cm, AC = 4 cm, DE = 3 cm, DF = 6 cm and ∠A = ∠D = 50°. Establish similarity and find BC/EF. [3 marks]

1. The two side ratios are AB/DE = 2/3 and AC/DF = 4/6 = 2/3, so the stated side pairs are proportional.
2. The equal angles ∠A and ∠D are included between those pairs of sides. Therefore △ABC ∼ △DEF by SAS.
3. Now the remaining corresponding sides must have the same ratio. Thus BC/EF = 2/3, although neither individual side length has been supplied.

Q5. In triangle ABC, D lies on AB, E lies on AC and DE ∥ BC. Given AD = 1.5 cm, DB = 3 cm and AE = 1 cm, find EC. [3 marks]

1. Since DE is parallel to BC and intersects the other sides at D and E, BPT gives AD/DB = AE/EC.
2. Substituting the given lengths gives 1.5/3 = 1/EC. Cross-multiplication therefore gives 1.5 × EC = 3 × 1.
3. Dividing by 1.5 gives EC = 2 cm. The proportion compares the segment next to A with the remaining segment on each side.

Q6. E and F lie on PQ and PR of triangle PQR. PE = 4 cm, EQ = 4.5 cm, PF = 8 cm and FR = 9 cm. Decide whether EF is parallel to QR, giving a reason. [2 marks]

1. The side-division ratios are PE/EQ = 4/4.5 = 8/9 and PF/FR = 8/9, so they are equal.
2. By the converse of the Basic Proportionality Theorem, EF is parallel to QR.

Q7. A girl 90 cm tall starts at the base of a lamp-post and walks away on level ground at 1.2 m/s. The lamp is 3.6 m above the ground. Both girl and post are vertical. A straight light ray through her head reaches the shadow tip on the ground. Find her shadow length after 4 seconds. [5 marks]

1. Convert the girl's height to metres: 90 cm = 0.9 m. Her distance from the post after 4 seconds is 1.2 × 4 = 4.8 m.
2. Let x be the shadow length in metres. The larger ground base, from the post to the shadow tip, is therefore 4.8 + x metres.
3. The lamp-post triangle and the girl triangle have equal right angles and share the angle at the shadow tip. They are similar by AA.
4. Corresponding bases and heights give (4.8 + x)/x = 3.6/0.9 = 4. Multiplying by x gives 4.8 + x = 4x.
5. Subtracting x gives 3x = 4.8, so x = 1.6. The girl's shadow after 4 seconds is therefore 1.6 m long.

Q8. A vertical pole 6 m high casts a shadow 4 m long. At the same time, a vertical tower casts a shadow 28 m long on level ground. Taking sunlight rays as parallel, find the tower's height using similarity. [3 marks]

1. The pole and tower form right triangles with their shadows. Parallel sunlight rays give equal angles with the ground, so these triangles are similar by AA.
2. Let h be the tower's height in metres. Corresponding heights and shadows give h/6 = 28/4, with tower measurements compared to pole measurements.
3. Hence h = 6 × 28/4 = 42. The tower is 42 m high.

Key takeaways

• Similar figures have the same shape but not necessarily the same size; every pair of congruent figures is also similar.
• Write corresponding vertices in matching positions before forming side ratios or identifying equal angles.
• AA uses two equal angle pairs, SSS uses three proportional side pairs, and SAS requires an equal included angle.
• The Basic Proportionality Theorem converts a parallel-line condition into equal ratios of divided sides.
• The converse of the Basic Proportionality Theorem converts equal side-division ratios into a parallel-line conclusion.
• Areas of similar triangles have the ratio of the squares of corresponding sides, with triangle order kept consistent.
• Map and model scales compare corresponding lengths in common units; a triangular area ratio uses the square of the length scale.
• For a lamp-post shadow, the larger ground base includes both the distance to the person and the person's shadow.

Test yourself

Can two similar figures also be congruent?

Yes. Similarity allows equal sizes, while congruency requires both the same shape and the same size.

If △ABC ∼ △DEF, which side corresponds to CA?

FD corresponds to CA because C matches F and A matches D in the similarity statement.

Why do two pairs of equal angles suffice for triangle similarity?

The angle sum property makes the remaining pair equal too, so AA establishes similarity.

What is the included angle between sides AB and AC?

It is ∠BAC, the angle at their common endpoint A, also written as ∠A.

In triangle ABC, D lies on AB, E lies on AC and DE ∥ BC. State the BPT ratio.

AD/DB = AE/EC, comparing the part next to A with the remaining part on each side.

Similar triangles have corresponding side ratio 2:3. What is their area ratio in the same order?

Square the side ratio: the corresponding area ratio is 4:9.

What must be done before dividing a map length by the corresponding ground length?

Express both lengths in the same unit and keep map length as the numerator.

In the lamp-post diagram, D lies between B and E. Why is BE not just BD?

BE also includes DE, the girl's shadow, so the full base is BD + DE.