ICSE Class 10 Maths Similarity Solutions Chapter 15

Written by Greya LakshmiPublished on 19 May 2026Updated on 9 June 2026

ICSE Class 10 Maths Similarity Chapter 15 overview

ICSE Class 10 Maths Chapter 15, Similarity, teaches how to prove that two triangles have the same shape and then use proportional sides to find unknown lengths. In Selina Concise Mathematics Class 10 ICSE, this chapter also connects similarity with maps and models, where a scale drawing represents a larger or smaller real object.

Concept snapshot: similar triangles are scaled copies

Think of a photo printed in two sizes. The smaller and larger photos have matching angles, but the lengths are multiplied by a fixed scale factor. Similar triangles work in the same way: corresponding angles are equal, and corresponding sides are in the same ratio.

A useful classroom check is this: same shape first, ratio second. Do not write side ratios until you have proved the triangle correspondence.

Formula and method reference for Similarity

Use this table before starting the worked solutions. The key task is to name the correct correspondence, not just to say that two triangles are similar.

Idea	When to use it	Mathematical form
AA or AAA similarity	Two or three pairs of corresponding angles are equal	If two angles match, the triangles are similar.
SAS similarity	Two pairs of sides around an included angle are proportional and the included angle is equal	$\frac{AB}{DE}=\frac{AC}{DF}$ and $\angle A=\angle D \Rightarrow \triangle ABC \sim \triangle DEF$
SSS similarity	All three pairs of corresponding sides are proportional	$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$
Corresponding sides	After proving similarity	If $\triangle ABC \sim \triangle DEF$ , then $\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$
Area ratio	When the question asks about areas of similar triangles	$ \frac{ $\triangle ABC$ }{ $\triangle DEF$ }=\left(\frac{AB}{DE}\right)^2 $
Scale drawing	Maps and models	For scale $1:n$ , actual length $=n\times$ drawing length.

How to solve Similarity questions step by step

Most Chapter 15 proofs follow a fixed route. Write the equal angles or side ratios first, then name the similarity criterion, and only then use proportional sides.

Worked example 1: Prove similarity from a side ratio and included angle

Question: In $\triangle ABC$ and $\triangle DEF$ , $\angle A=\angle D$ and $\frac{AB}{AC}=\frac{DE}{DF}$ . Prove that $\triangle ABC \sim \triangle DEF$ .

Step 1: Rewrite the given proportion so that corresponding sides are compared in the same order.

$\frac{AB}{AC}=\frac{DE}{DF}$

Step 2: Cross-arrange the ratios.

$\frac{AB}{DE}=\frac{AC}{DF}$

Step 3: The included angles are equal, since $\angle A=\angle D$ .

Step 4: Two pairs of sides including the equal angle are proportional.

Hence proved: $\triangle ABC \sim \triangle DEF$ by SAS similarity.

For related practice after this chapter, students may revise ICSE Class 10 Selina Maths solutions and then return to similarity proofs with a clearer idea of how geometry solutions are written.

Exercise 15(A): similarity criteria and proportional sides

Exercise 15(A) type questions usually test whether you can identify similar triangles from a diagram and then use the correct corresponding sides. The examples below follow the school-style steps expected in ICSE Class 10 Maths.

Worked example 2: Find $OB$ when $OC=1.5\,OA$

Question: In a figure, $AB \parallel CD$ , lines $AC$ and $BD$ meet at $O$ , and $OC=1.5\,OA$ . Find $OB$ in terms of $OD$ .

Step 1: Compare $\triangle OAB$ and $\triangle OCD$ .

Step 2: Since $AB \parallel CD$ , alternate interior angles are equal.

$\angle OAB=\angle OCD,\qquad \angle OBA=\angle ODC$

Step 3: Vertically opposite angles are equal.

$\angle AOB=\angle COD$

Step 4: Therefore the triangles are similar.

$\triangle OAB \sim \triangle OCD$

Step 5: Convert $OC=1.5\,OA$ into a ratio.

$\frac{OA}{OC}=\frac{1}{1.5}=\frac{2}{3}$

Step 6: Use corresponding sides of similar triangles.

$\frac{OB}{OD}=\frac{OA}{OC}=\frac{2}{3}$

Final answer: $OB=\frac{2}{3}\,OD$ .

Worked example 3: Find $x$ using proportional sides

Question: In a figure, $AB \parallel CD$ , lines $AC$ and $BD$ meet at $O$ , $AO=18$ , $CO=12$ , $OD=20$ , and $OB=x$ . Find $x$ .

Step 1: From parallel lines and vertically opposite angles, $\triangle AOB \sim \triangle COD$ .

Step 2: Write the corresponding side ratio.

$\frac{AO}{CO}=\frac{OB}{OD}$

Step 3: Substitute the given values.

$\frac{18}{12}=\frac{x}{20}$

Step 4: Cross-multiply.

$12x=18\times 20$

$x=\frac{18\times 20}{12}=30$

Final answer: $x=30$ .

Worked example 4: Prove a median ratio in similar triangles

Question: In $\triangle ABC$ and $\triangle DEF$ , the corresponding angles are equal and $AL$ , $DM$ are medians. Prove that $\frac{BC}{EF}=\frac{AL}{DM}$ .

Step 1: Since corresponding angles are equal, the triangles are similar.

$\triangle ABC \sim \triangle DEF$

Step 2: Therefore corresponding sides are proportional.

$\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$

Step 3: Since $AL$ and $DM$ are medians, $L$ and $M$ are midpoints of $BC$ and $EF$ .

$BL=\frac{1}{2}BC,\qquad EM=\frac{1}{2}EF$

Step 4: Hence the halves are also in the same ratio.

$\frac{BL}{EM}=\frac{BC}{EF}$

Step 5: In $\triangle ABL$ and $\triangle DEM$ , $\angle B=\angle E$ , and

$\frac{AB}{DE}=\frac{BL}{EM}$

Step 6: Therefore $\triangle ABL \sim \triangle DEM$ by SAS similarity.

Step 7: Use corresponding sides of these similar triangles.

$\frac{AL}{DM}=\frac{AB}{DE}$

Step 8: From Step 2, $\frac{AB}{DE}=\frac{BC}{EF}$ .

Hence proved: $\frac{BC}{EF}=\frac{AL}{DM}$ .

Exercise 15(B): parallel lines, right triangles and lengths

Exercise 15(B) type questions often combine parallel lines, perpendiculars and intersecting transversals. The important point is to mark equal angles before using ratios.

Worked example 5: Intersecting lines with $AC \parallel BD$

Question: Straight lines $AB$ and $CD$ intersect at $P$ , and $AC \parallel BD$ . Prove $\triangle APC \sim \triangle BPD$ . If $BD=2.4\text{ cm}$ , $AC=3.6\text{ cm}$ , $PD=4.0\text{ cm}$ , and $PB=3.2\text{ cm}$ , find $PA$ and $PC$ .

Step 1: Vertically opposite angles are equal.

$\angle APC=\angle BPD$

Step 2: Since $AC \parallel BD$ , alternate interior angles are equal.

$\angle ACP=\angle BDP$

Step 3: Therefore the triangles are similar.

$\triangle APC \sim \triangle BPD$

Step 4: Use corresponding sides.

$\frac{AC}{BD}=\frac{PA}{PB}=\frac{PC}{PD}$

Step 5: Find the common ratio.

$\frac{AC}{BD}=\frac{3.6}{2.4}=\frac{3}{2}$

Step 6: Find $PA$ .

$\frac{PA}{3.2}=\frac{3}{2}$

$PA=\frac{3}{2}\times 3.2=4.8\text{ cm}$

Step 7: Find $PC$ .

$\frac{PC}{4.0}=\frac{3}{2}$

$PC=\frac{3}{2}\times 4.0=6.0\text{ cm}$

Final answer: $PA=4.8\text{ cm}$ and $PC=6.0\text{ cm}$ .

Worked example 6: Find $BP\times DO$

Question: In a figure with $PB \parallel DQ$ , $BO=6\text{ cm}$ , and $DQ=8\text{ cm}$ . If $\triangle DOQ \sim \triangle BOP$ , find $BP\times DO$ .

Step 1: From similarity, write the ratio of corresponding sides.

$\frac{DO}{BO}=\frac{DQ}{BP}$

Step 2: Substitute $BO=6\text{ cm}$ and $DQ=8\text{ cm}$ .

$\frac{DO}{6}=\frac{8}{BP}$

Step 3: Cross-multiply.

$DO\times BP=6\times 8$

$DO\times BP=48$

Final answer: $BP\times DO=48\text{ cm}^2$ .

Worked example 7: Prove $\angle BAC=90^\circ$

Question: In $\triangle ABC$ , $AD\perp BC$ and $AD^2=BD\times CD$ . Show that $\angle BAC=90^\circ$ .

Step 1: Convert the given product into a side ratio.

$AD^2=BD\times CD$

$\frac{BD}{AD}=\frac{AD}{CD}$

Step 2: Since $AD\perp BC$ , both angles at $D$ are right angles.

$\angle BDA=\angle ADC=90^\circ$

Step 3: In $\triangle BDA$ and $\triangle ADC$ , the sides including the equal angle are proportional.

$\frac{BD}{AD}=\frac{AD}{CD}$

Step 4: Therefore the triangles are similar by SAS similarity.

$\triangle BDA \sim \triangle ADC$

Step 5: Hence corresponding angles are equal.

$\angle BAD=\angle ACD,\qquad \angle ABD=\angle DAC$

Step 6: Add the two parts of $\angle A$ .

$\angle BAC=\angle BAD+\angle DAC$

$\angle BAC=\angle C+\angle B$

Step 7: Use the angle-sum property of $\triangle ABC$ .

$\angle A+\angle B+\angle C=180^\circ$

$\angle A+\angle A=180^\circ$

$2\angle A=180^\circ$

Final answer: $\angle BAC=90^\circ$ .

Similarity applications to maps and models

Maps and models use the same idea as similar figures. A scale of $1:n$ means $1$ unit on the drawing represents $n$ units in real life. Keep the units consistent before writing the final answer.

Worked example 8: Map scale calculation

Question: A map uses the scale $1:25000$ . The distance between two points on the map is $7.2\text{ cm}$ . Find the actual distance in kilometres.

Step 1: Write the meaning of the scale.

$1\text{ cm on map}=25000\text{ cm in reality}$

Step 2: Multiply by the map distance.

$7.2\text{ cm on map}=7.2\times 25000\text{ cm}$

$=180000\text{ cm}$

Step 3: Convert centimetres to kilometres.

$100000\text{ cm}=1\text{ km}$

$180000\text{ cm}=1.8\text{ km}$

Final answer: The actual distance is $1.8\text{ km}$ .

Worked example 9: Model scale calculation

Question: A model tower is $18\text{ cm}$ high. The scale is $1:75$ . Find the actual height of the tower in metres.

Step 1: Write the scale meaning.

$1\text{ cm on model}=75\text{ cm in reality}$

Step 2: Multiply by $18$ .

$18\text{ cm on model}=18\times 75\text{ cm}$

$=1350\text{ cm}$

Step 3: Convert centimetres to metres.

$100\text{ cm}=1\text{ m}$

$1350\text{ cm}=13.5\text{ m}$

Final answer: The actual height is $13.5\text{ m}$ .

For more Class 10 practice, use the ICSE Class 10 Maths solutions hub. Mensuration problems also use scale and unit conversion, so revise Selina Class 10 Mensuration solutions after maps and models.

Examiner’s mindset for ICSE Similarity proofs

In geometry answers, a correct final ratio is not enough. The working should show three things clearly:

Equal angles or proportional sides: state why they are equal, such as vertically opposite angles, alternate interior angles, common angle or right angle.
Similarity criterion: write AA, SAS or SSS similarity after the evidence, not before it.
Correct correspondence: write $\triangle ABC \sim \triangle DEF$ in the order that matches the equal angles and side ratios.

A board-style answer usually loses clarity when the student writes only $\frac{AB}{DE}=\frac{BC}{EF}$ without proving $\triangle ABC \sim \triangle DEF$ first.

Common mistakes students make in Similarity

Mistake	Correction
Writing side ratios before proving the triangles similar.	First prove similarity using AA, SAS or SSS; then use corresponding sides.
Changing the order of triangle names.	If $\triangle ABC \sim \triangle DEF$ , keep $A\leftrightarrow D$ , $B\leftrightarrow E$ , $C\leftrightarrow F$ .
Using $\frac{AB}{AC}$ and $\frac{DE}{DF}$ without checking the included angle.	For SAS similarity, the equal angle must be between the two proportional sides.
Forgetting units in maps and models.	Convert $\text{cm}$ to $\text{m}$ or $\text{km}$ only after using the scale factor.
Using area ratio as the side ratio.	For similar triangles, area ratio is the square of the side ratio: $\frac{A_1}{A_2}=k^2$ .

Quick answer index

Section	Question type	Answer / result
15(A)	Parallel lines with $OC=1.5\,OA$	$OB=\frac{2}{3}\,OD$
15(A)	Find $x$ from $\frac{18}{12}=\frac{x}{20}$	$x=30$
15(A)	Medians in similar triangles	$\frac{BC}{EF}=\frac{AL}{DM}$
15(B)	Intersecting lines with $AC \parallel BD$	$PA=4.8\text{ cm}$ , $PC=6.0\text{ cm}$
15(B)	Product from similar triangles	$BP\times DO=48\text{ cm}^2$
15(B)	Altitude condition $AD^2=BD\times CD$	$\angle BAC=90^\circ$
Maps and models	Scale $1:25000$ , map distance $7.2\text{ cm}$	$1.8\text{ km}$
Maps and models	Scale $1:75$ , model height $18\text{ cm}$	$13.5\text{ m}$

Coordinate geometry also uses proportion and ratios in another form. For practice that connects well with similarity, use Selina Class 10 Coordinate Geometry solutions.

Frequently Asked Questions

Which similarity criterion should I use first in ICSE Class 10 Maths Similarity problems?

Start by checking equal angles. In many ICSE Class 10 Maths Similarity problems, AA or AAA is the shortest route because parallel lines, vertically opposite angles and right angles give angle equality directly.

How do I avoid writing the wrong side ratio in Chapter 15 Similarity?

Write the triangle correspondence before using ratios. If $\triangle APC \sim \triangle BPD$ , then $AP$ corresponds to $BP$ , $PC$ corresponds to $PD$ , and $AC$ corresponds to $BD$ .

Are maps and models part of Similarity in Selina Class 10?

Yes. The maps and models part uses scale factor. If the scale is $1:n$ , then the actual length is $n$ times the drawing length, after converting units correctly.

Why is the order of letters important in similar triangles?

The order shows corresponding vertices. Writing $\triangle ABC \sim \triangle DEF$ means $A\leftrightarrow D$ , $B\leftrightarrow E$ , and $C\leftrightarrow F$ . The side ratios must follow this order.

What is the usual error in right-triangle altitude questions?

The common error is to use $AD^2=AB\times AC$ . For an altitude from the right angle to the hypotenuse, the correct relation used in this chapter is $AD^2=BD\times CD$ , with the correct triangle correspondence.

Sources and further study

This page is aligned with the standard ICSE Class 10 Mathematics treatment of similar triangles, proportional sides, maps and models, and school-style geometry proof writing. For official syllabus reference, use the CISCE Regulations and Syllabuses publications page.

Concise Mathematics, Selina Publishers, Class 10, Chapter 15: Similarity with Applications to Maps and Models.
CISCE official Regulations and Syllabuses for ICSE Class 10 Mathematics.
Standard school geometry results on triangle similarity, proportional sides and scale factor.