Similarity Class 10 ICSE — Das Gupta Chapter 17

Written by Greya LakshmiPublished on 28 September 2026

Similarity Class 10 ICSE from A. Das Gupta’s Mathematics covers when two triangles are similar, the three axioms that prove similarity, and how the areas of similar triangles compare. This chapter builds one core idea — the area ratio equals the square of the side ratio — and extends it to altitudes, medians, and angle bisector segments. Exercise 17 applies these results to parallel-line figures and trapeziums.

Download the Similarity of Triangles Class 10 ICSE Mathematics chapter PDF from the A. Das Gupta textbook to revise offline or print for practice. It is the full chapter as printed in the book — 12 pages, 3 sections, 15 questions.

What is inside Chapter 17

12 pages · 3 sections · 15 exercise questions

Section	Title
1	Similar Triangles
2	Three Similarity Axioms For Triangles
3	Results on Area of Similar Triangles (Theorems)

EXERCISE 17: 15 questions

1 Similar Triangles

Two triangles are similar when both conditions hold together: each pair of corresponding angles is equal, and each pair of corresponding sides is in the same ratio. For $\Delta ABC$ and $\Delta DEF$ , that means $\angle A = \angle D$ , $\angle B = \angle E$ , $\angle C = \angle F$ , and $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$ .

The symbol $\sim$ is read as ‘is similar to’, so we write $\Delta ABC \sim \Delta DEF$ .

The order of the vertices matters. Writing $\Delta ABC \sim \Delta DEF$ names the correspondence: A pairs with D, B with E, C with F. If you wrote $\Delta ABC \sim \Delta EFD$ , then side AB would pair with EF and angle B with angle F — a different correspondence, so the ratio legs change. Always match letters in the same positions.

Why both conditions? Equal angles alone gives the same shape. Proportional sides alone gives the same scale. Similarity needs both together, which is what the $\sim$ symbol promises.

2 Three Similarity Axioms For Triangles

The axioms give you three working tests. You do not need to check every angle and every side ratio each time.

SAS-Axiom: if one pair of corresponding angles is equal and the sides including those angles are proportional, the triangles are similar. For $\angle A = \angle D$ and $\frac{AB}{DE} = \frac{AC}{DF}$ , you get $\Delta ABC \sim \Delta DEF$ . The equal angle must sit between the two proportional sides — that is what ‘including’ means.

SAS works because the included angle fixes how the two sides open. Once both sides scale by the same factor and the angle matches, the third side scales too.

AA-Axiom: two pairs of equal corresponding angles suffice. The third pair follows because the angle sum of any triangle is 180°, so once two pairs match, the third must match as well. AAA states the same test; you only verify two pairs in practice.

SSS-Axiom: when all three pairs of corresponding sides are proportional, the triangles are similar without checking any angle. Three fixed side ratios determine the shape completely, so the angles follow.

Axiom	Condition needed	Conclusion
SAS	$\angle A = \angle D$ and $\frac{AB}{DE} = \frac{AC}{DF}$	$\Delta ABC \sim \Delta DEF$
AA / AAA	$\angle A = \angle D$ and $\angle B = \angle E$	$\Delta ABC \sim \Delta DEF$
SSS	$\frac{AB}{DE} = \frac{AC}{DF} = \frac{BC}{EF}$	$\Delta ABC \sim \Delta DEF$

Choose the test by what the question hands you: two angle pairs means AA; three side ratios means SSS; one angle with two sides means SAS, provided the angle lies between them. In this Class 10 ICSE chapter, most proofs in Exercise 17 run on AA, since parallel lines and shared angles appear everywhere.

3 Results on Area of Similar Triangles (Theorems)

Theorem 1 states the core result: the areas of two similar triangles are proportional to the squares of their corresponding sides.

$\frac{\text{Area of } \Delta ABC}{\text{Area of } \Delta DEF} = \frac{AB^2}{DE^2} = \frac{BC^2}{EF^2} = \frac{AC^2}{DF^2}$

The proof starts from area = $\frac{1}{2} \times$ base $\times$ height. Drawing altitudes $AL \perp BC$ and $DM \perp EF$ , the area ratio becomes $\frac{BC}{EF} \times \frac{AL}{DM}$ . The right triangles $\Delta ALB$ and $\Delta DME$ are similar — each has a right angle and $\angle B = \angle E$ from the given similarity — so $\frac{AL}{DM} = \frac{AB}{DE}$ .

Since the original triangles are similar, $\frac{AB}{DE} = \frac{BC}{EF}$ . Substituting gives the squared ratio.

Why does the ratio square? The side ratio appears once in the base ratio and once in the height ratio. Two copies of the same factor multiply, producing the square.

Theorem 2 extends the result to altitudes: the area ratio equals $\frac{AL^2}{DM^2}$ , the squares of the corresponding altitudes. The proof reuses Theorem 1 and the similarity of $\Delta ALB$ and $\Delta DME$ to swap the base ratio for the altitude ratio.

Theorem 3 does the same for medians: the area ratio equals $\frac{AP^2}{DQ^2}$ . The key step proves $\Delta APB \sim \Delta DQE$ by SAS, using the median property that $BP = \frac{1}{2} BC$ and $EQ = \frac{1}{2} EF$ .

Theorem 4 completes the set for angle bisector segments: the area ratio equals $\frac{AX^2}{DY^2}$ . Since $\angle BAX = \frac{1}{2} \angle A$ and $\angle EDY = \frac{1}{2} \angle D$ , and the original triangles give $\angle A = \angle D$ , the half-angles match. So $\Delta ABX \sim \Delta DEY$ by AA.

Theorem	Corresponding length	Area ratio formula
Theorem 1	Sides	$\frac{AB^2}{DE^2} = \frac{BC^2}{EF^2} = \frac{AC^2}{DF^2}$
Theorem 2	Altitudes	$\frac{AL^2}{DM^2}$
Theorem 3	Medians	$\frac{AP^2}{DQ^2}$
Theorem 4	Angle bisector segments	$\frac{AX^2}{DY^2}$

Worked example. Two similar triangles have areas $27\ \text{cm}^2$ and $48\ \text{cm}^2$ . The smaller triangle has a side of $6\ \text{cm}$ . Find the corresponding side of the larger triangle.

$\frac{27}{48} = \frac{9}{16} = \left(\frac{3}{4}\right)^2$

Step 1: The area ratio is $\frac{9}{16}$ , so the side ratio is its square root: $\frac{3}{4}$ .

Step 2: Let the larger side be $x$ . Set $\frac{6}{x} = \frac{3}{4}$ and solve.

$x = \frac{6 \times 4}{3} = 8\ \text{cm}$

Final answer: The corresponding side of the larger triangle is $8\ \text{cm}$ .

The common mistakes in this chapter cluster around the square. Mixing up the area ratio with the side ratio is the most frequent error — the area ratio is the side ratio squared, not the ratio itself. Forgetting to square the side ratio gives answers off by the ratio factor. Pairing non-corresponding sides also breaks the answer, because the theorem only holds for sides in the same position.

A recurring pattern in Exercise 17 asks for the area of a small triangle compared with the surrounding trapezium when a line is parallel to the base. The method: prove the small triangle similar to the whole triangle, find the area ratio from the side ratio, then subtract the small area from the whole to get the trapezium. Questions 7, 8 and 15 in this exercise all follow this structure.

For the preceding chapter on similarity as a size transformation, see Similarity as a Size Transformation.

What to practise from Similarity Class 10 ICSE

Exercise 17 is the only exercise block in this chapter, with 15 questions. The questions group into four patterns you can practise separately.

Parallel-line side ratios (Q1, Q6, Q7, Q8, Q12, Q15): a line parallel to one side cuts off a smaller similar triangle, and the side ratio follows from the proportion.
Triangle-to-trapezium area ratios (Q7, Q8, Q15): find the small triangle’s area ratio, then subtract from the whole triangle to get the trapezium.
Altitudes and medians (Q3, Q10, Q14): apply Theorem 2 or Theorem 3, or prove a relation between altitude and median ratios.
Proof questions with A.A. (Q5, Q9, Q11, Q13): identify vertically opposite or alternate angles to prove similarity, then set up the side proportion.

Use the area theorems in reverse too: when a question gives two areas and one side, take the square root of the area ratio to recover the side ratio. For the full set of Class 10 ICSE Maths chapters, visit the Class 10 Mathematics books hub. You can also verify the Class 10 syllabus and exam format on the official CISCE website.

Similarity Class 10 ICSE frequently asked questions

What topics come under similarity Class 10 ICSE?

This chapter covers the definition of similar triangles, the three similarity axioms (SAS, AA/AAA, SSS), and four theorems that relate the area ratio of similar triangles to the squares of their corresponding sides, altitudes, medians, and angle bisector segments. Exercise 17 applies all of these to parallel-line figures and trapeziums.

How do I decide which similarity axiom to use — SAS, AA or SSS?

Look at what the question gives you. Two pairs of equal angles means AA. Three pairs of proportional sides means SSS. One equal angle with the two sides around it proportional means SAS — but only when the equal angle sits between those two sides.

Why is the area of similar triangles proportional to the square of their sides?

The area of a triangle is $\frac{1}{2} \times$ base $\times$ height. In similar triangles, both the base and the height scale by the same factor, the side ratio. Two copies of the same factor multiply together, so the area ratio becomes the side ratio squared.

How do I find the ratio of a small triangle to the trapezium in Class 10 ICSE when a line is parallel to the base?

Prove the small triangle similar to the whole triangle using AA, with a common angle and corresponding angles from the parallel line. Find the area ratio from the side ratio, then subtract the small triangle’s area from the whole triangle’s area to get the trapezium.

Do the area theorems in this ICSE chapter work for altitudes, medians and angle bisectors as well?

Yes. Theorems 2, 3 and 4 extend Theorem 1 to altitudes, medians and angle bisector segments respectively. In each case, the area ratio equals the square of the ratio of the corresponding lengths.

Reference: ICSE Class 10 Mathematics textbook material, chapter 17: Similarity of Triangles — CISCE-aligned.

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