Triangles Class 10 Formulas

Triangles Class 10 formulas — this sheet collects the similarity results of NCERT Chapter 6: when two figures are similar, the scale factor, the Basic Proportionality Theorem (Thales’ Theorem), its converse, and the five similarity criteria — AAA, AA, SSS, SAS and RHS. Each result below comes with the condition that makes it valid and the meaning of every symbol.

Formulas are grouped by topic, followed by a symbol table, per-formula “when to use” guidance and three worked examples with original numbers. For the derivations and full reasoning behind each theorem, see the Triangles Class 10 notes — this page deliberately keeps only the formulas, not the proofs.

Formulas at a Glance

Each row above is explained in the grouped list below. To verify any statement against the official source, open the NCERT Class 10 Mathematics textbook; for other chapters, see the Class 10 maths formula sheets.

All Formulas, Grouped by Topic

Similar Figures

Two polygons with the same number of sides are similar only when both conditions hold together: every pair of corresponding angles is equal, and every pair of corresponding sides is in the same ratio (NCERT, p. 75).

\[ \angle A = \angle A’,\ \angle B = \angle B’,\ \angle C = \angle C’,\ \angle D = \angle D’ \]

and \[ \frac{AB}{A’B’} = \frac{BC}{B’C’} = \frac{CD}{C’D’} = \frac{DA}{D’A’} \]

The common value of all the side ratios is the scale factor, also called the Representative Fraction (NCERT, p. 76):

\[ k = \frac{AB}{A’B’} = \frac{BC}{B’C’} = \frac{CD}{C’D’} = \frac{DA}{D’A’} \]

Figure 6.1 shows the simplest cases: all circles are similar, all squares are similar and all equilateral triangles are similar, because the shape matches even when the size changes. A circle and a square are not similar.

One consequence worth memorising: all congruent figures are similar, but similar figures need not be congruent (NCERT, p. 75) — the scale factor of congruent figures is exactly 1.

Basic Proportionality Theorem

Theorem 6.1 (Basic Proportionality Theorem, or Thales’ Theorem): in triangle ABC, if the line DE is parallel to side BC and meets AB at D and AC at E, then the other two sides are divided in the same ratio (NCERT, p. 80):

\[ DE \parallel BC \quad \Rightarrow \quad \frac{AD}{DB} = \frac{AE}{EC} \]

Figure 6.10 shows this setting. The proof of the theorem compares the areas of the small triangles formed: triangles on the same base and between the same parallels have equal areas, which is why the two ratios must match.

Whole-side form (derived in Example 1): when the whole sides are wanted instead of the separate segments (NCERT, p. 82):

\[ \frac{AD}{AB} = \frac{AE}{AC} \]

Theorem 6.2 (converse of the BPT): if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side (NCERT, p. 82):

\[ \frac{AD}{DB} = \frac{AE}{EC} \quad \Rightarrow \quad DE \parallel BC \]

Trapezium result (derived in Example 2): in a trapezium ABCD with AB parallel to DC, if EF is drawn parallel to AB, meeting AD at E and BC at F, then (NCERT, p. 83):

\[ \frac{AE}{ED} = \frac{BF}{FC} \]

The converse also proves quick results such as: if \( \frac{PS}{SQ} = \frac{PT}{TR} \) in triangle PQR, then ST is parallel to QR (Example 3, NCERT, p. 84).

Criteria for Similarity of Triangles

Two triangles ABC and DEF are similar, written \( \triangle ABC \sim \triangle DEF \), when their corresponding angles are equal and their corresponding sides are in the same ratio (NCERT, p. 85):

\[ \angle A = \angle D,\ \angle B = \angle E,\ \angle C = \angle F \]

and \[ \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} \]

The order of the vertices in the symbol matters: \( \triangle ABC \sim \triangle DEF \) already declares that A corresponds to D, B to E and C to F (NCERT, p. 86). Figure 6.22 shows this correspondence.

Unlike general polygons, a triangle needs only one of the two conditions — each condition forces the other. The chapter gives five criteria for spotting this quickly.

AAA criterion (Theorem 6.3): if the three corresponding angles are equal, the corresponding sides are automatically in the same ratio (NCERT, p. 87):

\[ \angle A = \angle D,\ \angle B = \angle E,\ \angle C = \angle F \quad \Rightarrow \quad \triangle ABC \sim \triangle DEF \]

AA criterion: two equal pairs of angles are enough for triangles, because the angle sum property then forces the third pair to be equal too (NCERT, p. 88):

\[ \angle A = \angle D,\ \angle B = \angle E \quad \Rightarrow \quad \triangle ABC \sim \triangle DEF \]

SSS criterion (Theorem 6.4): if the three sides of one triangle are proportional to the three sides of the other, the corresponding angles are equal (NCERT, p. 88):

\[ \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} \quad \Rightarrow \quad \triangle ABC \sim \triangle DEF \]

SAS criterion (Theorem 6.5): if one angle of a triangle equals one angle of the other and the sides forming that angle are proportional, the triangles are similar (NCERT, p. 90). The angle must be the one included between the two proportional sides:

\[ \frac{AB}{DE} = \frac{AC}{DF} \ \text{and}\ \angle A = \angle D \quad \Rightarrow \quad \triangle ABC \sim \triangle DEF \]

RHS criterion (Note to the Reader): in two right triangles, if the hypotenuse and one side of one triangle are proportional to the hypotenuse and one side of the other, the triangles are similar (NCERT, p. 98). With the right angles at B and E:

\[ \angle B = \angle E = 90^\circ \ \text{and}\ \frac{AC}{DF} = \frac{AB}{DE} \quad \Rightarrow \quad \triangle ABC \sim \triangle DEF \]

Medians in similar triangles (derived in Example 8): if \( \triangle ABC \sim \triangle PQR \) and CM, RN are the medians to the sides AB, PQ, then the medians are in the same ratio as the corresponding sides (NCERT, p. 94):

\[ \frac{CM}{RN} = \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} \]

What Each Symbol Means

When to Use Each Formula

If the formula you need belongs to another chapter, start from the maths formulas index.

Worked Examples

Three uses of the sheet’s formulas, with original numbers. After these, practise on the textbook’s own questions in the NCERT solutions for Chapter 6 Triangles.

Worked Example 1: Find EC using the Basic Proportionality Theorem

Worked Example 2: Test whether EF is parallel to QR (converse of the BPT)

Worked Example 3: Find a tower’s height from shadows (AA criterion)

Figure 6.32 shows the same set-up with a lamp-post and a girl — the two similar right triangles give the shadow length by the same ratio method.

Common Mistakes to Avoid

The errors below are the most frequent when applying these ratios and criteria.

Frequently Asked Questions

What is the difference between congruent and similar figures?

Congruent figures have the same shape and the same size; similar figures have the same shape but not necessarily the same size. Every congruent pair is also similar (their scale factor is 1), but a similar pair need not be congruent (NCERT, p. 75).

Why is AA enough to prove similarity? Don’t all three angles have to match?

For triangles, two matches force the third. If \( \angle A = \angle D \) and \( \angle B = \angle E \), the angle sum property gives \( \angle C = 180^\circ – \angle A – \angle B = 180^\circ – \angle D – \angle E = \angle F \). So AA is AAA with less work (NCERT, p. 88).

When do I use the Basic Proportionality Theorem and when its converse?

Use Theorem 6.1 when a line is known to be parallel to a side and you need a missing segment or a proportion. Use Theorem 6.2 when the ratios of division are given and you must prove the line is parallel to the third side.

In SAS similarity, can the equal angle be any angle of the triangle?

No. The angle must be the one included between the two sides whose ratio you compare. That is why SAS needs an angle, while SSS uses only sides.

Reference: NCERT Class 10 Mathematics textbook, chapter Triangles.