Triangles Class 10: Notes, Theorems and Important Questions

This guide covers Triangles Class 10, chapter 6 of NCERT Class 10 Mathematics, which is all about similar triangles. You will learn the Basic Proportionality Theorem with its proof, the AA, SSS and SAS similarity criteria, and how to use them to find heights and shadow lengths. Important questions with answers and common mistakes come at the end.

Similar figures

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

• All congruent figures are similar, but similar figures need not be congruent.
• All circles are similar. They are congruent only if their radii are equal.
• All squares are similar: every angle is 90° and the side ratios are always equal.

Two polygons with the same number of sides are similar if:

1. all corresponding angles are equal, and
2. all corresponding sides are in the same ratio (the scale factor, or representative fraction).

Both conditions are needed. A square and a rectangle both have four 90° angles, but their sides are not in the same ratio, so they are not similar.

Similarity allows indirect measurement: using small, known measurements to find large distances such as the height of Mount Everest or the distance to the moon.

Basic Proportionality Theorem (Thales Theorem)

Theorem 6.1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

In triangle ABC, with D on AB, E on AC and DE ∥ BC: AD/DB = AE/EC.

Proof idea (using areas):

1. Triangles ADE and BDE have bases AD and DB on line AB and share the height from E. So area(ADE)/area(BDE) = AD/DB.
2. Triangles ADE and CDE share the height from D to AC. So area(ADE)/area(CDE) = AE/EC.
3. Triangles BDE and CDE have the same base DE and lie between the same parallels DE and BC, so their areas are equal.
4. Therefore AD/DB = AE/EC.

Theorem 6.2 (converse): If a line divides any two sides of a triangle in the same ratio, it is parallel to the third side.

Proof idea: suppose AD/DB = AE/EC. Draw through D a line parallel to BC meeting AC at E′. By Theorem 6.1, AD/DB = AE′/E′C. So AE/EC = AE′/E′C. The same ratio picks out the same point on AC, so E and E′ coincide, and DE ∥ BC.

Example: In triangle ABC, DE ∥ BC, AD = 1.5 cm, DB = 3 cm and AE = 1 cm. Then 1.5/3 = 1/EC, so 1/2 = 1/EC and EC = 2 cm.

AAA and AA similarity

Theorem 6.3 (AAA): If the corresponding angles of two triangles are equal, their corresponding sides are in the same ratio, so the triangles are similar. This is special to triangles; for other polygons, equal angles alone are not enough.

AA criterion: The angles of a triangle add up to 180°. So if two angles of one triangle equal two angles of another, the third angles are equal too. Two pairs of equal angles are enough.

Writing the statement: ΔABC ~ ΔDEF means ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB/DE = BC/EF = CA/FD. The order of letters shows which vertices match.

SSS and SAS similarity

Theorem 6.4 (SSS): If the sides of one triangle are proportional to the sides of another, their corresponding angles are equal and the triangles are similar. Check all three ratios. If even one differs, they are not similar.

Tip: pair the smallest side with the smallest, the middle with the middle and the largest with the largest.

Example: sides 3.8, 6, 3√3 and 7.6, 12, 6√3. Ratios: 3.8/7.6 = 1/2, 6/12 = 1/2, 3√3/6√3 = 1/2. All equal, so the triangles are similar by SSS.

Theorem 6.5 (SAS): If one angle of a triangle equals one angle of another, and the sides including these angles are proportional, the triangles are similar. The equal angle must be the included angle, between the two sides you compare.

Example: lines AB and CD cross at O, with OA = 4, OC = 8, OD = 5, OB = 10. OA/OC = 4/8 = 1/2 and OD/OB = 5/10 = 1/2. The vertically opposite angles ∠AOD and ∠COB are equal. So ΔAOD ~ ΔCOB by SAS.

Summary of criteria and applications

RHS similarity needs proportional sides, unlike the RHS congruence test, which needs equal lengths.

Heights and shadows: at the same moment, the sun's rays meet a pole and a tower at the same angle, so the triangles formed by each object, its shadow and the ray are similar.

Medians: a median divides the opposite side into two equal parts. In similar triangles, corresponding medians are in the same ratio as corresponding sides.

Steps for any problem: name the two triangles, list the equal angles or side ratios, state the criterion, write the proportion with matching sides, solve, and check the units.

Remember this

• Similar: same shape; equal angles and proportional sides.
• BPT: DE ∥ BC gives AD/DB = AE/EC. The converse also holds.
• For triangles, AA is enough.
• SSS: check all three ratios. SAS: the angle must be included.
• The order of letters in ΔABC ~ ΔDEF tells you which sides match.

Important questions with answers

1. Are all squares similar? Yes. All angles are 90°, and since each square has four equal sides, the ratio of corresponding sides is always the same.

2. If two polygons have equal corresponding angles, must they be similar? No. A square and a rectangle have equal angles but their sides are not proportional.

3. In ΔABC, DE ∥ BC, AD = 4 cm, DB = 6 cm and AE = 6 cm. Find EC. By BPT, 4/6 = 6/EC. So 4 × EC = 36 and EC = 9 cm.

4. In ΔPQR, S is on PQ and T on PR with PS/SQ = 3/4 and PT/TR = 3/4. What can you say about ST? ST divides two sides in the same ratio, so by the converse of BPT, ST ∥ QR.

5. ΔABC has angles 60° and 80°; ΔPQR has angles 80° and 40°. Are they similar? The third angle of ABC is 180° − 140° = 40°, and of PQR is 180° − 120° = 60°. Both have angles 40°, 60°, 80°, so they are similar by AA.

6. ΔABC has sides 2, 3, 4 and ΔPQR has sides 4, 6, 9. Are they similar? 2/4 = 0.5 and 3/6 = 0.5, but 4/9 is about 0.44. The ratios are not all equal, so no.

7. ∠A = ∠D = 50°, AB = 2 cm, AC = 4 cm, DE = 3 cm, DF = 6 cm. Are ΔABC and ΔDEF similar? AB/DE = 2/3 and AC/DF = 4/6 = 2/3, and the included angles are equal. Similar by SAS.

8. A 6 m pole casts a 4 m shadow. At the same time a tower casts a 28 m shadow. Find the tower's height. By AA similarity, h/6 = 28/4 = 7, so h = 42 m.

9. A girl 90 cm tall walks away from the base of a 3.6 m lamp-post at 1.2 m/s. Find her shadow's length after 4 s. She walks 1.2 × 4 = 4.8 m. Let the shadow be x m. Similar triangles give 3.6/0.9 = (4.8 + x)/x. So 4x = 4.8 + x, 3x = 4.8 and x = 1.6 m.

Common mistakes to avoid

• Matching sides in the wrong order. Follow the vertex order in the similarity statement.
• Using SAS when the equal angle is not between the two sides.
• Checking only two ratios for SSS.
• Mixing units, such as 90 cm with 3.6 m. Convert first.
• Assuming equal angles make any two polygons similar.

Draw a neat labelled diagram for every problem, and study this chapter with Joy whenever you want extra practice.