NCERT Solutions for Class 10 Maths Chapter 6 Triangles

NCERT Solutions for Class 10 Maths Chapter 6 Triangles help students understand one of the most important parts of geometry in a clear and organised way. The chapter begins with the idea of similar figures and explains how two shapes can have the same form even when their sizes are different. Students learn that similar polygons have equal corresponding angles and proportional corresponding sides. They also study examples involving circles, squares, equilateral triangles, and other figures to understand the meaning of similarity in a practical way.

A major part of the chapter is the Basic Proportionality Theorem, also called BPT, and its converse. These ideas show how a line drawn parallel to one side of a triangle divides the other two sides in the same ratio. The exercises teach students how to find unknown lengths, check whether two lines are parallel, and prove results about midpoints and proportional sides. Step-by-step solutions make it easier to understand why each ratio is used and how each conclusion is reached.

The chapter also explains the main criteria used to prove that two triangles are similar. These include AA or AAA similarity, SSS similarity, and SAS similarity. Students must identify equal angles, compare corresponding sides, and write the names of triangles in the correct order. This is important because the correct order shows which sides and angles match. The solutions guide students through each proof in a simple sequence, helping them avoid common mistakes.

Another important idea covered in this chapter is the relation between the areas of similar triangles. When two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. Students also use similarity to solve questions based on altitudes, medians, trapeziums, parallel lines, and geometric constructions. These questions build logical thinking and improve proof-writing skills.

Download Class 10 Chapter 6 Triangles NCERT Solutions PDF

Download the Class 10 Maths Chapter 6 Triangles NCERT Solutions PDF to study all exercises in an easy and clear way. The PDF includes step-by-step answers, important concepts, and simple explanations to help students understand similarity, proportionality, and triangle-based questions. It is useful for homework, revision, exam preparation, and checking answers anytime.

Access NCERT Solutions for Class 10 Maths Chapter 6 Triangles Question and Answer

Exercise 6.1

These questions check the meaning of similarity and the conditions needed for two polygons to be similar.

Solution:

Answers: (i) similar, (ii) similar, (iii) equilateral, (iv) equal and proportional.

Solution:

Examples of similar figures:

1. Any two circles of different radii.
2. Any two squares of different side lengths.

Examples of non-similar figures:

1. A square and a non-square rectangle.
2. An equilateral triangle and a right-angled triangle.

Similarity depends on shape, not merely on having the same number of sides. A square and a rectangle both have four sides, but their side ratios generally differ.

Solution:

The ratio of every side of the rhombus to the corresponding side of the square is

1.5/3 = 1/2.

Thus, corresponding sides are proportional. However, the square has four right angles, whereas the slanting rhombus does not have four right angles. Therefore, corresponding angles are not equal.

The quadrilaterals are not similar.

Exercise 6.2

Solution:

Because DE is parallel to BC, BPT gives

AD/DB = AE/EC.

(i) Finding EC

1.5/3 = 1/EC. After simplifying, 1/2 = 1/EC. Therefore, EC = 2 cm.

(ii) Finding AD

AD/7.2 = 1.8/5.4 = 1/3. Therefore, AD = 7.2/3 = 2.4 cm.

(i) EC = 2 cm; (ii) AD = 2.4 cm.

Solution:

By the converse of BPT, EF is parallel to QR exactly when PE/EQ = PF/FR.

(i) No, (ii) Yes, (iii) Yes.

Solution:

Here M lies on AB, L lies on AC, and N lies on AD.

1. In triangle ABC, LM is parallel to CB. Hence, AM/AB = AL/AC.
2. In triangle ACD, LN is parallel to CD. Hence, AN/AD = AL/AC.
3. Both ratios equal AL/AC. Therefore, AM/AB = AN/AD.

Solution:

1. In triangle BAE, DF is parallel to AE. By BPT, BD/DA = BF/FE. (1)
2. In triangle BAC, DE is parallel to AC. By BPT, BD/DA = BE/EC. (2)
3. From (1) and (2), BF/FE = BE/EC.

Solution:

1. In triangle POQ, DE is parallel to OQ. Therefore, PD/DO = PE/EQ. (1)
2. In triangle POR, DF is parallel to OR. Therefore, PD/DO = PF/FR. (2)
3. From (1) and (2), PE/EQ = PF/FR.
4. Thus, E and F divide sides PQ and PR in the same ratio. By the converse of BPT, EF is parallel to QR.

Solution:

1. From AB is parallel to PQ in triangle OPQ, OB/OQ = OA/OP. (1)
2. From AC is parallel to PR in triangle OPR, OC/OR = OA/OP. (2)
3. Hence, OB/OQ = OC/OR. Using OQ = OB + BQ and OR = OC + CR, OB/(OB + BQ) = OC/(OC + CR).
4. Cross-multiplying and cancelling OB × OC, OB × CR = OC × BQ. Therefore, OB/BQ = OC/CR.
5. By the converse of BPT in triangle OQR, BC is parallel to QR.

Solution:

In triangle ABC, let D be the midpoint of AB. Through D, draw DE is parallel to BC, meeting AC at E.

1. Since D is the midpoint, AD = DB, so AD/DB = 1.
2. By BPT, AD/DB = AE/EC.
3. Therefore, AE/EC = 1, giving AE = EC.

Hence E is the midpoint of AC; the parallel line bisects the third side.

Solution:

In triangle ABC, let D and E be the midpoints of AB and AC, respectively.

AD = DB, AE = EC.

Therefore,

AD/DB = 1 = AE/EC.

The two sides are divided in the same ratio. By the converse of BPT,

DE is parallel to BC.

Solution:

Consider triangle AOB and triangle COD.

1. angle AOB = angle COD, because they are vertically opposite.
2. angle ABO = angle CDO, because AB is parallel to DC and BD is a transversal.
3. Hence triangle AOB is similar to triangle COD by AA similarity.
4. Corresponding sides are proportional: AO/CO = BO/DO.
5. Cross-multiplying, AO × DO = BO × CO. Dividing by BO × DO, AO/BO = CO/DO.

Solution:

Given

AO/BO = CO/DO.

Cross-multiplying gives

AO × DO = BO × CO. Therefore, AO/CO = BO/DO. (1)

Also, angle AOB = angle COD, being vertically opposite. Thus, in triangle AOB and triangle COD, the two sides including the equal angle are proportional. Therefore, triangle AOB is similar to triangle COD by SAS similarity.

Hence angle ABO = angle CDO. These are alternate interior angles made by transversal BD, so AB is parallel to CD.

Thus ABCD has one pair of opposite sides parallel and is a trapezium.

Exercise 6.3 — Similarity Criteria and Applications

Apply AA/AAA, SSS, or SAS carefully. Always write the vertices in the correct corresponding order.

Solution:

Solution: 

1. D,O,B are collinear, so angle DOC and angle BOC form a linear pair: angle DOC = 180° - 125° = 55°.
2. In triangle DOC, angle DCO = 180° - 70° - 55° = 55°.
3. From triangle ODC is similar to triangle OBA, the correspondence is O corresponds to O, D corresponds to B, C corresponds to A. Therefore, angle OAB = angle DCO = 55°.

angle DOC = 55°, angle DCO = 55°, and angle OAB = 55°.

Solution:

Compare triangle AOB and triangle COD.

1. angle AOB = angle COD, vertically opposite angles.
2. angle ABO = angle CDO, alternate interior angles because AB is parallel to DC.
3. Hence triangle AOB is similar to triangle COD by AA.
4. Therefore, corresponding sides are proportional: OA/OC = OB/OD.

Solution:

1. The marked equality is angle PQR = angle PRQ. Therefore, in triangle PQR, the sides opposite these equal angles are equal: PQ = PR. (1)
2. The given proportion is QR/QS = QT/PR. Using PR = PQ, QR/QS = QT/PQ. Therefore, PQ/QT = QS/QR. (2)
3. Since P,Q,T are collinear and Q,S,R are collinear, angle PQS = angle TQR. (3)
4. From (2) and (3), the two sides including the equal angle are proportional. Hence, triangle PQS is similar to triangle TQR by SAS similarity.

Solution: 

1. Given: angle RPQ = angle RTS.
2. Since S lies on PR and T lies on QR, rays RS and RP are the same, and rays RT and RQ are the same. Therefore, angle PRQ = angle TRS.
3. Two corresponding angles are equal, so triangle RPQ is similar to triangle RTS by AA similarity.

Solution:

From triangle ABE is congruent to triangle ACD, corresponding sides give

AB = AC, AE = AD.

Therefore,

AD/AB = AE/AC.

Also, angle DAE = angle BAC, because AD lies along AB and AE lies along AC. Thus, two pairs of sides including an equal angle are proportional.

triangle ADE is similar to triangle ABC, by SAS similarity.

Solution:

Since AD is perpendicular to BC and CE is perpendicular to AB, every angle formed at D by AD and BC, and every angle formed at E by CE and AB, is 90°.

(i) triangle AEP is similar to triangle CDP

• angle AEP = angle CDP = 90°.
• angle APE = angle CPD, vertically opposite.
• Therefore, the triangles are similar by AA.

(ii) triangle ABD is similar to triangle CBE

• angle ADB = angle CEB = 90°.
• angle ABD = angle CBE, both being the angle at B.
• Therefore, the triangles are similar by AA.

(iii) triangle AEP is similar to triangle ADB

• angle AEP = angle ADB = 90°.
• angle PAE = angle DAB, because AP lies on AD and AE lies on AB.
• Therefore, the triangles are similar by AA.

(iv) triangle PDC is similar to triangle BEC

• angle PDC = angle BEC = 90°.
• angle PCD = angle BCE, because CP lies on CE and CD lies on CB.
• Therefore, the triangles are similar by AA.

Solution:

In a parallelogram, AB is parallel to CD and AD is parallel to BC.

1. Since AE lies on AD, AE is parallel to BC. Hence angle AEB = angle CBF.
2. Since CF lies on CD, CF is parallel to AB. Hence angle ABE = angle CFB.
3. Therefore, triangle ABE is similar to triangle CFB by AA similarity.

Solution:

(i) Similarity

1. angle ABC = angle AMP = 90°.
2. angle BAC = angle PAM, because AB and AP are on one line, while AC and AM are on one line.
3. Hence, triangle ABC is similar to triangle AMP by AA similarity.

(ii) Required ratio

The correspondence is A corresponds to A, B corresponds to M, C corresponds to P. Therefore,

CA/PA = BC/MP.

Solution:

From triangle ABC is similar to triangle FEG, the correspondence is A corresponds to F, B corresponds to E, C corresponds to G. Thus,

angle A = angle F, angle B = angle E, angle C = angle G.

Since CD and GH are angle bisectors,

angle ACD = angle DCB = tfrac12angle C, angle FGH = angle HGE = tfrac12angle G.

Therefore, angle ACD = angle FGH and angle DCB = angle HGE.

(iii) First prove triangle DCA is similar to triangle HGF

• angle DCA = angle HGF.
• angle DAC = angle HFG, because angle A = angle F.
• Hence triangle DCA is similar to triangle HGF by AA.

(i) Use corresponding sides from the above similarity

CD/GH = AC/FG.

(ii) Prove triangle DCB is similar to triangle HGE

• angle DCB = angle HGE.
• angle DBC = angle HEG, because angle B = angle E.
• Hence triangle DCB is similar to triangle HGE by AA.

Solution:

1. AD is perpendicular to BC and EF is perpendicular to AC, so angle ADB = angle EFC = 90°.
2. Since AB = AC, the base angles of isosceles triangle ABC are equal: angle ABC = angle BCA.
3. Because D lies on BC, angle ABD = angle ABC. Because E,C,B are collinear and F lies on AC, angle ECF = angle BCA.
4. Therefore, angle ABD = angle ECF. Along with the right angles, triangle ABD is similar to triangle ECF by AA similarity.

Solution:

Let the common ratio be k. Since AD and PM are medians, D and M are midpoints:

BD = BC/2, QM = QR/2.

Therefore,

BD/QM = BC/2/QR/2 = BC/QR = k.

We already have AB/PQ = k and AD/PM = k. Hence the three sides of triangle ABD are proportional to the three sides of triangle PQM:

triangle ABD is similar to triangle PQM (SSS).

Thus angle ABD = angle PQM. Since BD lies on BC and QM lies on QR, angle ABC = angle PQR.

Also, AB/PQ = BC/QR. The sides including the equal angle are proportional, so

triangle ABC is similar to triangle PQR, by SAS similarity.

Solution:

Compare triangle ADC and triangle BAC.

1. angle ADC = angle BAC, given.
2. Since D lies on BC, angle ACD = angle BCA.
3. Hence triangle ADC is similar to triangle BAC by AA.
4. The correspondence gives AC/BC = CD/AC.
5. Cross-multiplying, AC² = BC × CD.

Solution:

Let

AB/PQ = AC/PR = AD/PM = k.

Since D and M are midpoints of BC and QR, BD = BC/2 and QM = QR/2.

Apply Apollonius' theorem to triangle ABC and triangle PQR:

AB² + AC² = 2(AD² + BD²), (1) PQ² + PR² = 2(PM² + QM²). (2)

Multiply (2) by k²:

k²PQ² + k²PR² = 2(k²PM² + k²QM²).

Using AB = kPQ, AC = kPR, and AD = kPM,

AB² + AC² = 2(AD² + k²QM²). (3)

Comparing (1) and (3),

BD² = k²QM². Therefore, BD = kQM.

Therefore,

BC = 2BD = 2kQM = kQR. Therefore, BC/QR = k.

Thus,

AB/PQ = AC/PR = BC/QR.

Hence triangle ABC is similar to triangle PQR, by SSS similarity.

Solution:

The pole and tower are vertical, and the sunlight makes the same angle with the ground at the same time. Therefore, the two right-angled triangles formed by height and shadow are similar.

Let the tower's height be h metres.

h/28 = 6/4.

h = 28 × 6/4 = 7 × 6 = 42.

The height of the tower is 42 m.

Solution:

Since triangle ABC is similar to triangle PQR,

AB/PQ = BC/QR. (1)

As AD and PM are medians, BD = BC/2 and QM = QR/2. Hence,

BD/QM = BC/2/QR/2 = BC/QR = AB/PQ. (2)

Also, corresponding angles of similar triangles are equal: angle ABC = angle PQR. Since D lies on BC and M lies on QR,

angle ABD = angle PQM.

In triangle ABD and triangle PQM, the two sides around the equal included angle are proportional by (2). Therefore, triangle ABD is similar to triangle PQM by SAS similarity.

Thus, corresponding sides give

AB/PQ = AD/PM.