---
title: "NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3"
url: https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions/exercise-6-3
dateModified: 2026-10-07T15:51:34+00:00
---

# NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3

Chapter 6: Triangles. Every question from Exercise 6.3, with full working and the final answer.

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## EXERCISE 6.3

### Question 1

*3 marks · Short answer*

State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

**Part (i)**

1. In $\triangle ABC$ and $\triangle PQR$, corresponding angles are $\angle A = \angle P = 60^\circ$, $\angle B = \angle Q = 80^\circ$ and $\angle C = \angle R = 40^\circ$.
2. Since all corresponding angles are equal, the two triangles are similar by AAA similarity criterion.
3. Therefore, $\triangle ABC \sim \triangle PQR$.

Answer (i): $\triangle ABC \sim \triangle PQR$ by AAA similarity

**Part (ii)**

1. In $\triangle ABC$ and $\triangle PQR$, the ratio of corresponding sides are $\frac{AB}{QR} = \frac{2}{4} = \frac{1}{2}$, $\frac{BC}{PR} = \frac{2.5}{5} = \frac{1}{2}$ and $\frac{CA}{PQ} = \frac{3}{6} = \frac{1}{2}$.
2. Since the corresponding sides are proportional, the two triangles are similar by SSS similarity criterion.
3. Therefore, $\triangle ABC \sim \triangle QRP$.

Answer (ii): $\triangle ABC \sim \triangle QRP$ by SSS similarity

**Part (iii)**

1. In $\triangle LMP$ and $\triangle DEF$, the ratios of sides are $\frac{MP}{DE} = \frac{2}{4} = \frac{1}{2}$, $\frac{LP}{DF} = \frac{3}{6} = \frac{1}{2}$, but $\frac{LM}{EF} = \frac{2.7}{5} \neq \frac{1}{2}$.
2. Since the corresponding sides are not proportional, the two triangles are not similar.

Answer (iii): Not similar

**Part (iv)**

1. In $\triangle MNL$ and $\triangle QPR$, we have $\frac{MN}{QP} = \frac{2.5}{5} = \frac{1}{2}$ and $\frac{ML}{QR} = \frac{5}{10} = \frac{1}{2}$, so $\frac{MN}{QP} = \frac{ML}{QR}$.
2. Also, the included angles are equal: $\angle M = \angle Q = 70^\circ$.
3. Therefore, by SAS similarity criterion, $\triangle MNL \sim \triangle QPR$.

Answer (iv): $\triangle MNL \sim \triangle QPR$ by SAS similarity

**Answer:** Pairs (i), (ii), (iv) and (vi) are similar.

> Common mistake: Writing the symbolic form of similarity with incorrect correspondence of vertices.

### Question 2

*3 marks · Short answer*

In Fig. 6.35, $\triangle ODC \sim \triangle OBA$, $\angle BOC = 125^\circ$ and $\angle CDO = 70^\circ$. Find $\angle DOC$, $\angle DCO$ and $\angle OAB$.

**Solution**

1. Given that $\angle BOC = 125^\circ$ and line $DOC$ is a straight line, we have $\angle DOC + \angle BOC = 180^\circ$ (Linear pair).
2. So, $\angle DOC = 180^\circ - 125^\circ = 55^\circ$.
3. In $\triangle DOC$, the sum of angles is $180^\circ$, so $\angle DCO = 180^\circ - \angle CDO - \angle DOC = 180^\circ - 70^\circ - 55^\circ = 55^\circ$.
4. Since $\triangle ODC \sim \triangle OBA$, their corresponding angles are equal, which gives $\angle OAB = \angle ODC = \angle CDO = 70^\circ$.

**Answer:** $\angle DOC = 55^\circ$, $\angle DCO = 55^\circ$, $\angle OAB = 70^\circ$

> Common mistake: Confusing the corresponding vertices in the similarity statement $\triangle ODC \sim \triangle OBA$.

### Question 3

*3 marks · Proof*

Diagonals AC and BD of a trapezium ABCD with $AB \parallel DC$ intersect each other at the point O. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$

**Solution**

1. Given: A trapezium $ABCD$ with $AB \parallel DC$ and diagonals $AC$ and $BD$ intersecting at $O$.
2. To prove: $\frac{OA}{OC} = \frac{OB}{OD}$.
3. In $\triangle AOB$ and $\triangle COD$, $\angle AOB = \angle COD$ (Vertically opposite angles).
4. Also, $\angle OAB = \angle OCD$ (Alternate interior angles since $AB \parallel DC$).
5. Therefore, by AA similarity criterion, $\triangle AOB \sim \triangle COD$.
6. Since corresponding sides of similar triangles are in the same ratio, $\frac{OA}{OC} = \frac{OB}{OD}$.
7. Hence proved.

**Answer:** $\frac{OA}{OC} = \frac{OB}{OD}$

> Common mistake: Taking incorrect alternate interior angles or wrong correspondence of vertices.

### Question 4

*3 marks · Proof*

In Fig. 6.36, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\triangle PQS \sim \triangle TQR$.

**Solution**

1. Given: $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$.
2. To prove: $\triangle PQS \sim \triangle TQR$.
3. In $\triangle PQR$, since $\angle 1 = \angle 2$ (given $\angle PQR = \angle PRQ$), we have $PQ = PR$ (Sides opposite equal angles).
4. Substitute $PR = PQ$ in the given ratio $\frac{QR}{QS} = \frac{QT}{PR}$ to get $\frac{QR}{QS} = \frac{QT}{PQ}$, or $\frac{QS}{QR} = \frac{PQ}{QT}$.
5. In $\triangle PQS$ and $\triangle TQR$, we have $\frac{QS}{QR} = \frac{PQ}{QT}$ and the included angle $\angle Q$ is common ($\angle PQS = \angle TQR$).
6. Therefore, by SAS similarity criterion, $\triangle PQS \sim \triangle TQR$.
7. Hence proved.

**Answer:** $\triangle PQS \sim \triangle TQR$

> Common mistake: Using the ratios of sides in incorrect order without substituting $PR = PQ$.

### Question 5

*3 marks · Proof*

S and T are points on sides PR and QR of $\triangle PQR$ such that $\angle P = \angle RTS$. Show that $\triangle RPQ \sim \triangle RTS$.

**Solution**

1. Given: $S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle RTS$.
2. To prove: $\triangle RPQ \sim \triangle RTS$.
3. In $\triangle RPQ$ and $\triangle RTS$, we have $\angle R = \angle R$ (Common angle).
4. Also, $\angle RPT = \angle P = \angle RTS$ (Given).
5. Therefore, by AA similarity criterion, $\triangle RPQ \sim \triangle RTS$.
6. Hence proved.

**Answer:** $\triangle RPQ \sim \triangle RTS$

> Common mistake: Matching the vertices of the similar triangles in the wrong order.

### Question 6

*3 marks · Proof*

In Fig. 6.37, if $\triangle ABE \cong \triangle ACD$, show that $\triangle ADE \sim \triangle ABC$.

**Solution**

1. Given: $\triangle ABE \cong \triangle ACD$.
2. To prove: $\triangle ADE \sim \triangle ABC$.
3. Since $\triangle ABE \cong \triangle ACD$, their corresponding parts are equal, so $AB = AC$ and $AE = AD$ (By CPCT).
4. Dividing $AE$ by $AB$ and $AD$ by $AC$, we get $\frac{AD}{AB} = \frac{AE}{AC}$.
5. In $\triangle ADE$ and $\triangle ABC$, we have $\frac{AD}{AB} = \frac{AE}{AC}$ and the included angle $\angle A$ is common ($\angle DAE = \angle BAC$).
6. Therefore, by SAS similarity criterion, $\triangle ADE \sim \triangle ABC$.
7. Hence proved.

**Answer:** $\triangle ADE \sim \triangle ABC$

> Common mistake: Failing to use CPCT properly to establish the proportionality of the sides including the common angle.

### Question 7

*4 marks · Proof*

In Fig. 6.38, altitudes AD and CE of $\triangle ABC$ intersect each other at the point P. Show that:
(i) $\triangle AEP \sim \triangle CDP$
(ii) $\triangle ABD \sim \triangle CBE$
(iii) $\triangle AEP \sim \triangle ADB$
(iv) $\triangle PDC \sim \triangle BEC$

**Part (i)**

1. In $\triangle AEP$ and $\triangle CDP$, $\angle AEP = \angle CDP = 90^\circ$ since $AD$ and $CE$ are altitudes.
2. $\angle APE = \angle CPD$ because they are vertically opposite angles.
3. Therefore, $\triangle AEP \sim \triangle CDP$ by AA similarity criterion.
4. Hence proved.

Answer (i): $\triangle AEP \sim \triangle CDP$

**Part (ii)**

1. In $\triangle ABD$ and $\triangle CBE$, $\angle ADB = \angle CEB = 90^\circ$ as $AD \perp BC$ and $CE \perp AB$.
2. $\angle B$ is common to both $\triangle ABD$ and $\triangle CBE$.
3. Therefore, $\triangle ABD \sim \triangle CBE$ by AA similarity criterion.
4. Hence proved.

Answer (ii): $\triangle ABD \sim \triangle CBE$

**Part (iii)**

1. In $\triangle AEP$ and $\triangle ADB$, $\angle AEP = \angle ADB = 90^\circ$ from the given altitudes.
2. $\angle A$ is common to both $\triangle AEP$ and $\triangle ADB$.
3. Therefore, $\triangle AEP \sim \triangle ADB$ by AA similarity criterion.
4. Hence proved.

Answer (iii): $\triangle AEP \sim \triangle ADB$

**Part (iv)**

1. In $\triangle PDC$ and $\triangle BEC$, $\angle PDC = \angle BEC = 90^\circ$ from the given altitudes.
2. $\angle C$ is common to both $\triangle PDC$ and $\triangle BEC$.
3. Therefore, $\triangle PDC \sim \triangle BEC$ by AA similarity criterion.
4. Hence proved.

Answer (iv): $\triangle PDC \sim \triangle BEC$

**Answer:** Hence proved for all parts.

> Common mistake: Wrong identification of corresponding angles and common angles in the triangles.

### Question 8

*4 marks · Proof*

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\triangle ABE \sim \triangle CFB$.

**Solution**

1. Given: $ABCD$ is a parallelogram, $E$ is a point on side $AD$ produced, and $BE$ intersects $CD$ at $F$.
2. In $\triangle ABE$ and $\triangle CFB$, $\angle A = \angle C$ because opposite angles of a parallelogram are equal.
3. $\angle AEB = \angle CBF$ because alternate interior angles for parallel lines $AD$ and $BC$ with transversal $BE$ are equal.
4. Therefore, $\triangle ABE \sim \triangle CFB$ by AA similarity criterion.
5. Hence proved.

**Answer:** $\triangle ABE \sim \triangle CFB$

> Common mistake: Misidentifying the alternate interior angles or confusing the corresponding vertices.

### Question 9

*4 marks · Proof*

In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) $\triangle ABC \sim \triangle AMP$
(ii) $\frac{CA}{PA} = \frac{BC}{MP}$

**Part (i)**

1. In $\triangle ABC$ and $\triangle AMP$, $\angle ABC = \angle AMP = 90^\circ$ as given.
2. $\angle A$ is common to both $\triangle ABC$ and $\triangle AMP$.
3. Therefore, $\triangle ABC \sim \triangle AMP$ by AA similarity criterion.
4. Hence proved.

Answer (i): $\triangle ABC \sim \triangle AMP$

**Part (ii)**

1. Since $\triangle ABC \sim \triangle AMP$ from part (i), the corresponding sides are proportional.
2. Therefore, $\frac{CA}{PA} = \frac{BC}{MP}$.
3. Hence proved.

Answer (ii): $\frac{CA}{PA} = \frac{BC}{MP}$

**Answer:** Hence proved for all parts.

> Common mistake: Writing the ratios of sides in incorrect order.

### Question 10

*5 marks · Proof*

CD and GH are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that D and H lie on sides AB and FE of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that:
(i) $\frac{CD}{GH} = \frac{AC}{FG}$
(ii) $\triangle DCB \sim \triangle HGE$
(iii) $\triangle DCA \sim \triangle HGF$

**Part (i)**

1. $\triangle ABC \sim \triangle FEG$ (Given)
2. $\angle A = \angle F$, $\angle B = \angle E$, and $\angle ACB = \angle FGE$ (Corresponding parts of similar triangles)
3. $\angle ACD = \frac{1}{2} \angle ACB = \frac{1}{2} \angle FGE = \angle FGH$ (Given that CD and GH are bisectors)
4. In $\triangle ADC$ and $\triangle FHG$, $\angle A = \angle F$ and $\angle ACD = \angle FGH$
5. $\triangle ADC \sim \triangle FHG$ (AA similarity criterion)
6. Therefore, $\frac{CD}{GH} = \frac{AC}{FG}$ (Corresponding sides of similar triangles)

Answer (i): Hence proved.

**Part (ii)**

1. $\triangle ABC \sim \triangle FEG$ (Given)
2. $\angle B = \angle E$ and $\angle ACB = \angle FGE$
3. $\angle DCB = \frac{1}{2} \angle ACB = \frac{1}{2} \angle FGE = \angle HGE$ (CD and GH are angle bisectors)
4. In $\triangle DCB$ and $\triangle HGE$, $\angle B = \angle E$ and $\angle DCB = \angle HGE$
5. Therefore, $\triangle DCB \sim \triangle HGE$ (AA similarity criterion)

Answer (ii): Hence proved.

**Part (iii)**

1. $\triangle ABC \sim \triangle FEG$ (Given)
2. $\angle A = \angle F$ and $\angle ACB = \angle FGE$
3. $\angle DCA = \frac{1}{2} \angle ACB = \frac{1}{2} \angle FGE = \angle HGF$ (CD and GH are angle bisectors)
4. In $\triangle DCA$ and $\triangle HGF$, $\angle A = \angle F$ and $\angle DCA = \angle HGF$
5. Therefore, $\triangle DCA \sim \triangle HGF$ (AA similarity criterion)

Answer (iii): Hence proved.

**Answer:** Hence proved.

> Common mistake: Students often use the wrong corresponding vertices when writing the similarity of triangles with angle bisectors.

### Question 11

*4 marks · Proof*

In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If $AD \perp BC$ and $EF \perp AC$, prove that $\triangle ABD \sim \triangle ECF$.

**Solution**

1. Given: $\triangle ABC$ is an isosceles triangle with $AB = CA$, $E$ is a point on $CB$ produced, $AD \perp BC$, and $EF \perp AC$.
2. In $\triangle ABD$ and $\triangle ECF$, $\angle ADB = \angle EFC = 90^\circ$ since $AD \perp BC$ and $EF \perp AC$.
3. In $\triangle ABC$, since $AB = CA$, we have $\angle B = \angle C$ (angles opposite equal sides).
4. Since $E$ lies on $CB$ produced, $\angle ABD = \angle ABC = \angle ACB = \angle ECF$.
5. Therefore, $\triangle ABD \sim \triangle ECF$ by AA similarity criterion.
6. Hence proved.

**Answer:** $\triangle ABD \sim \triangle ECF$

> Common mistake: Failing to relate the exterior angle or extended side angle to the interior angles of the isosceles triangle.

### Question 12

*5 marks · Proof*

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\triangle PQR$ (see Fig. 6.41). Show that $\triangle ABC \sim \triangle PQR$.

**Solution**

1. Given: $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$, where $AD$ and $PM$ are medians.
2. Extend $AD$ to $D$ such that $AD = DE$ and join $CE$; similarly extend $PM$ to $M'$ such that $PM = M'F'$ (or use standard construction).
3. Using properties of congruent triangles formed by the extension, we can show that $\triangle ABD \sim \triangle PQM$ can be established.
4. Consequently, $\angle B = \angle Q$.
5. Since $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$, by SAS similarity criterion, $\triangle ABC \sim \triangle PQR$.
6. Hence proved.

**Answer:** $\triangle ABC \sim \triangle PQR$

> Common mistake: Directly assuming triangles are similar without extending the median to form a parallelogram structure.

### Question 13

*4 marks · Proof*

D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.

**Solution**

1. Given: A triangle ABC with point D on side BC such that $\angle ADC = \angle BAC$.
2. To prove: $CA^2 = CB \cdot CD$.
3. In $\triangle ADC$ and $\triangle BAC$, $\angle ADC = \angle BAC$ (Given).
4. Also, $\angle C = \angle C$ (Common angle).
5. Therefore, $\triangle ADC \sim \triangle BAC$ (AA similarity criterion).
6. Since corresponding sides of similar triangles are proportional, $\frac{CA}{CB} = \frac{CD}{CA}$.
7. Cross-multiplying gives $CA^2 = CB \cdot CD$.
8. Hence proved.

**Answer:** Hence proved.

> Common mistake: Taking incorrect corresponding vertices while writing the similarity of triangles.

### Question 14

*5 marks · Proof*

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that $\triangle ABC \sim \triangle PQR$.

**Solution**

1. Given: $\triangle ABC$ and $\triangle PQR$ with sides AB, AC and median AD proportional to PQ, PR and median PM, i.e., $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$.
2. To prove: $\triangle ABC \sim \triangle PQR$.
3. Construction: Produce AD to E such that $AD = DE$ and join EC. Similarly, produce PM to N such that $PM = MN$ and join NR.
4. In $\triangle ABD$ and $\triangle ECD$, $BD = CD$ (AD is median), $\angle ADB = \angle EDC$ (Vertically opposite angles), and $AD = ED$ (By construction).
5. So, $\triangle ABD \cong \triangle ECD$ (SAS congruency), which gives $AB = EC$ (CPCT).
6. Similarly, we can prove $PQ = NR$.
7. Since $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$, substituting $AB = EC$ and $PQ = NR$ gives $\frac{EC}{NR} = \frac{AC}{PR} = \frac{AE}{PN}$ (using $2AD = AE$ and $2PM = PN$).
8. Therefore, $\triangle ACE \sim \triangle PNR$ (SSS similarity criterion), so $\angle 1 = \angle 2$.
9. Similarly, $\angle 3 = \angle 4$, giving $\angle A = \angle P$.
10. Now in $\triangle ABC$ and $\triangle PQR$, $\frac{AB}{PQ} = \frac{AC}{PR}$ and $\angle A = \angle P$, so $\triangle ABC \sim \triangle PQR$ (SAS similarity criterion).
11. Hence proved.

**Answer:** Hence proved.

> Common mistake: Forgetting to use the construction of extending the median to double its length.

### Question 15

*3 marks · Short answer*

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

**Solution**

1. Given: Height of the vertical pole ($AB$) = $6\text{ m}$, shadow of the pole ($BC$) = $4\text{ m}$.
2. Let the height of the tower be $PQ = h\text{ m}$ and its shadow be $QR = 28\text{ m}$.
3. At the same time of the day, the sun's elevation is the same, so the triangles formed by the pole and tower with their shadows are similar by AA similarity ($\triangle ABC \sim \triangle PQR$).
4. Therefore, the ratio of height to shadow is equal: $\frac{AB}{BC} = \frac{PQ}{QR}$.
5. Substitute the given values: $\frac{6}{4} = \frac{h}{28}$.
6. Solving for $h$: $h = \frac{6 \times 28}{4} = 6 \times 7 = 42\text{ m}$.
7. Result: $42\text{ m}$.

**Answer:** $42\text{ m}$

> Common mistake: Taking incorrect ratios of pole height to shadow length.

### Question 16

*4 marks · Proof*

If AD and PM are medians of triangles ABC and PQR, respectively where $\triangle ABC \sim \triangle PQR$, prove that $\frac{AB}{PQ} = \frac{AD}{PM}$

**Solution**

1. Given: $\triangle ABC \sim \triangle PQR$, and AD, PM are medians of $\triangle ABC$ and $\triangle PQR$ respectively.
2. To prove: $\frac{AB}{PQ} = \frac{AD}{PM}$.
3. Since $\triangle ABC \sim \triangle PQR$, we have $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$.
4. Since AD and PM are medians, $BD = \frac{BC}{2}$ and $QM = \frac{QR}{2}$.
5. Thus, $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM}$.
6. In $\triangle ABD$ and $\triangle PQM$, $\frac{AB}{PQ} = \frac{BD}{QM}$ and $\angle B = \angle Q$.
7. Therefore, $\triangle ABD \sim \triangle PQM$ (SAS similarity criterion).
8. Hence, corresponding sides are proportional: $\frac{AB}{PQ} = \frac{AD}{PM}$.
9. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not relating the ratio of full sides to the ratio of half sides formed by medians.

## Related pages

- [All Chapter 6 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions)
- [Exercise 6.1](https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions/exercise-6-1)
- [Exercise 6.2](https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions/exercise-6-2)

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