NCERT Class 10 Maths – Exercise 6.3 Solutions

NCERT Class 10 Maths

Chapter 6 – Triangles | Exercise 6.3

(Rationalized Syllabus 2025-26)

💡 Similarity Criteria

AAA (or AA): Corresponding angles are equal.
SSS: Ratio of corresponding sides is equal.
SAS: One angle equal and including sides proportional.

State which pairs of triangles in Figure are similar. Write the similarity criterion used and write the pairs of similar triangles in symbolic form.

(i) $\Delta ABC$ and $\Delta PQR$

$\angle A = \angle P = 60^\circ$, $\angle B = \angle Q = 80^\circ$, $\angle C = \angle R = 40^\circ$.

✔️ $\Delta ABC \sim \Delta PQR$ (AAA Criterion)

(ii) $\Delta ABC$ and $\Delta QRP$

Ratios: $\frac{AB}{QR} = \frac{2}{4} = \frac{1}{2}$, $\frac{BC}{RP} = \frac{2.5}{5} = \frac{1}{2}$, $\frac{CA}{PQ} = \frac{3}{6} = \frac{1}{2}$.

✔️ $\Delta ABC \sim \Delta QRP$ (SSS Criterion)

(iii) $\Delta LMP$ and $\Delta DEF$

Ratios: $\frac{MP}{DE} = \frac{2}{4} = \frac{1}{2}$, $\frac{LP}{DF} = \frac{3}{6} = \frac{1}{2}$, but $\frac{LM}{EF} = \frac{2.7}{5} \neq \frac{1}{2}$.

✖️ Not Similar

(iv) $\Delta MNL$ and $\Delta QPR$

$\angle M = \angle Q = 70^\circ$.

Ratios of containing sides: $\frac{MN}{QP} = \frac{2.5}{5} = \frac{1}{2}$, $\frac{ML}{QR} = \frac{5}{10} = \frac{1}{2}$.

✔️ $\Delta MNL \sim \Delta QPR$ (SAS Criterion)

(v) $\Delta ABC$ and $\Delta FDE$

Given $\angle A = 80^\circ$, but side AC is not known to check proportionality with including angle.

In $\Delta FDE$, $\angle F = 80^\circ$. Proportional sides are not including the angle in the first triangle.

✖️ Not Similar

(vi) $\Delta DEF$ and $\Delta PQR$

In $\Delta DEF$, $\angle F = 180 – (70+80) = 30^\circ$.

In $\Delta PQR$, $\angle P = 180 – (80+30) = 70^\circ$.

So, $\angle D = \angle P$, $\angle E = \angle Q$, $\angle F = \angle R$.

✔️ $\Delta DEF \sim \Delta PQR$ (AAA Criterion)

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

1. $\angle DOC + \angle BOC = 180^\circ$ (Linear Pair)

$$\angle DOC + 125^\circ = 180^\circ \Rightarrow \angle DOC = 55^\circ$$

2. In $\Delta ODC$, sum of angles is $180^\circ$:

$$70^\circ + 55^\circ + \angle DCO = 180^\circ$$

$$125^\circ + \angle DCO = 180^\circ \Rightarrow \angle DCO = 55^\circ$$

3. Since $\Delta ODC \sim \Delta OBA$, corresponding angles are equal:

$$\angle OAB = \angle OCD \Rightarrow \angle OAB = 55^\circ$$

✔️ Angles are 55°, 55°, 55°

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

In $\Delta DOC$ and $\Delta BOA$:

1. $\angle CDO = \angle ABO$ (Alt. Interior Angles, DC || AB)

2. $\angle DCO = \angle BAO$ (Alt. Interior Angles)

3. $\angle DOC = \angle BOA$ (Vertically Opposite Angles)

$$\therefore \Delta DOC \sim \Delta BOA \quad \text{(AAA Similarity)}$$

Corresponding sides are proportional:

$$\frac{DO}{BO} = \frac{OC}{OA} \Rightarrow \frac{OA}{OC} = \frac{OB}{OD}$$

✔️ Hence Proved

In the figure, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\Delta PQS \sim \Delta TQR$.

Given: $\angle 1 = \angle 2 \Rightarrow PQ = PR$ (Sides opposite to equal angles in $\Delta PQR$).

Substitute $PR = PQ$ in the given ratio:

$$\frac{QR}{QS} = \frac{QT}{PQ}$$

Taking reciprocal: $\frac{QS}{QR} = \frac{PQ}{QT}$

In $\Delta PQS$ and $\Delta TQR$:

1. $\frac{PQ}{QT} = \frac{QS}{QR}$ (Proved above)

2. $\angle PQS = \angle TQR = \angle 1$ (Common Angle)

✔️ $\Delta PQS \sim \Delta TQR$ (SAS Similarity)

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

In $\Delta RPQ$ and $\Delta RTS$:

1. $\angle RPQ = \angle RTS$ (Given)

2. $\angle R = \angle R$ (Common Angle)

✔️ $\Delta RPQ \sim \Delta RTS$ (AA Similarity)

In figure, if $\Delta ABE \cong \Delta ACD$, show that $\Delta ADE \sim \Delta ABC$.

Since $\Delta ABE \cong \Delta ACD$ (CPCT):

$$AB = AC \quad \text{and} \quad AE = AD$$

Therefore:

$$\frac{AD}{AB} = \frac{AE}{AC}$$

In $\Delta ADE$ and $\Delta ABC$:

1. $\frac{AD}{AB} = \frac{AE}{AC}$ (Proved above)

2. $\angle A = \angle A$ (Common)

✔️ $\Delta ADE \sim \Delta ABC$ (SAS Similarity)

In figure, altitudes AD and CE of $\Delta ABC$ intersect each other at the point P. Show that:

(i) $\Delta AEP \sim \Delta CDP$

$\angle AEP = \angle CDP = 90^\circ$

$\angle APE = \angle CPD$ (Vertically Opposite)

✔️ AA Similarity

(ii) $\Delta ABD \sim \Delta CBE$

$\angle ADB = \angle CEB = 90^\circ$

$\angle B = \angle B$ (Common)

✔️ AA Similarity

(iii) $\Delta AEP \sim \Delta ADB$

$\angle AEP = \angle ADB = 90^\circ$

$\angle A = \angle A$ (Common)

✔️ AA Similarity

(iv) $\Delta PDC \sim \Delta BEC$

$\angle PDC = \angle BEC = 90^\circ$

$\angle C = \angle C$ (Common)

✔️ AA Similarity

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

In $\Delta ABE$ and $\Delta CFB$:

1. $\angle A = \angle C$ (Opposite angles of parallelogram)

2. $\angle AEB = \angle CBF$ (Alt. Interior Angles, AE || BC)

✔️ $\Delta ABE \sim \Delta CFB$ (AA Similarity)

In figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) $\Delta ABC \sim \Delta AMP$

1. $\angle ABC = \angle AMP = 90^\circ$

2. $\angle A = \angle A$ (Common)

✔️ AA Similarity

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

Since $\Delta ABC \sim \Delta AMP$, corresponding sides are proportional.

✔️ Hence Proved

Q10

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 $\Delta ABC$ and $\Delta EFG$ respectively. If $\Delta ABC \sim \Delta FEG$, show that:

(i) $\frac{CD}{GH} = \frac{AC}{FG}$

In $\Delta ACD$ and $\Delta FGH$:

1. $\angle A = \angle F$ (Given similar triangles)

2. $\angle ACD = \angle FGH$ (Halves of equal angles C and G)

So $\Delta ACD \sim \Delta FGH$ (AA). Sides are proportional.

✔️ Proved

(ii) $\Delta DCB \sim \Delta HGE$

1. $\angle B = \angle E$

2. $\angle BCD = \angle EGH$ (Halves of equal angles)

✔️ AA Similarity

(iii) $\Delta DCA \sim \Delta HGF$

Proved in part (i) logic.

✔️ AA Similarity

Q11-Q13

Additional Problems

Q11. In an isosceles triangle ABC (AB=AC), E is on BC produced. AD $\perp$ BC, EF $\perp$ AC. Prove $\Delta ABD \sim \Delta ECF$.

1. $\angle ADB = \angle EFC = 90^\circ$

2. $\angle B = \angle C$ (Angles opposite to equal sides)

In $\Delta ECF$, angle is actually $\angle ECF$ which is same as $\angle C$.

✔️ AA Similarity

Q12. Sides AB, BC and median AD are proportional to PQ, QR and median PM. Prove $\Delta ABC \sim \Delta PQR$.

Given $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$.

Since AD, PM are medians, $BC = 2BD$ and $QR = 2QM$.

Ratio becomes $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM} \Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$.

So $\Delta ABD \sim \Delta PQM$ (SSS). Thus $\angle B = \angle Q$.

Now in main triangles: $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$.

✔️ SAS Similarity

Q13. D is a point on BC such that $\angle ADC = \angle BAC$. Show $CA^2 = CB \cdot CD$.

In $\Delta ADC$ and $\Delta BAC$:

1. $\angle ADC = \angle BAC$ (Given)

2. $\angle C = \angle C$ (Common)

$\Delta ADC \sim \Delta BAC$ (AA).

$$\frac{CA}{CB} = \frac{CD}{CA} \Rightarrow CA^2 = CB \cdot CD$$

✔️ Proved

Q14-Q16

Advanced Problems

Q14. Sides AB, AC and median AD are proportional to PQ, PR and median PM. Prove $\Delta ABC \sim \Delta PQR$.

Construction: Extend AD to E such that AD=DE. Extend PM to L such that PM=ML. Join diagonals to form parallelograms ABEC and PQLR.

Since diagonals bisect, ABEC and PQLR are parallelograms.

So $AC = BE$ and $PR = QL$.

Given: $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$.

Substitute: $\frac{AB}{PQ} = \frac{BE}{QL} = \frac{2AD}{2PM} = \frac{AE}{PL}$.

By SSS, $\Delta ABE \sim \Delta PQL$. So $\angle BAE = \angle QPL$.

Similarly, $\angle CAE = \angle RPL$.

Adding them, $\angle BAC = \angle QPR$.

By SAS (Sides proportional and included angle equal):

✔️ $\Delta ABC \sim \Delta PQR$

Q15. A vertical pole of length 6 m casts a shadow 4 m long. At the same time, a tower casts a shadow 28 m long. Find height of tower.

Triangles formed by pole/tower and shadows are similar (Sun’s elevation is same).

$$\frac{\text{Height of Pole}}{\text{Shadow of Pole}} = \frac{\text{Height of Tower}}{\text{Shadow of Tower}}$$

$$\frac{6}{4} = \frac{h}{28} \Rightarrow h = \frac{6 \times 28}{4} = 6 \times 7 = 42$$

✔️ Height = 42 m

Q16. If AD and PM are medians of similar triangles ABC and PQR, prove $\frac{AB}{PQ} = \frac{AD}{PM}$.

Since $\Delta ABC \sim \Delta PQR$, $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$.

Since medians, $BC = 2BD$ and $QR = 2QM$.

So $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM}$.

By SAS, $\Delta ABD \sim \Delta PQM$.

Corresponding sides: $\frac{AB}{PQ} = \frac{AD}{PM}$.

✔️ Proved

🎉 Exercise 6.3 Completed | Chapter 6 Finished!