NCERT Exemplar Class 10 Maths Chapter 6 Triangles Solutions (CBSE 2026-27)

Angles in the same segment of a circle are equal: \(\angle OCA = \angle OBA\) and \(\angle OAC = \angle ODB\). Also, \(\angle AOC = \angle DOB\) (vertically opposite angles).

By AAA similarity criterion, \(\Delta OAC \sim \Delta ODB\).

By the intersecting chords theorem, \(OA \cdot OB = OC \cdot OD\). Since \(OB = OD\), we have \(OA = OC\). Thus, \(\Delta OAC\) is isosceles (\(OA = OC\)) and \(\Delta ODB\) is isosceles (\(OB = OD\)). Hence, the triangles are isosceles and similar.

Since \(DE \parallel BC\), \(\Delta ADE \sim \Delta ABC\) (by AA similarity):

\[\frac{DE}{BC} = \frac{AD}{AB} = \frac{AD}{AD + BD}\] \[\frac{DE}{7.5} = \frac{2}{2 + 3} = \frac{2}{5} \implies DE = \frac{2 \times 7.5}{5} = 3\text{ cm}\]

In right-angled \(\Delta BAC\) with \(AD \perp BC\), \(\Delta DBA \sim \Delta DAC\):

\[\angle BDA = \angle ADC = 90^\circ \quad \text{and} \quad \angle BAD = 90^\circ – \angle B = \angle C\]

By AA similarity, \(\Delta DBA \sim \Delta DAC\). Taking corresponding ratios:

\[\frac{BD}{AD} = \frac{AD}{CD} \implies BD \cdot CD = AD^2\]

Since \(\Delta ABC \sim \Delta EDF\), corresponding sides are proportional:

\[\frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF}\]

• From \(\frac{BC}{DF} = \frac{AC}{EF} \implies BC \cdot EF = AC \cdot DF\) (Option A is true).

• From \(\frac{AB}{ED} = \frac{AC}{EF} \implies AB \cdot EF = AC \cdot DE\) (Option B is true).

• From \(\frac{AB}{ED} = \frac{BC}{DF} \implies BC \cdot DE = AB \cdot DF\) (Option D is true).

Therefore, \(BC \cdot DE = AB \cdot EF\) is not true.

From \(\frac{AB}{QR} = \frac{BC}{RP} = \frac{CA}{PQ}\), matching vertices gives: \(A \leftrightarrow Q\), \(B \leftrightarrow R\), \(C \leftrightarrow P\). Reversing the mapping: \(P \leftrightarrow C\), \(Q \leftrightarrow A\), \(R \leftrightarrow B\).

Hence, \(\Delta PQR \sim \Delta CAB\).

In \(\Delta APB\) and \(\Delta DPC\):

\[\frac{AP}{PD} = \frac{6}{5} = 1.2, \quad \frac{BP}{PC} = \frac{3}{2.5} = 1.2 \implies \frac{AP}{PD} = \frac{BP}{PC}\]

\(\angle APB = \angle DPC = 50^\circ\) (vertically opposite angles). By SAS similarity criterion, \(\Delta APB \sim \Delta DPC\).

Therefore, corresponding angles are equal: \(\angle A = \angle D = 30^\circ\).

In \(\Delta APB\), by angle sum property:

\[\angle PBA = 180^\circ – (\angle APB + \angle A) = 180^\circ – (50^\circ + 30^\circ) = 100^\circ\]

Given \(\angle D = \angle Q\) and \(\angle E = \angle R\), by AA similarity criterion: \(\Delta DEF \sim \Delta QRP\).

\[\frac{DE}{QR} = \frac{EF}{RP} = \frac{DF}{QP}\]

Option (B) states \(\frac{DE}{PQ} = \frac{EF}{RP}\), which incorrectly pairs \(DE\) with \(PQ\) instead of \(QR\). Hence, (B) is not true.

Since two pairs of corresponding angles are equal (\(\angle B = \angle E\) and \(\angle C = \angle F\)), the triangles are similar by AA similarity.

However, \(AB = 3\,DE \neq DE\), meaning corresponding sides are not equal. Hence, the triangles are similar but not congruent.

In \(\Delta ABC\), the included angle between sides \(AB\) and \(BC\) is \(\angle B\).

In \(\Delta DEF\), the included angle between sides \(DE\) and \(FD\) is \(\angle D\).

For SAS similarity criterion to hold, the included angles must be equal: \(\mathbf{\angle B = \angle D}\).

\[PE = DE – DP = 15 – 5 = 10\text{ cm}\] \[\frac{DP}{PE} = \frac{5}{10} = \frac{1}{2}\] \[\frac{DQ}{QF} = \frac{6}{18} = \frac{1}{3}\]

Since \(\frac{DP}{PE} \neq \frac{DQ}{QF}\), by the converse of Basic Proportionality Theorem, \(PQ\) is not parallel to \(EF\).

The correspondence of vertices is \(F \leftrightarrow S\), \(E \leftrightarrow T\), and \(D \leftrightarrow U\). The correct ratios of corresponding sides are:

\[\frac{FE}{ST} = \frac{ED}{TU} = \frac{FD}{SU}\]

Hence, \(\frac{DE}{ST} = \frac{EF}{TU}\) is false.

From the similarity \(\Delta DEF \sim \Delta RPQ\), the correspondence of vertices is \(D \leftrightarrow R\), \(E \leftrightarrow P\), and \(F \leftrightarrow Q\). Therefore, the correct corresponding angles are:

\[\angle D = \angle R, \quad \angle E = \angle P, \quad \angle F = \angle Q\]

Hence, \(\angle D = \angle R\) is true, but \(\angle F = \angle P\) is false (since \(\angle F = \angle Q\)).

\[AQ = PQ – PA = 12.5 – 5 = 7.5\text{ cm}\] \[\frac{PA}{AQ} = \frac{5}{7.5} = \frac{50}{75} = \frac{2}{3}\] \[\frac{PB}{BR} = \frac{4}{6} = \frac{2}{3}\]

Since \(\frac{PA}{AQ} = \frac{PB}{BR} = \frac{2}{3}\), by the converse of Basic Proportionality Theorem, \(AB \parallel QR\).

\[\frac{PB}{PD} = \frac{5}{10} = \frac{1}{2}, \quad \frac{PC}{PE} = \frac{6}{12} = \frac{1}{2} \implies \frac{PB}{PD} = \frac{PC}{PE}\]

\(\angle BPC = \angle DPE\) (vertically opposite angles). Therefore, by SAS similarity criterion:

\[\Delta PBC \sim \Delta PDE \text{ is not correct; the correct order is } \Delta PBC \sim \Delta PDB\]

Comparing corresponding sides: \(\frac{PB}{PD} = \frac{PC}{PE}\) with included angle \(\angle BPC = \angle DPE\) gives \(\Delta PBC \sim \Delta PDE\) only if vertices correspond correctly as \(\Delta PBC \sim \Delta PDE\), but here side \(PB\) corresponds to \(PD\) and \(PC\) to \(PE\), so \(\Delta PBC \sim \Delta PDE\) is false (it is \(\Delta PBC \sim \Delta PDE\) under \(B \leftrightarrow D, C \leftrightarrow E\), which gives \(\Delta PBC \sim \Delta PDE\) incorrect naming).

In \(\Delta PQR\):

\[\angle R = 180^\circ – (55^\circ + 25^\circ) = 100^\circ\]

In \(\Delta MST\):

\[\angle T = 180^\circ – (100^\circ + 25^\circ) = 55^\circ\]

Comparing angles between \(\Delta QPR\) and \(\Delta TSM\):

\[\angle Q = 25^\circ \neq \angle T = 55^\circ\]

Since corresponding angles are not equal, the statement \(\Delta QPR \sim \Delta TSM\) is false (the correct similarity is \(\Delta PQR \sim \Delta TMS\)).

For polygons with more than three sides, equality of corresponding angles alone does not guarantee similarity; their corresponding sides must also be in proportion. For example, a square and a rectangle have all corresponding angles equal (\(90^\circ\)), but their corresponding sides are not proportional, so they are not similar.

Let the sides of the second triangle be \(a, b, c\) with perimeter \(P = a + b + c\). The sides of the first triangle are \(3a, 3b\), and let the third side be \(x\), with perimeter \(3P\):

\[3a + 3b + x = 3P = 3(a + b + c) \implies x = 3c\]

Thus, all three sides of the first triangle are proportional to the corresponding sides of the second triangle (ratio \(= 3\)). By SSS similarity criterion, the two triangles are similar.

Both triangles have a right angle (\(90^\circ\)) and one pair of equal acute angles. By the AA similarity criterion, the two triangles must be similar.

In \(\Delta ADE\) and \(\Delta ACB\), \(\angle A = \angle A\) (common angle). But \(\angle D\) is given equal to \(\angle C\), which means vertex \(D\) corresponds to vertex \(C\), and vertex \(E\) corresponds to vertex \(B\).

Hence, the correct similarity relation is \(\mathbf{\Delta ADE \sim \Delta ACD}\), not \(\Delta ADE \sim \Delta ACB\).

For two triangles to be similar by the SAS criterion, the equal angle must be the included angle between the proportional sides. If the angle is not included between those sides, the triangles are not necessarily similar.

Given: \(\frac{AD}{DB} = \frac{AE}{EC}\).

By the converse of Basic Proportionality Theorem:

\[DE \parallel BC\]

Corresponding angles are equal:

\[\angle D = \angle B \quad \text{and} \quad \angle E = \angle C \quad \text{— (1)}\]

It is given that \(\angle D = \angle E\). Therefore, from (1):

\[\angle B = \angle C\]

In \(\Delta ABC\), since angles opposite to sides \(AB\) and \(AC\) are equal, we have \(AB = AC\). Hence, \(\Delta BAC\) is an isosceles triangle.

By Basic Proportionality Theorem, if \(DE \parallel AB\):

\[\frac{CD}{DA} = \frac{CE}{EB}\] \[\frac{x + 3}{3x + 19} = \frac{x}{3x + 4}\] \[(x + 3)(3x + 4) = x(3x + 19)\] \[3x^2 + 4x + 9x + 12 = 3x^2 + 19x\] \[13x + 12 = 19x \implies 6x = 12 \implies x = 2\]

Answer: \(x = 2\).

In \(\Delta PST\), since \(\angle 1 = \angle 2\), the sides opposite to equal angles are equal:

\[PS = PT \quad \text{— (1)}\]

Given \(\Delta NSQ \cong \Delta MTR\), their corresponding parts are congruent (CPCT):

\[SQ = TR \quad \text{— (2)}\]

Adding (1) and (2):

\[PS + SQ = PT + TR \implies PQ = PR \quad \text{— (3)}\]

Dividing (1) by (3):

\[\frac{PS}{PQ} = \frac{PT}{PR}\]

In \(\Delta PTS\) and \(\Delta PRQ\):

\[\frac{PS}{PQ} = \frac{PT}{PR} \quad \text{and} \quad \angle P = \angle P \text{ (common angle)}\]

By SAS similarity criterion, \(\mathbf{\Delta PTS \sim \Delta PRQ}\).

Since \(AB \parallel DC\) (or line segment \(AP \parallel CQ\)):

\[\angle PAO = \angle QCO \quad \text{(alternate interior angles)}\] \[\angle APO = \angle CQO \quad \text{(alternate interior angles)}\] \[\angle AOP = \angle COQ \quad \text{(vertically opposite angles)}\]

By AAA similarity criterion, \(\Delta AOP \sim \Delta COQ\). Taking corresponding ratios:

\[\frac{OA}{OC} = \frac{AP}{CQ} \implies OA \cdot CQ = OC \cdot AP\]

Perimeter of \(\Delta DEF = DE + EF + FD = 6 + 9 + 12 = 27\text{ cm}\).

For similar triangles, the ratio of their perimeters equals the ratio of their corresponding sides:

\[\frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta DEF} = \frac{AB}{DE}\] \[\frac{\text{Perimeter of } \Delta ABC}{27} = \frac{4}{6} = \frac{2}{3} \implies \text{Perimeter of } \Delta ABC = \frac{2 \times 27}{3} = 18\text{ cm}\]

Join diagonal \(BD\) intersecting \(PQ\) at point \(G\).

In \(\Delta BDC\), \(GQ \parallel DC\). By Basic Proportionality Theorem:

\[\frac{BG}{GD} = \frac{BQ}{QC} = \frac{35}{15} = \frac{7}{3}\]

In \(\Delta DAB\), \(PG \parallel AB\). By BPT:

\[\frac{AP}{PD} = \frac{BG}{GD} \implies \frac{AP}{18} = \frac{7}{3} \implies AP = \frac{18 \times 7}{3} = 42\text{ cm}\] \[AD = AP + PD = 42 + 18 = 60\text{ cm}\]

In \(\Delta ACB\) and \(\Delta ADC\):

\[\angle A = \angle A \quad \text{(common angle)}\] \[\angle ACB = \angle ADC \quad \text{(given)}\]

By AA similarity criterion, \(\Delta ACB \sim \Delta ADC\). Corresponding sides are proportional:

\[\frac{AB}{AC} = \frac{AC}{AD} \implies AC^2 = AB \cdot AD\] \[8^2 = AB \cdot 3 \implies 64 = 3\,AB \implies AB = \frac{64}{3}\text{ cm}\] \[BD = AB – AD = \frac{64}{3} – 3 = \frac{64 – 9}{3} = \frac{55}{3}\text{ cm}\]

At the same time, the angle of elevation of the sun is the same for both objects. Since both the tower and pole stand vertically (\(90^\circ\)), the triangles formed by the objects and their shadows are similar by AA similarity.

\[\frac{\text{Height of pole}}{\text{Height of tower}} = \frac{\text{Shadow of pole}}{\text{Shadow of tower}}\] \[\frac{h}{15} = \frac{16}{24} = \frac{2}{3} \implies h = \frac{2 \times 15}{3} = 10\text{ metres}\]

Since \(\Delta ABC \sim \Delta DEF\):

\[\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} \implies \frac{2x – 1}{18} = \frac{2x + 2}{3x + 9} = \frac{3x}{6x}\]

Since \(\frac{3x}{6x} = \frac{1}{2}\):

\[\frac{2x – 1}{18} = \frac{1}{2} \implies 4x – 2 = 18 \implies 4x = 20 \implies x = 5\]

Substituting \(x = 5\) into side expressions:

\[AB = 2(5) – 1 = 9\text{ cm}, \quad BC = 2(5) + 2 = 12\text{ cm}, \quad CA = 3(5) = 15\text{ cm}\] \[DE = 18\text{ cm}, \quad EF = 3(5) + 9 = 24\text{ cm}, \quad FD = 6(5) = 30\text{ cm}\]

In \(\Delta APB\) and \(\Delta CPD\):

\[\angle A = \angle C \quad \text{(given)}\] \[\angle APB = \angle CPD \quad \text{(vertically opposite angles)}\]

By AA similarity criterion, \(\Delta APB \sim \Delta CPD\). Taking corresponding ratios:

\[\frac{AP}{CP} = \frac{BP}{DP} = \frac{AB}{CD}\] \[\frac{12}{4} = \frac{15}{PD} \implies 3 = \frac{15}{PD} \implies PD = \frac{15}{3} = 5\text{ cm}\] \[3 = \frac{6}{CD} \implies CD = \frac{6}{3} = 2\text{ cm}\]

Answer: \(PD = 5\text{ cm}\) and \(CD = 2\text{ cm}\).

Since \(\Delta ABC \sim \Delta EDF\), corresponding sides are proportional:

\[\frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF}\] \[\frac{5}{12} = \frac{BC}{15} = \frac{7}{EF}\]

Solving for \(BC\):

\[BC = \frac{5 \times 15}{12} = \frac{75}{12} = \frac{25}{4} = 6.25\text{ cm}\]

Solving for \(EF\):

\[EF = \frac{7 \times 12}{5} = \frac{84}{5} = 16.8\text{ cm}\]

Answer: \(BC = 6.25\text{ cm}\) and \(EF = 16.8\text{ cm}\).

Let the pole be \(AB = 6\text{ m}\), woman be \(CD = 1.5\text{ m}\), shadow be \(DE = 3\text{ m}\), and distance from pole base be \(BD = x\text{ m}\).

In \(\Delta ABE\) and \(\Delta CDE\), \(\angle B = \angle D = 90^\circ\) and \(\angle E = \angle E\) (common angle). By AA similarity:

\[\Delta ABE \sim \Delta CDE \implies \frac{AB}{CD} = \frac{BE}{DE}\] \[\frac{6}{1.5} = \frac{x + 3}{3} \implies 4 = \frac{x + 3}{3} \implies x + 3 = 12 \implies x = 9\text{ m}\]

Answer: The woman is standing \(9\text{ m}\) away from the base of the pole.

Since line \(l \parallel m\), alternate interior angles formed by transversals are equal:

1. In \(\Delta PAE\) and \(\Delta PBF\): \(\angle PAE = \angle PBF\), \(\angle PEA = \angle PFB\), and \(\angle APE = \angle BPF\). By AAA similarity:

\[\Delta PAE \sim \Delta PBF \implies \frac{AE}{BF} = \frac{AP}{BP} \quad \text{— (1)}\]

2. In \(\Delta PAC\) and \(\Delta PBD\): \(\angle PAC = \angle PBD\) and \(\angle PCA = \angle PDB\). By AA similarity:

\[\Delta PAC \sim \Delta PBD \implies \frac{AC}{BD} = \frac{AP}{BP} \quad \text{— (2)}\]

3. In \(\Delta PCE\) and \(\Delta PDF\): \(\angle PCE = \angle PDF\) and \(\angle PEC = \angle PFD\). By AA similarity:

\[\Delta PCE \sim \Delta PDF \implies \frac{CE}{FD} = \frac{CP}{DP} = \frac{AP}{BP} \quad \text{— (3)}\]

From equations (1), (2), and (3):

\[\frac{AE}{BF} = \frac{AC}{BD} = \frac{CE}{FD}\]

Since all lines are perpendicular to the same line \(l\), they are mutually parallel: \(PA \parallel QB \parallel RC \parallel SD\).

By the intercept theorem (generalized Thales theorem), parallel lines cut off proportional intercepts on any transversals:

\[\frac{PQ}{AB} = \frac{QR}{BC} = \frac{RS}{CD} = \frac{PQ + QR + RS}{AB + BC + CD}\] \[AB + BC + CD = 6 + 9 + 12 = 27\text{ cm}, \quad SP = PQ + QR + RS = 36\text{ cm}\] \[\text{Ratio } k = \frac{36}{27} = \frac{4}{3}\] \[PQ = \frac{4}{3} \times AB = \frac{4}{3} \times 6 = 8\text{ cm}\] \[QR = \frac{4}{3} \times BC = \frac{4}{3} \times 9 = 12\text{ cm}\] \[RS = \frac{4}{3} \times CD = \frac{4}{3} \times 12 = 16\text{ cm}\]

Answer: \(PQ = 8\text{ cm}, \, QR = 12\text{ cm}, \, RS = 16\text{ cm}\).

In \(\Delta ABD\), since \(PO \parallel AB\), by Basic Proportionality Theorem:

\[\frac{PO}{AB} = \frac{DP}{AD} \quad \text{— (1)}\]

In \(\Delta ADC\), \(PO \parallel DC\), so \(\frac{AP}{PD} = \frac{AO}{OC}\). Adding \(1\) to both sides gives:

\[\frac{AD}{PD} = \frac{AC}{OC} \implies \frac{DP}{AD} = \frac{OC}{AC} \quad \text{— (2)}\]

Substituting (2) into (1):

\[\frac{PO}{AB} = \frac{OC}{AC} \implies PO = AB \times \frac{OC}{AC} \quad \text{— (3)}\]

In \(\Delta ABC\), since \(OQ \parallel AB\), by BPT:

\[\frac{OQ}{AB} = \frac{CQ}{BC} = \frac{OC}{AC} \implies OQ = AB \times \frac{OC}{AC} \quad \text{— (4)}\]

Comparing (3) and (4):

\[PO = QO\]