NCERT Class 10 Maths Triangles Ex 6.3 Solutions 🔥 Chapter 6 Step-by-Step Answers

Concept Used

Two triangles are similar if:
- AAA (or AA): Corresponding angles are equal
- SSS: Corresponding sides are proportional
- SAS: Two sides are proportional and included angle is equal
Correct order of vertices must be maintained while writing similarity
Ratios must be consistent across all corresponding sides

Solution Roadmap

Step 1: Identify given angles or sides
Step 2: Match corresponding vertices carefully
Step 3: Check one similarity criterion (AAA / SSS / SAS)
Step 4: Verify ALL required conditions
Step 5: Write symbolic form correctly

In ΔABC and ΔPQR:

\[\angle A = \angle P\] \[\angle B = \angle Q\] \[\angle C = \angle R\]

All three corresponding angles are equal.

\[\therefore \triangle ABC \sim \triangle PQR\]

Criterion: AAA similarity

Given sides:

\[\frac{AB}{PQ} = \frac{2}{4} = \frac{1}{2}\] \[\frac{BC}{QR} = \frac{2.5}{5} = \frac{1}{2}\] \[\frac{AC}{PR} = \frac{3}{6} = \frac{1}{2}\]

All three ratios are equal.

\[\therefore \triangle ABC \sim \triangle PQR\]

Criterion: SSS similarity

Correction Applied: Earlier side pairing was incorrect (BC matched with PR). Now corrected.

\[\frac{MP}{DE} = \frac{2}{4} = \frac{1}{2}\] \[\frac{LP}{DF} = \frac{3}{6} = \frac{1}{2}\] \[\frac{LM}{EF} = \frac{2.7}{5} \neq \frac{1}{2}\]

Since all three ratios are NOT equal:

\[\therefore \triangle LMP \not\sim \triangle DEF\]

Reason: SSS condition fails

\[\frac{MN}{PQ} = \frac{5}{10} = \frac{1}{2}\] \[\frac{ML}{QR} = \frac{5}{10} = \frac{1}{2}\] \[\angle M = \angle Q\]

Two sides are proportional and included angle is equal.

\[\therefore \triangle MNL \sim \triangle QRP\]

Criterion: SAS similarity

Correction Applied: Earlier expression "2·5/5" was invalid and removed.

\[\angle A = \angle F\] \[\frac{AB}{FD} = \frac{2}{5}\]

Only one angle and one side ratio is known.

Second side or angle is missing.

Conclusion: Similarity cannot be established.

\[\angle D = \angle P = 70^\circ\] \[\angle E = \angle Q = 80^\circ\] \[\angle F = \angle R = 30^\circ\]

All corresponding angles are equal.

\[\therefore \triangle DEF \sim \triangle PQR\]

Criterion: AAA similarity

Exam Significance

Very important for CBSE Board Exams (2–4 marks direct questions)
Forms base for height-distance and trigonometry applications
Frequently used in:
- NTSE
- Olympiads
- JEE Foundation level
Common mistakes tested:
- Wrong correspondence of vertices
- Incorrect ratio pairing
- Using incomplete conditions

Solution

\[\triangle ODC \sim \triangle OBA \quad \text{(Given)}\]

From similarity, corresponding angles are equal:

\[\angle ODC = \angle OBA\] \[\angle CDO = 70^\circ \Rightarrow \angle OBA = 70^\circ\]

Now consider angles around point O:

\[\angle BOC = 125^\circ\] \[\angle DOA = 125^\circ \quad \text{(Vertically opposite)}\]

Let $ \angle DOC = x $

\[\text{Sum of angles around point } O = 360^\circ\] \[125^\circ + 125^\circ + x + x = 360^\circ\] \[250^\circ + 2x = 360^\circ\] \[2x = 110^\circ\] \[x = 55^\circ\] \[\therefore \angle DOC = 55^\circ\]

Now in $ \triangle ODC $:

\[\angle DOC + \angle ODC + \angle DCO = 180^\circ\] \[55^\circ + 70^\circ + \angle DCO = 180^\circ\] \[\angle DCO = 180^\circ - 125^\circ\] \[\angle DCO = 55^\circ\]

Now in $ \triangle OBA $:

\[\angle OAB + \angle OBA + \angle AOB = 180^\circ\] \[\angle OAB + 70^\circ + 55^\circ = 180^\circ\] \[\angle OAB = 180^\circ - 125^\circ\] \[\angle OAB = 55^\circ\]

Final Answers:

\[\angle DOC = 55^\circ\] \[\angle DCO = 55^\circ\] \[\angle OAB = 55^\circ\]

Exam Significance

Classic CBSE Board question combining similarity + angle properties
Tests multi-step reasoning (not direct formula)
Frequently appears in 3–5 mark questions
Important for:
- NTSE Stage 1
- Olympiad geometry
- JEE Foundation
Common traps:
- Wrong angle correspondence
- Forgetting angles around a point
- Skipping triangle angle sum step

Concept Used

If two lines are parallel, alternate interior angles are equal
Vertically opposite angles are equal
If two triangles are similar, corresponding sides are proportional
AA similarity is sufficient to prove triangles similar

Solution Roadmap

Identify triangles formed by diagonals
Use parallel lines to establish equal angles
Apply AA similarity
Use proportional sides property of similar triangles

Trapezium ABCD with AB ∥ DC

Solution

Given:

\[AB \parallel DC\]

Consider triangles $ \triangle AOB $ and $ \triangle COD $

Step 1: Establish angle equality

\[\angle ABO = \angle CDO \quad \text{(Alternate interior angles)}\] \[\angle BAO = \angle DCO \quad \text{(Alternate interior angles)}\] \[\angle AOB = \angle COD \quad \text{(Vertically opposite angles)}\]

Step 2: Apply similarity

Since two angles are equal, triangles are similar by AA criterion:

\[\therefore \triangle AOB \sim \triangle COD\]

Step 3: Use proportional sides

\[\frac{AO}{CO} = \frac{BO}{DO}$$

Final Result:

\[\boxed{\dfrac{OA}{OC} = \dfrac{OB}{OD}}\]

Exam Significance

Highly important proof question in CBSE Board (3–4 marks)
Tests understanding of:
- Parallel lines
- Angle properties
- Triangle similarity
Frequently used as base concept in:
- Coordinate geometry
- Mensuration proofs
- Competitive exams (NTSE, Olympiads)
Common mistakes:
- Wrong triangle selection
- Incorrect angle correspondence
- Using BPT unnecessarily (not needed here)

Concept Used

If two sides of one triangle are proportional to two sides of another triangle and the included angle is equal, triangles are similar (SAS)
Careful identification of corresponding sides is essential
No assumption should be made unless given or derived

Solution Roadmap

Use given ratio and rewrite in usable form
Identify triangles to compare
Match corresponding sides correctly
Use given angle equality as included angle
Apply SAS similarity

Fig. 6.36 (Illustration)

Solution

Given:

\[\frac{QR}{QS} = \frac{QT}{PR}\] \[\angle 1 = \angle 2\]

Step 1: Rearrange the given ratio

\[\frac{QS}{QR} = \frac{PR}{QT}\]

Step 2: Identify corresponding sides

In triangles $ \triangle PQS $ and $ \triangle TQR $:

QS corresponds to QR
PS corresponds to QT

Step 3: Use given angle equality

\[\angle PQS = \angle TQR \quad \text{(Given: ∠1 = ∠2)}\]

Step 4: Apply SAS similarity

Two sides are proportional and included angle is equal: \[\therefore \triangle PQS \sim \triangle TQR\]

Conclusion:

\[\boxed{\triangle PQS \sim \triangle TQR}\]

Exam Significance

Important proof question based on SAS similarity
Tests ability to:
- Manipulate ratios
- Identify correct triangle pairs
- Avoid invalid assumptions
Frequently asked in CBSE (3–4 marks)
Foundation concept for:
- Trigonometry proofs
- Coordinate geometry
- Olympiad problems
Common mistakes:
- Assuming equal sides without proof
- Wrong correspondence of triangles
- Using incorrect angle for SAS

Concept Used

If two angles of one triangle are equal to two angles of another triangle, triangles are similar (AA)
Points lying on the same line form equal straight-line angles
Collinearity helps in identifying equal angles

Solution Roadmap

Identify triangles to compare
Use given angle equality
Use collinearity (S on PR, T on QR) to find second angle pair
Apply AA similarity

Triangle PQR with S on PR and T on QR

Solution

Given:

\[\angle P = \angle RTS\]

Step 1: Identify triangles

Consider triangles $ \triangle RPQ $ and $ \triangle RTS $

Step 2: Use given angle

\[\angle RPQ = \angle RTS \quad \text{(Given)}\]

Step 3: Use collinearity

Since:

S lies on PR
T lies on QR

Therefore: \[\angle RQP = \angle RST\]

(Angles formed by the same lines QR and PR)

Step 4: Apply AA similarity

Two corresponding angles are equal: \[\therefore \triangle RPQ \sim \triangle RTS\]

Conclusion:

\[\boxed{\triangle RPQ \sim \triangle RTS}\]

Exam Significance

Classic AA similarity proof (frequently asked in CBSE)
Tests understanding of:
- Collinearity
- Angle correspondence
- Triangle matching
Used in advanced geometry proofs and trigonometry
Common mistakes:
- Calling angles “common” when they are not
- Ignoring straight-line relationships
- Wrong triangle order

Concept Used

If two triangles are congruent, their corresponding parts are equal (CPCT)
If two sides are proportional and included angle is equal, triangles are similar (SAS)
Correct identification of corresponding parts is essential

Solution Roadmap

Use congruence to extract equal sides
Form ratios of corresponding sides
Identify included angle
Apply SAS similarity

Fig. 6.37 (Illustration)

Solution

Given:

\[\triangle ABE \cong \triangle ACD\]

Step 1: Use CPCT

From congruence: \[AE = AD \quad \text{(Corresponding sides)}\] \[AB = AC \quad \text{(Corresponding sides)}\]

Step 2: Form ratio

\[\frac{AD}{AB} = \frac{AE}{AC}\]

Step 3: Identify included angle

\[\angle DAE = \angle BAC\]

(Same angle at vertex A)

Step 4: Apply SAS similarity

Two sides are proportional and included angle is equal: \[\therefore \triangle ADE \sim \triangle ABC\]

Conclusion:

\[\boxed{\triangle ADE \sim \triangle ABC}\]

Exam Significance

Important mixed concept question (Congruence + Similarity)
Tests ability to:
- Use CPCT correctly
- Convert equality into ratios
- Apply SAS logically
Common in CBSE 3–5 mark questions
Also relevant for:
- Olympiad geometry
- NTSE
- JEE foundation
Common mistakes:
- Forgetting included angle in SAS
- Wrong correspondence from CPCT
- Skipping ratio formation step

Concept Used

Altitude of a triangle is perpendicular to the opposite side
Perpendicular lines form right angles (90°)
Vertically opposite angles are equal
If two angles are equal, triangles are similar (AA)

Solution Roadmap

Use AD ⟂ BC and CE ⟂ AB
Identify right angles
Use vertical angles at intersection P
Apply AA similarity carefully with correct correspondence

Altitudes AD and CE intersect at P

$ \triangle AEP \sim \triangle CDP $
$ \triangle ABD \sim \triangle CBE $
$ \triangle AEP \sim \triangle ADB $
$ \triangle PDC \sim \triangle BEC $

Solution

(i) In ΔAEP and ΔCDP

\[\angle AEP = 90^\circ \quad (\text{CE ⟂ AB})\] \[\angle CDP = 90^\circ \quad (\text{AD ⟂ BC})\] \[\angle APE = \angle CPD \quad (\text{Vertically opposite})\] \[\therefore \triangle AEP \sim \triangle CDP \quad (\text{AA})\]

(ii) In ΔABD and ΔCBE

\[\angle ADB = 90^\circ \quad (\text{AD ⟂ BC})\] \[\angle CEB = 90^\circ \quad (\text{CE ⟂ AB})\] \[\angle ABD = \angle CBE \quad (\text{Common angle at B})\] \[\therefore \triangle ABD \sim \triangle CBE \quad (\text{AA})\]

(iii) In ΔAEP and ΔADB

\[\angle AEP = 90^\circ\] \[\angle ADB = 90^\circ\] \[\angle EAP = \angle DAB \quad (\text{Same angle at A})\] \[\therefore \triangle AEP \sim \triangle ADB \quad (\text{AA})\]

(iv) In ΔPDC and ΔBEC

\[\angle PDC = 90^\circ\] \[\angle BEC = 90^\circ\] \[\angle PCD = \angle BCE \quad (\text{Same angle at C})\] \[\therefore \triangle PDC \sim \triangle BEC \quad (\text{AA})\]

Exam Significance

Highly important multi-part similarity question (CBSE 4–5 marks)
Tests deep understanding of:
- Altitudes
- Perpendicular geometry
- Angle relationships
- Triangle similarity
Common in:
- Board exams
- NTSE
- Olympiads
Common mistakes:
- Not using perpendicular property
- Wrong angle identification
- Incorrect triangle correspondence

Concept Used

In a parallelogram: AB ∥ CD and AD ∥ BC
Alternate interior angles are equal when a transversal cuts parallel lines
If two angles of one triangle are equal to two angles of another triangle, triangles are similar (AA)

Solution Roadmap

Use properties of parallelogram (parallel sides)
Identify correct alternate angle pairs
Match corresponding vertices carefully
Apply AA similarity

Parallelogram ABCD with E on AD produced

Solution

Given:

ABCD is a parallelogram \[AB \parallel CD, \quad AD \parallel BC\]

Step 1: Identify angle pairs

Since AB ∥ CD and BE is a transversal: \[\angle ABE = \angle CFB \quad \text{(Alternate interior angles)}\]
Since AD ∥ BC and AE is a transversal: \[\angle BAE = \angle BCF \quad \text{(Alternate interior angles)}\]

Step 2: Apply AA similarity

Two pairs of corresponding angles are equal: \[\therefore \triangle ABE \sim \triangle CFB\]

Conclusion:

\[\boxed{\triangle ABE \sim \triangle CFB}\]

Exam Significance

Important CBSE proof question (3–4 marks)
Tests:
- Parallelogram properties
- Parallel lines and transversals
- Triangle similarity (AA)
Common mistakes:
- Wrong angle pairing
- Ignoring parallel sides
- Incorrect triangle correspondence
Useful for coordinate geometry and advanced proofs

Concept Used

If two angles are equal, triangles are similar (AA)
All right angles are equal (90°)
In similar triangles, corresponding sides are proportional
BPT is NOT used here

Solution Roadmap

Use right angle equality
Use common angle at A
Apply AA similarity
Write correct correspondence of vertices
Use proportionality of sides

Two right triangles ABC and AMP

Solution

Given:

\[\angle ABC = 90^\circ,\quad \angle AMP = 90^\circ\]

(i) Prove similarity

\[\angle ABC = \angle AMP \quad (90^\circ)\] \[\angle A = \angle A \quad \text{(Common)}\]

Two angles are equal:

\[\therefore \triangle ABC \sim \triangle AMP \quad \text{(AA)}\]

Correct correspondence:

\[A \leftrightarrow A,\quad B \leftrightarrow M,\quad C \leftrightarrow P\]

(ii) Use proportional sides

From similarity: \[\frac{AC}{AP} = \frac{BC}{MP} = \frac{AB}{AM}\] Taking required ratio: \[\frac{CA}{PA} = \frac{BC}{MP}\]

Conclusion:

\[\boxed{\frac{CA}{PA} = \frac{BC}{MP}}\]

Exam Significance

Very common CBSE 3–4 mark question
Tests:
- AA similarity
- Right triangle properties
- Correct correspondence of vertices
Common mistakes:
- Using BPT incorrectly
- Wrong side correspondence
- Skipping similarity justification
Important for trigonometry foundation

Concept Used

If two triangles are similar, corresponding angles are equal and sides are proportional
Angle bisector divides an angle into two equal parts
AA similarity is sufficient to prove triangles similar
Use similarity → then proportional sides (NOT BPT here)

Solution Roadmap

Use similarity of ΔABC and ΔFEG to match angles
Use angle bisector property to split angles
Form smaller triangles and apply AA similarity
Use proportional sides from similarity

Two similar triangles with angle bisectors

Given:

\[\triangle ABC \sim \triangle FEG\]

So, corresponding angles are equal:

\[\angle ACB = \angle EGF\] \[\angle ABC = \angle FEG\] \[\angle BAC = \angle EFG\]

(i) In ΔCDA and ΔGHF

Since CD and GH are angle bisectors: \[\angle ACD = \frac{1}{2}\angle ACB\] \[\angle FGH = \frac{1}{2}\angle EGF\] \[\Rightarrow \angle ACD = \angle FGH\]
Also: \[\angle CAD = \angle GFH\]

(From similarity of ΔABC and ΔFEG)

\[\therefore \triangle CDA \sim \triangle GHF \quad (\text{AA})\]
From similarity: \[\frac{CD}{GH} = \frac{AC}{FG}\]

(ii) In ΔDCB and ΔHGE

\[\angle DCB = \frac{1}{2}\angle ACB\] \[\angle HGE = \frac{1}{2}\angle EGF\] \[\Rightarrow \angle DCB = \angle HGE\]
\[\angle DBC = \angle HEG\]

(From similarity of main triangles)

\[\therefore \triangle DCB \sim \triangle HGE \quad (\text{AA})\]

(iii) In ΔDCA and ΔHGF

\[\angle DCA = \angle HGF \quad (\text{angle bisector})\] \[\angle DAC = \angle HFG \quad (\text{from similarity})\]
\[\therefore \triangle DCA \sim \triangle HGF \quad (\text{AA})\]

Exam Significance

Advanced CBSE proof (4–5 marks)
Tests:
- Similarity + angle bisector concepts
- Breaking triangles into sub-triangles
- Logical multi-step reasoning
Common mistakes:
- Using BPT incorrectly
- Missing half-angle reasoning
- Wrong correspondence
Very useful for Olympiad geometry

Concept Used

In an isosceles triangle, angles opposite equal sides are equal
Perpendicular lines form right angles (90°)
If two angles of one triangle are equal to two angles of another triangle, triangles are similar (AA)

Solution Roadmap

Use isosceles property to relate base angles
Use perpendicular condition to identify right angles
Match correct corresponding angles
Apply AA similarity

Isosceles triangle with perpendiculars

Solution

Given:

\[AB = AC \quad (\text{Isosceles triangle})\] \[AD \perp BC,\quad EF \perp AC\]

Step 1: Right angles

\[\angle ADB = 90^\circ\] \[\angle EFC = 90^\circ\] \[\Rightarrow \angle ADB = \angle EFC\]

Step 2: Base angles of isosceles triangle

Since $ AB = AC $: \[\angle ABC = \angle BCA\]
Since D lies on BC and E lies on CB produced: \[\angle ABD = \angle ABC\] \[\angle ECF = \angle BCA\]
\[\Rightarrow \angle ABD = \angle ECF\]

Step 3: Apply AA similarity

\[\angle ADB = \angle EFC\] \[\angle ABD = \angle ECF\] \[\therefore \triangle ABD \sim \triangle ECF \quad (\text{AA})\]

Conclusion:

\[\boxed{\triangle ABD \sim \triangle ECF}\]

Exam Significance

Important CBSE similarity proof (3–4 marks)
Tests:
- Isosceles triangle properties
- Perpendicular geometry
- Angle identification
Common mistakes:
- Not using base angles equality
- Incorrect angle pairing
- Overcomplicating with unnecessary steps
Useful for trigonometry and coordinate geometry basics

Concept Used

If three corresponding sides are proportional, triangles are similar (SSS)
If two sides are proportional and included angle is equal, triangles are similar (SAS)
A median divides a side into two equal parts
Similarity of smaller triangles can help prove similarity of larger triangles

Solution Roadmap

Use medians to relate BD and QM
Prove ΔABD ∼ ΔPQM using SSS
Use this to establish angle equality
Apply SAS in ΔABC and ΔPQR

Triangles with medians AD and PM

Solution

Given:

\[\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}\]

Also, AD and PM are medians:

\[BD = \frac{BC}{2}, \quad QM = \frac{QR}{2}\]

Step 1: Ratio of BD and QM

\[\frac{BD}{QM} = \frac{BC/2}{QR/2} = \frac{BC}{QR}\]

Step 2: Compare ΔABD and ΔPQM

\[\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}\] \[\therefore \triangle ABD \sim \triangle PQM \quad (\text{SSS})\]

Step 3: Use similarity to get angle equality

\[\angle ABD = \angle PQM\]
Since: \[\angle ABC = \angle ABD,\quad \angle PQR = \angle PQM\] \[\Rightarrow \angle ABC = \angle PQR\]

Step 4: Apply SAS similarity

\[\frac{AB}{PQ} = \frac{BC}{QR}\] \[\angle ABC = \angle PQR\] \[\therefore \triangle ABC \sim \triangle PQR \quad (\text{SAS})\]

Conclusion:

\[\boxed{\triangle ABC \sim \triangle PQR}\]

Exam Significance

Very important CBSE 4–5 mark proof
Tests:
- Use of medians
- Multi-step similarity reasoning
- Transition from SSS → SAS
Common mistakes:
- Skipping intermediate triangle similarity
- Assuming angle equality without proof
- Incorrect median handling
High relevance in Olympiads and advanced geometry

Concept Used

If two angles are equal, triangles are similar (AA)
In similar triangles, corresponding sides are proportional
Correct correspondence of vertices is essential

Solution Roadmap

Identify triangles to compare
Use given angle equality
Use common angle at C
Apply AA similarity
Write correct ratio and simplify

Triangle ABC with point D on BC

Solution

Given:

\[\angle ADC = \angle BAC\]

Step 1: Identify triangles

Consider triangles $ \triangle BAC $ and $ \triangle ADC $

Step 2: Use angle equality

\[\angle ADC = \angle BAC \quad (\text{Given})\] \[\angle ACD = \angle BCA \quad (\text{Common angle at C})\]

Step 3: Apply AA similarity

\[\therefore \triangle BAC \sim \triangle ADC\]

Correct correspondence:

\[B \leftrightarrow A,\quad A \leftrightarrow D,\quad C \leftrightarrow C\]

Step 4: Write proportional sides

\[\frac{CA}{CB} = \frac{CD}{CA}\]

Step 5: Simplify

\[CA^2 = CB \cdot CD\]

Conclusion:

\[\boxed{CA^2 = CB \cdot CD}\]

Exam Significance

Very important CBSE 3–4 mark question
Tests:
- AA similarity
- Correct correspondence
- Algebraic manipulation of ratios
Common mistakes:
- Wrong final product (AD instead of CD)
- Incorrect triangle order
- Skipping correspondence
Important for coordinate geometry and trigonometry

Concept Used

Median divides a side into two equal parts
Construction can help convert median problems into side proportionality
If three sides are proportional, triangles are similar (SSS)
If two sides are proportional and included angle is equal, triangles are similar (SAS)

Solution Roadmap

Extend medians to create equal segments
Use congruence to relate sides
Convert given ratios into new triangle ratios
Prove similarity of constructed triangles
Use angle equality to prove final similarity

Construction on medians AD and PM

Solution

Given:

\[\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}\]

Construction:

Produce AD to E such that: \[AD = DE\] Produce PM to S such that: \[PM = MS\]

Step 1: Use median property

\[BD = DC,\quad QM = MR\]

Step 2: Prove congruence

In $ \triangle ADC $ and $ \triangle BDE $: \[AD = DE \quad (\text{Construction})\] \[BD = DC \quad (\text{Median})\] \[\angle ADC = \angle BDE \quad (\text{Vertically opposite})\] \[\therefore \triangle ADC \cong \triangle BDE \quad (\text{SAS})\] \[\Rightarrow AC = BE \quad (1)\]
In $ \triangle PRM $ and $ \triangle QSM $: \[PM = MS \quad (\text{Construction})\] \[QM = MR \quad (\text{Median})\] \[\angle PMR = \angle QMS \quad (\text{Vertically opposite})\] \[\therefore \triangle PRM \cong \triangle QSM \quad (\text{SAS})\] \[\Rightarrow PR = QS \quad (2)\]

Step 3: Form ratios

From given: \[\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}\] Using (1) and (2): \[\frac{AB}{PQ} = \frac{BE}{QS} = \frac{AE}{PS}\]

Step 4: Similarity of constructed triangles

\[\therefore \triangle ABE \sim \triangle PQS \quad (\text{SSS})\]
\[\Rightarrow \angle BAE = \angle QPS\]

Step 5: Final angle relation

\[\angle BAC = \angle QPR\]

Step 6: Apply SAS

\[\frac{AB}{PQ} = \frac{AC}{PR}\] \[\angle BAC = \angle QPR\] \[\therefore \triangle ABC \sim \triangle PQR \quad (\text{SAS})\]

Conclusion:

\[\boxed{\triangle ABC \sim \triangle PQR}\]

Exam Significance

Most advanced question of Exercise 6.3 (5 marks level)
Tests:
- Creative construction
- Congruence + similarity integration
- Multi-step logical reasoning
Common mistakes:
- Wrong construction labeling
- Skipping congruence step
- Incorrect ratio substitution
High-level geometry (Olympiad & NTSE relevance)

Concept Used

Objects casting shadows at the same time form similar triangles
Angle of elevation of the sun is the same
Hence, corresponding sides are proportional

Solution Roadmap

Represent pole and tower as right triangles
Use same sun angle → triangles are similar
Apply proportionality of height and shadow

Pole and tower forming similar triangles

Solution

Given:

Height of pole = 6 m Shadow of pole = 4 m Shadow of tower = 28 m

Let height of tower = x m

Step 1: Similarity of triangles

Since both shadows are formed at the same time:

Angle of elevation of sun is same
Both form right triangles

\[\therefore \text{Triangles are similar}\]

Step 2: Apply proportionality

\[\frac{\text{Height of pole}}{\text{Shadow of pole}} = \frac{\text{Height of tower}}{\text{Shadow of tower}}\] \[\frac{6}{4} = \frac{x}{28}\]

Step 3: Solve

\[x = \frac{28 \times 6}{4}\] \[x = 42 \text{ m}\]

Conclusion:

\[\boxed{\text{Height of the tower} = 42 \text{ m}}\]

Exam Significance

Very common CBSE application question (2–3 marks)
Tests real-life use of similarity
Frequently asked in:
- Board exams
- NTSE
- Foundation exams
Common mistakes:
- Wrong ratio (shadow/height confusion)
- Skipping similarity justification

Concept Used

If two triangles are similar, corresponding sides are proportional
A median divides a side into two equal parts
Ratios of corresponding segments remain equal

Solution Roadmap

Use similarity of ΔABC and ΔPQR
Relate midpoints using medians
Form ratios BD/QM
Use proportionality to prove required result

Similar triangles with medians AD and PM

Solution

Given:

\[\triangle ABC \sim \triangle PQR\]

Step 1: From similarity

\[\frac{AB}{PQ} = \frac{BC}{QR} \tag{1}\]

Step 2: Use median property

Since AD and PM are medians: \[BD = DC = \frac{BC}{2}\] \[QM = MR = \frac{QR}{2}\]

Step 3: Form ratio

\[\frac{BD}{QM} = \frac{BC/2}{QR/2} = \frac{BC}{QR} \tag{2}\]

Step 4: Compare (1) and (2)

\[\frac{AB}{PQ} = \frac{BD}{QM}\]

Step 5: Similarity of smaller triangles

In triangles $ \triangle ABD $ and $ \triangle PQM $: \[\frac{AB}{PQ} = \frac{BD}{QM}\] \[\angle ABD = \angle PQM \quad (\text{from main similarity})\] \[\therefore \triangle ABD \sim \triangle PQM \quad (\text{SAS})\]

Step 6: Final result

From similarity: \[\frac{AD}{PM} = \frac{AB}{PQ}\]

Conclusion:

\[\boxed{\frac{AB}{PQ} = \frac{AD}{PM}}\]

Exam Significance

Important CBSE 4-mark proof question
Tests:
- Similarity properties
- Median concepts
- Multi-step reasoning
Common mistakes:
- Confusing median with angle bisector
- Using BPT incorrectly
- Skipping intermediate triangle similarity
Important for Olympiad-level geometry

Chapter 6 — TRIANGLES

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Chapter Complete!

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