Class 10-NCERT Solutions-Chapter-06-Triangles

Answer:

(i) similar

Reason: Circles of different radii have the same shape but different sizes, so they are similar, not congruent.

(ii) similar

Reason: Squares always have angles of \( 90^{\circ} \) and sides in proportion, but their sizes can differ.

(iii) equilateral

Reason: Equilateral triangles always have all angles as \( 60^{\circ} \) and all sides equal, ensuring similarity. Isosceles triangles may have different angles.

(iv) (a) equal, (b) proportional

Reason: This is the standard definition of similarity for polygons.

Solution:

(i) Two examples of similar figures:

1. Two equilateral triangles with sides 2 cm and 5 cm respectively. (Same shape, different size).
2. Two circles with radii 3 cm and 5 cm respectively. (All circles are similar).

(ii) Two examples of non-similar figures:

1. A square and a rectangle. (Corresponding angles are equal, but sides are not proportional).
2. A triangle and a parallelogram. (Different shapes entirely).

Solution:

The two quadrilaterals are not similar.

Step-by-step Explanation:

1. Check corresponding sides:
   \( \frac{PQ}{AB} = \frac{1.5}{3} = \frac{1}{2} \)
   \( \frac{QR}{BC} = \frac{1.5}{3} = \frac{1}{2} \)
   The sides are proportional.
2. Check corresponding angles:
   Quadrilateral ABCD is a square, so all its angles are \( 90^{\circ} \).
   Quadrilateral PQRS is a rhombus (not a square), so its angles are not \( 90^{\circ} \).
   Therefore, \( \angle P \neq \angle A \).

Since the corresponding angles are not equal, the figures are not similar.

Solution:

(i) To find EC:

We are given that \( DE \parallel BC \).

According to the Basic Proportionality Theorem (BPT) (Thales Theorem), if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

Therefore, \( \frac{AD}{DB} = \frac{AE}{EC} \).

Given values: \( AD = 1.5 \) cm, \( DB = 3 \) cm, \( AE = 1 \) cm.

Substituting the values:

\[ \frac{1.5}{3} = \frac{1}{EC} \]

\[ \frac{1}{2} = \frac{1}{EC} \]

\[ EC = 2 \text{ cm} \]

(ii) To find AD:

We are given \( DE \parallel BC \). Using BPT again:

\[ \frac{AD}{DB} = \frac{AE}{EC} \]

Given values: \( DB = 7.2 \) cm, \( AE = 1.8 \) cm, \( EC = 5.4 \) cm.

Substituting the values:

\[ \frac{AD}{7.2} = \frac{1.8}{5.4} \]

\[ \frac{AD}{7.2} = \frac{1}{3} \]

\[ AD = \frac{7.2}{3} \]

\[ AD = 2.4 \text{ cm} \]

Solution:

According to the Converse of Basic Proportionality Theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

(i)

Calculate ratio \( \frac{PE}{EQ} = \frac{3.9}{3} = 1.3 \).

Calculate ratio \( \frac{PF}{FR} = \frac{3.6}{2.4} = \frac{36}{24} = 1.5 \).

Since \( 1.3 \neq 1.5 \), the sides are not divided proportionally.

Therefore, EF is not parallel to QR.

(ii)

Calculate ratio \( \frac{PE}{QE} = \frac{4}{4.5} = \frac{40}{45} = \frac{8}{9} \).

Calculate ratio \( \frac{PF}{RF} = \frac{8}{9} \).

Since \( \frac{PE}{QE} = \frac{PF}{RF} \), the sides are divided proportionally.

Therefore, \( EF \parallel QR \).

(iii)

We need to find EQ and FR first.

\( EQ = PQ - PE = 1.28 - 0.18 = 1.10 \) cm.

\( FR = PR - PF = 2.56 - 0.36 = 2.20 \) cm.

Calculate ratio \( \frac{PE}{EQ} = \frac{0.18}{1.10} = \frac{18}{110} = \frac{9}{55} \).

Calculate ratio \( \frac{PF}{FR} = \frac{0.36}{2.20} = \frac{36}{220} = \frac{9}{55} \).

(Alternatively, checking \( \frac{PE}{PQ} = \frac{PF}{PR} \) is also valid: \( \frac{0.18}{1.28} = \frac{9}{64} \) and \( \frac{0.36}{2.56} = \frac{9}{64} \)).

Since the ratios are equal, \( EF \parallel QR \).

Solution:

In \( \Delta ABC \):

Given that \( LM \parallel CB \).

By Basic Proportionality Theorem (BPT), we have:

\[ \frac{AM}{AB} = \frac{AL}{AC} \quad \dots(1) \]

In \( \Delta ADC \):

Given that \( LN \parallel CD \).

By Basic Proportionality Theorem (BPT), we have:

\[ \frac{AN}{AD} = \frac{AL}{AC} \quad \dots(2) \]

Conclusion:

From equations (1) and (2), both ratios are equal to \( \frac{AL}{AC} \).

Therefore, \[ \frac{AM}{AB} = \frac{AN}{AD} \]

Hence proved.

Solution:

Step 1: In \( \Delta ABE \):

We are given \( DF \parallel AE \).

By Basic Proportionality Theorem (BPT):

\[ \frac{BF}{FE} = \frac{BD}{DA} \quad \dots(1) \]

Step 2: In \( \Delta ABC \):

We are given \( DE \parallel AC \).

By Basic Proportionality Theorem (BPT):

\[ \frac{BE}{EC} = \frac{BD}{DA} \quad \dots(2) \]

Step 3: Compare results

From (1) and (2), both fractions equal \( \frac{BD}{DA} \).

Therefore, \[ \frac{BF}{FE} = \frac{BE}{EC} \]

Hence proved.

Solution:

Step 1: In \( \Delta POQ \):

Given \( DE \parallel OQ \).

By BPT: \[ \frac{PE}{EQ} = \frac{PD}{DO} \quad \dots(1) \]

Step 2: In \( \Delta POR \):

Given \( DF \parallel OR \).

By BPT: \[ \frac{PF}{FR} = \frac{PD}{DO} \quad \dots(2) \]

Step 3: Compare and Conclude

From (1) and (2), we get:

\[ \frac{PE}{EQ} = \frac{PF}{FR} \]

Now consider \( \Delta PQR \). Since the line segment EF divides the sides PQ and PR in the same ratio, by the Converse of Basic Proportionality Theorem:

\[ EF \parallel QR \]

Hence proved.

Solution:

Step 1: In \( \Delta OPQ \), since \( AB \parallel PQ \), by BPT:

\[ \frac{OA}{AP} = \frac{OB}{BQ} \quad \dots(1) \]

Step 2: In \( \Delta OPR \), since \( AC \parallel PR \), by BPT:

\[ \frac{OA}{AP} = \frac{OC}{CR} \quad \dots(2) \]

Step 3: From (1) and (2), we equate the common ratio:

\[ \frac{OB}{BQ} = \frac{OC}{CR} \]

Step 4: In \( \Delta OQR \), the sides OQ and OR are divided in the same ratio by the line segment BC.

Therefore, by the Converse of BPT:

\[ BC \parallel QR \]

Hence proved.

Solution:

Given: A triangle ABC where D is the midpoint of AB (so \( AD = DB \)) and a line \( DE \parallel BC \) intersects AC at E.

To Prove: E is the midpoint of AC (i.e., \( AE = EC \)).

Proof:

Since \( DE \parallel BC \), by Basic Proportionality Theorem:

\[ \frac{AD}{DB} = \frac{AE}{EC} \]

Since D is the midpoint of AB, \( AD = DB \). Therefore:

\[ \frac{AD}{DB} = 1 \]

Substituting this in the BPT equation:

\[ 1 = \frac{AE}{EC} \]

\[ \Rightarrow AE = EC \]

Thus, E bisects AC.

Solution:

Given: Triangle ABC where D is the midpoint of AB (\( AD=DB \)) and E is the midpoint of AC (\( AE=EC \)).

To Prove: \( DE \parallel BC \).

Proof:

Since D is the midpoint of AB:

\[ \frac{AD}{DB} = 1 \quad \dots(1) \]

Since E is the midpoint of AC:

\[ \frac{AE}{EC} = 1 \quad \dots(2) \]

From (1) and (2):

\[ \frac{AD}{DB} = \frac{AE}{EC} \]

Since the line segment DE divides the sides AB and AC in the same ratio, by the Converse of Basic Proportionality Theorem:

\[ DE \parallel BC \]

Solution:

Construction: Draw a line \( EO \parallel DC \) through O, meeting AD at E. (Since \( DC \parallel AB \), therefore \( EO \parallel AB \) as well).

Step 1: In \( \Delta ADC \), \( EO \parallel DC \). By BPT:

\[ \frac{AE}{ED} = \frac{AO}{OC} \quad \dots(1) \]

Step 2: In \( \Delta ABD \), \( EO \parallel AB \). By BPT:

\[ \frac{DE}{EA} = \frac{DO}{OB} \]

Taking reciprocal:

\[ \frac{AE}{ED} = \frac{BO}{OD} \quad \dots(2) \]

Step 3: From (1) and (2):

\[ \frac{AO}{OC} = \frac{BO}{OD} \]

Rearranging the terms:

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

Hence shown.

Solution:

Given: \( \frac{AO}{BO} = \frac{CO}{DO} \). We can rewrite this as \( \frac{AO}{CO} = \frac{BO}{DO} \).

Construction: Draw a line \( EO \parallel AB \) meeting AD at E.

Proof:

In \( \Delta ABD \), since \( EO \parallel AB \), by BPT:

\[ \frac{AE}{ED} = \frac{BO}{OD} \quad \dots(1) \]

From given condition:

\[ \frac{AO}{CO} = \frac{BO}{DO} \quad \dots(2) \]

Comparing (1) and (2):

\[ \frac{AE}{ED} = \frac{AO}{CO} \]

Now consider \( \Delta ADC \). Since the line segment EO divides the sides AD and AC in the same ratio, by Converse of BPT:

\[ EO \parallel DC \]

Since we constructed \( EO \parallel AB \), and we proved \( EO \parallel DC \), it follows that:

\[ AB \parallel DC \]

Since a pair of opposite sides is parallel, quadrilateral ABCD is a trapezium.

Solution:

(i) In \( \Delta ABC \) and \( \Delta PQR \):

\[ \angle A = \angle P = 60^{\circ} \]

\[ \angle B = \angle Q = 80^{\circ} \]

\[ \angle C = \angle R = 40^{\circ} \]

Since corresponding angles are equal, the triangles are similar.

Result: \( \Delta ABC \sim \Delta PQR \) (By AAA similarity criterion).

(ii) In \( \Delta ABC \) and \( \Delta QRP \):

\[ \frac{AB}{QR} = \frac{2}{4} = \frac{1}{2} \]

\[ \frac{BC}{RP} = \frac{2.5}{5} = \frac{1}{2} \]

\[ \frac{CA}{PQ} = \frac{3}{6} = \frac{1}{2} \]

Since corresponding sides are proportional, the triangles are similar.

Result: \( \Delta ABC \sim \Delta QRP \) (By SSS similarity criterion).

(iii) In \( \Delta LMP \) and \( \Delta DEF \):

\[ \frac{MP}{DE} = \frac{2}{4} = \frac{1}{2} \]

\[ \frac{LP}{DF} = \frac{3}{6} = \frac{1}{2} \]

\[ \frac{LM}{EF} = \frac{2.7}{5} \neq \frac{1}{2} \]

The sides are not proportional.

Result: Not similar.

(iv) In \( \Delta MNL \) and \( \Delta QPR \):

\[ \frac{MN}{QP} = \frac{2.5}{5} = \frac{1}{2} \]

\[ \frac{ML}{QR} = \frac{5}{10} = \frac{1}{2} \]

\[ \angle M = \angle Q = 70^{\circ} \]

Two sides are proportional and the included angle is equal.

Result: \( \Delta MNL \sim \Delta QPR \) (By SAS similarity criterion).

(v) In \( \Delta ABC \) and \( \Delta FDE \):

\[ \angle A = 80^{\circ} \]

\[ \angle F = 80^{\circ} \]

However, in \( \Delta ABC \), the known angle \( \angle A \) is not included between the known sides AB and BC (BC side length is not given relative to AC). Therefore, we cannot establish proportionality for the included angle.

Result: Not similar.

(vi) In \( \Delta DEF \):

\[ \angle F = 180^{\circ} - (70^{\circ} + 80^{\circ}) = 30^{\circ} \]

In \( \Delta PQR \):

\[ \angle P = 180^{\circ} - (80^{\circ} + 30^{\circ}) = 70^{\circ} \]

Now comparing angles:

\[ \angle D = \angle P = 70^{\circ} \]

\[ \angle E = \angle Q = 80^{\circ} \]

\[ \angle F = \angle R = 30^{\circ} \]

Result: \( \Delta DEF \sim \Delta PQR \) (By AAA similarity criterion).

Solution:

1. Finding \( \angle DOC \):

DOB is a straight line. Thus, angles on a straight line add up to \( 180^{\circ} \).

\[ \angle DOC + \angle BOC = 180^{\circ} \]

\[ \angle DOC + 125^{\circ} = 180^{\circ} \]

\[ \angle DOC = 180^{\circ} - 125^{\circ} = 55^{\circ} \]

2. Finding \( \angle DCO \):

In \( \Delta ODC \), the sum of angles is \( 180^{\circ} \).

\[ \angle DCO + \angle CDO + \angle DOC = 180^{\circ} \]

\[ \angle DCO + 70^{\circ} + 55^{\circ} = 180^{\circ} \]

\[ \angle DCO + 125^{\circ} = 180^{\circ} \]

\[ \angle DCO = 55^{\circ} \]

3. Finding \( \angle OAB \):

It is given that \( \Delta ODC \sim \Delta OBA \).

Therefore, corresponding angles are equal.

\[ \angle OAB = \angle OCD \]

\[ \angle OAB = 55^{\circ} \]

Answer: \( \angle DOC = 55^{\circ} \), \( \angle DCO = 55^{\circ} \), \( \angle OAB = 55^{\circ} \).

Solution:

In \( \Delta DOC \) and \( \Delta BOA \):

1. \( AB \parallel DC \) implies alternate interior angles are equal:
   • \( \angle OCD = \angle OAB \) (Alternate interior angles)
   • \( \angle ODC = \angle OBA \) (Alternate interior angles)
2. \( \angle DOC = \angle BOA \) (Vertically opposite angles)

Therefore, by AAA similarity criterion:

\[ \Delta DOC \sim \Delta BOA \]

Since the triangles are similar, the ratio of their corresponding sides is equal:

\[ \frac{DO}{BO} = \frac{OC}{OA} \]

Rearranging the terms:

\[ \frac{OA}{OC} = \frac{OB}{OD} \]

Hence shown.

Solution:

Step 1: In \( \Delta PQR \), we are given \( \angle 1 = \angle 2 \).

This implies that sides opposite to equal angles are equal.

\[ PQ = PR \quad \dots(1) \]

Step 2: We are given the ratio:

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

Substitute \( PR = PQ \) from equation (1):

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

Taking the reciprocal:

\[ \frac{QS}{QR} = \frac{PQ}{QT} \quad \dots(2) \]

Step 3: In \( \Delta PQS \) and \( \Delta TQR \):

1. \( \frac{QS}{QR} = \frac{PQ}{QT} \) (From equation 2)
2. \( \angle PQS = \angle TQR \) (Common angle \( \angle Q \))

Therefore, by SAS similarity criterion:

\[ \Delta PQS \sim \Delta TQR \]

Hence proved.

Solution:

In \( \Delta RPQ \) and \( \Delta RTS \):

1. \( \angle RPQ = \angle RTS \) (Given in the question)
2. \( \angle PRQ = \angle TRS \) (Common angle \( \angle R \))

Since two angles are equal, the third angle must also be equal.

Therefore, by AA similarity criterion:

\[ \Delta RPQ \sim \Delta RTS \]

Hence shown.

Solution:

Step 1: Given \( \Delta ABE \cong \Delta ACD \).

By CPCT (Corresponding Parts of Congruent Triangles):

• \( AB = AC \quad \dots(1) \)
• \( AE = AD \quad \dots(2) \)

Step 2: Divide equation (2) by equation (1):

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

Step 3: In \( \Delta ADE \) and \( \Delta ABC \):

1. \( \frac{AD}{AB} = \frac{AE}{AC} \) (Proved above)
2. \( \angle DAE = \angle BAC \) (Common angle \( \angle A \))

Therefore, by SAS similarity criterion:

\[ \Delta ADE \sim \Delta ABC \]

Hence shown.

Solution:

(i) In \( \Delta AEP \) and \( \Delta CDP \):

• \( \angle AEP = \angle CDP = 90^{\circ} \) (Altitudes)
• \( \angle APE = \angle CPD \) (Vertically opposite angles)

Result: \( \Delta AEP \sim \Delta CDP \) (AA criterion).

(ii) In \( \Delta ABD \) and \( \Delta CBE \):

• \( \angle ADB = \angle CEB = 90^{\circ} \) (Altitudes)
• \( \angle ABD = \angle CBE \) (Common angle \( \angle B \))

Result: \( \Delta ABD \sim \Delta CBE \) (AA criterion).

(iii) In \( \Delta AEP \) and \( \Delta ADB \):

• \( \angle AEP = \angle ADB = 90^{\circ} \) (Altitudes)
• \( \angle PAE = \angle DAB \) (Common angle \( \angle A \))

Result: \( \Delta AEP \sim \Delta ADB \) (AA criterion).

(iv) In \( \Delta PDC \) and \( \Delta BEC \):

• \( \angle PDC = \angle BEC = 90^{\circ} \) (Altitudes)
• \( \angle PCD = \angle BCE \) (Common angle \( \angle C \))

Result: \( \Delta PDC \sim \Delta BEC \) (AA criterion).

Solution:

In \( \Delta ABE \) and \( \Delta CFB \):

1. \( \angle A = \angle C \) (Opposite angles of a parallelogram are equal).
2. \( \angle AEB = \angle CBF \) (Alternate interior angles, since \( AE \parallel BC \) and BE is the transversal).

Therefore, by AA similarity criterion:

\[ \Delta ABE \sim \Delta CFB \]

Hence shown.

Solution:

(i) In \( \Delta ABC \) and \( \Delta AMP \):

• \( \angle ABC = \angle AMP = 90^{\circ} \) (Given)
• \( \angle A = \angle A \) (Common angle)

Therefore, \( \Delta ABC \sim \Delta AMP \) (AA similarity criterion).

(ii)

Since \( \Delta ABC \sim \Delta AMP \), the ratio of their corresponding sides must be equal.

The sides opposite to equal angles correspond to each other:

• Hypotenuse AC corresponds to Hypotenuse AP.
• Side BC (opposite \( \angle A \)) corresponds to side MP (opposite \( \angle A \)).

Thus:

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

Hence proved.

Solution:

Given: \( \Delta ABC \sim \Delta FEG \). This implies:

• \( \angle A = \angle F \)
• \( \angle B = \angle E \)
• \( \angle C = \angle G \) (Since bisectors split these angles equally, \( \angle ACD = \angle FGH \) and \( \angle DCB = \angle HGE \)).

(i) & (iii) Prove \( \Delta DCA \sim \Delta HGF \):

In \( \Delta DCA \) and \( \Delta HGF \):

1. \( \angle A = \angle F \) (From similarity of large triangles)
2. \( \angle ACD = \angle FGH \) (Halves of equal angles \( \angle C \) and \( \angle G \))

Therefore, \( \Delta DCA \sim \Delta HGF \) (AA criterion).

Since these triangles are similar, their sides are proportional:

\[ \frac{CD}{GH} = \frac{AC}{FG} \]

(ii) Prove \( \Delta DCB \sim \Delta HGE \):

In \( \Delta DCB \) and \( \Delta HGE \):

1. \( \angle B = \angle E \) (From similarity of large triangles)
2. \( \angle DCB = \angle HGE \) (Halves of equal angles \( \angle C \) and \( \angle G \))

Therefore, \( \Delta DCB \sim \Delta HGE \) (AA criterion).

Solution:

Since \( \Delta ABC \) is isosceles with \( AB = AC \):

\[ \angle ABD = \angle ECF \quad (\text{Angles opposite equal sides}) \]

In \( \Delta ABD \) and \( \Delta ECF \):

1. \( \angle ADB = \angle EFC = 90^{\circ} \) (Given altitudes)
2. \( \angle ABD = \angle ECF \) (Proved above)

Therefore, \( \Delta ABD \sim \Delta ECF \) (AA similarity criterion).

Hence proved.

Solution:

Given:

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

Since AD and PM are medians, \( BD = \frac{1}{2}BC \) and \( QM = \frac{1}{2}QR \).

Substitute these into the proportion:

\[ \frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM} \]

\[ \Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM} \]

Since the sides of \( \Delta ABD \) and \( \Delta PQM \) are proportional:

\[ \Delta ABD \sim \Delta PQM \quad (\text{SSS criterion}) \]

From this similarity, corresponding angles are equal:

\[ \angle B = \angle Q \]

Now in \( \Delta ABC \) and \( \Delta PQR \):

1. \( \frac{AB}{PQ} = \frac{BC}{QR} \) (Given)
2. \( \angle B = \angle Q \) (Proved above)

Therefore, \( \Delta ABC \sim \Delta PQR \) (SAS similarity criterion).

Solution:

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

1. \( \angle ADC = \angle BAC \) (Given)
2. \( \angle ACD = \angle BCA \) (Common angle \( \angle C \))

Therefore, \( \Delta ADC \sim \Delta BAC \) (AA similarity criterion).

Since the triangles are similar, corresponding sides are proportional:

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

Cross-multiplying gives:

\[ CA \times CA = CB \times CD \]

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

Hence shown.

Solution:

Given: \( \frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM} \)

Construction: Extend AD to E such that \( AD = DE \) and join CE. Similarly, extend PM to L such that \( PM = ML \) and join RL.

Step 1: In quadrilateral ABEC, diagonals bisect each other at D. Thus, ABEC is a parallelogram. Therefore, \( AC = BE \) and \( AB = CE \). Similarly, \( PQ = RL \) and \( PR = QL \).

Step 2: From the given proportion:

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

Substitute \( AB = CE \), \( PQ = RL \), \( AC = BE \), \( PR = QL \):

Also, \( AD = \frac{1}{2}AE \) and \( PM = \frac{1}{2}PL \).

So, the ratio becomes:

\[ \frac{CE}{RL} = \frac{BE}{QL} = \frac{AE}{PL} \]

This implies \( \Delta BEC \sim \Delta QLR \) (SSS). Wait, let's simplify focusing on \( \Delta ABE \) and \( \Delta PQL \).

From construction, \( AB/PQ = BE/QL = AE/PL \) (since \( AC=BE, PR=QL \)).

Thus \( \Delta ABE \sim \Delta PQL \).

Consequently, \( \angle BAE = \angle QPM \).

Similarly, we can prove \( \Delta ACE \sim \Delta PRL \), giving \( \angle CAE = \angle RPM \).

Step 3: Adding the angles:

\[ \angle BAE + \angle CAE = \angle QPM + \angle RPM \]

\[ \angle BAC = \angle QPR \]

Step 4: In \( \Delta ABC \) and \( \Delta PQR \):

1. \( \frac{AB}{PQ} = \frac{AC}{PR} \) (Given)
2. \( \angle A = \angle P \) (Proved)

Therefore, \( \Delta ABC \sim \Delta PQR \) (SAS similarity).

Solution:

Let the height of the tower be \( h \) meters.

Since the shadows are cast at the same time, the angle of elevation of the sun is the same for both objects.

Therefore, the triangle formed by the pole and its shadow is similar to the triangle formed by the tower and its shadow (AA similarity: \( 90^{\circ} \) angle and Sun's angle).

\[ \frac{\text{Height of Pole}}{\text{Height of Tower}} = \frac{\text{Length of Pole Shadow}}{\text{Length of Tower Shadow}} \]

\[ \frac{6}{h} = \frac{4}{28} \]

\[ \frac{6}{h} = \frac{1}{7} \]

\[ h = 6 \times 7 \]

\[ h = 42 \text{ m} \]

Answer: The height of the tower is 42 m.

Solution:

Given \( \Delta ABC \sim \Delta PQR \). This implies:

1. Corresponding angles are equal: \( \angle B = \angle Q \).
2. Corresponding sides are proportional: \( \frac{AB}{PQ} = \frac{BC}{QR} \).

Since AD and PM are medians, D and M are midpoints of BC and QR respectively.

Therefore, \( BC = 2BD \) and \( QR = 2QM \).

Substituting this into the side ratio:

\[ \frac{AB}{PQ} = \frac{2BD}{2QM} \]

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

Now in \( \Delta ABD \) and \( \Delta PQM \):

1. \( \frac{AB}{PQ} = \frac{BD}{QM} \) (Proved above)
2. \( \angle B = \angle Q \) (Given from similarity)

Therefore, \( \Delta ABD \sim \Delta PQM \) (SAS similarity criterion).

Since these triangles are similar, their corresponding sides are proportional:

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

Hence proved.