# Class 10 NCERT Solutions Maths Chapter 6 - Triangles

## Triangles Exercise Ex. 6.1

### Solution 1

(i) All circles are SIMILAR.

(ii) All squares are SIMILAR.

(iii) All EQUILATERAL triangles are similar.

(iv) Two polygons of same number of sides are similar, if their corresponding angles are EQUAL and their corresponding sides are PROPORTIONAL.

### Solution 2

(i) Two equilateral triangles with sides 1 cm and 2 cm.

Two squares with sides 1 cm and 2 cm

(ii) Trapezium and Square

Triangle and Parallelogram

### Solution 3

Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional i.e. 1:2 but their corresponding angles are not equal.

## Triangles Exercise Ex. 6.2

### Solution 1

(i)

Let EC = x

Since DE || BC.

Therefore, by basic proportionality theorem,

(ii)

Let AD = x

Since DE || BC,

Therefore by basic proportionality theorem,

### Solution 2

(i)

Given that PE = 3.9, EQ = 3, PF = 3.6, FR = 2.4

Now,

(ii)

PE = 4, QE = 4.5, PF = 8, RF = 9

(iii)

PQ = 1.28, PR = 2.56, PE = 0.18, PF = 0.36

### Solution 3

In the given figure

Since LM || CB,

Therefore by basic proportionality theorem,

### Solution 4

In Δ ABC,

Since DE || AC

### Solution 5

In Δ POQ

Since DE || OQ

### Solution 6

### Solution 7

Consider the given figure

PQ is a line segment drawn through midpoint P of line AB such that PQ||BC

i.e. AP = PB

Now, by basic proportionality theorem

i.e. AQ = QC

Or, Q is midpoint of AC.

### Solution 8

Consider the given figure

PQ is a line segment joining midpoints P and Q of line AB and AC respectively.

i.e. AP = PB and AQ = QC

Now, we may observe that

And hence basic proportionality theorem is verified

So, PQ||BC

### Solution 9

### Solution 10

## Triangles Exercise Ex. 6.3

### Solution 1

(i) ∠A = ∠P = 60°

∠B = ∠Q = 80

∠C = ∠R = 40°

Therefore Δ ABC ~ Δ PQR [by AAA rule]

(iii) Triangles are not similar as the corresponding sides are not proportional.

(iv) Triangles are not similar as the corresponding sides are not proportional.

(v) Triangles are not similar as the corresponding sides are not proportional.

(vi) In Δ DEF

∠D + ∠E + ∠F = 180°

(Sum of measures of angles of a triangle is 180)

70° + 80° + ∠F = 180°

∠F = 30°

Similarly in ∠PQR

∠P + ∠Q + ∠R = 180°

(Sum of measures of angles of a triangle is 180)

∠P + 80° +30° = 180°

∠P = 70°

Now In Δ DEF and Δ PQR

∠D = ∠P = 70°

∠E = ∠Q = 80°

∠F = ∠R = 30°

Therefore Δ DEF ~ Δ PQR [by AAA rule]

### Solution 2

Since DOB is a straight line

Therefore ∠DOC + ∠COB = 180°

Therefore ∠DOC = 180° - 125°

= 55°

In ∠DOC,

∠DCO + ∠CDO + ∠DOC = 180°

∠DCO + 70° + 55° = 180°

∠DCO = 55°

Since Δ ODC ~ Δ OBA,

Therefore ∠OCD = ∠OAB [corresponding angles equal in similar triangles]

Therefore ∠OAB = 55°

### Solution 3

In Δ DOC and Δ BOA

AB || CD

Therefore ∠CDO = ∠ABO [Alternate interior angles]

∠DCO = ∠BAO [Alternate interior angles]

∠DOC = ∠BOA [Vertically opposite angles]

Therefore Δ DOC ~ Δ BOA [AAA rule)

### Solution 4

In Δ PQR

∠PQR = ∠PRQ

Therefore PQ = PR (i)

Given,

### Solution 5

In Δ RPQ and Δ RST

∠RTS = ∠QPS [given]

∠R = ∠R [common angle]

∠RST = ∠RQP [ Remaining angles]

Therefore Δ RPQ ~ Δ RTS [by AAA rule]

### Solution 6

Since Δ ABE ≅ Δ ACD

Therefore AB = AC (1)

AD = AE (2)

Now, in Δ ADE and Δ ABC,

Dividing equation (2) by (1)

### Solution 7

(i)

In Δ AEP and Δ CDP

Since ∠CDP = ∠AEP = 90°

∠CPD = ∠APE (vertically opposite angles)

∠PCD = ∠PAE (remaining angle)

Therefore by AAA rule,

Δ AEP ~ Δ CDP

(ii)

In Δ ABD and Δ CBE

∠ADB = ∠CEB = 90°

∠ABD = ∠CBE (common angle)

∠DAB = ∠ECB (remaining angle)

Therefore by AAA rule

Δ ABD ~ Δ CBE

(iii)

In Δ AEP and Δ ADB

∠AEP = ∠ADB = 90°

∠PAE = ∠DAB (common angle)

∠APE = ∠ABD (remaining angle)

Therefore by AAA rule

Δ AEP ~ Δ ADB

(iv)

In Δ PDC and Δ BEC

∠PDC = ∠BEC = 90°

∠PCD = ∠BCE (common angle)

∠CPD = ∠CBE

Therefore by AAA rule

Δ PDC ~ Δ BEC

### Solution 8

Δ ABE and Δ CFB

∠A = ∠C (opposite angles of a parallelogram)

∠AEB = ∠CBF (Alternate interior angles AE || BC)

∠ABE = ∠CFB (remaining angle)

Therefore Δ ABE ~ Δ CFB (by AAA rule)

### Solution 9

In Δ ABC and Δ AMP

∠ABC = ∠AMP = 90

∠A = ∠A (common angle)

∠ACB = ∠APM (remaining angle)

Therefore Δ ABC ~ Δ AMP (by AAA rule)

### Solution 10

Since Δ ABC ~ Δ FEG

Therefore ∠A = ∠F

∠B = ∠E

As, ∠ACB = ∠FGE

Therefore ∠ACD = ∠FGH (angle bisector)

And ∠DCB = ∠HGE (angle bisector)

Therefore Δ ACD ~ Δ FGH (by AAA rule)

And Δ DCB ~ Δ HGE (by AAA rule)

and ∠ACD = ∠FGH

Therefore Δ DCA ~ Δ HGF (by SAS rule)

### Solution 11

In Δ ABD and Δ ECF,

Given that AB = AC (isosceles triangles)

So, ∠ABD = ∠ECF

∠ADB = ∠EFC = 90

∠BAD = ∠CEF

Therefore ΔABD ~ ΔECF (by AAA rule)

### Solution 12

Median divides opposite side.

Therefore Δ ABD ~ Δ PQM (by SSS rule)

Therefore ∠ABD = ∠PQM (corresponding angles of similar triangles)

Therefore Δ ABC ~ Δ PQR (by SAS rule)

### Solution 13

In Δ ADC and Δ BAC

Given that ∠ADC = ∠BAC

∠ACD = ∠BCA (common angle)

∠CAD = ∠CBA (remaining angle)

Hence, Δ ADC ~ Δ BAC [by AAA rule]

So, corresponding sides of similar triangles will be proportional to each other

### Solution 14

### Solution 15

m

Let AB be a tower

CD be a pole

Shadow of AB is BE

Shadow of CD is DF

The light rays from sun will fall on tower and pole at same angle and at the same time.

So, ∠DCF = ∠BAE

And ∠DFC = ∠BEA

∠CDF = ∠ABE (tower and pole are vertical to ground)

Therefore Δ ABE ~ Δ CDF

So, height of tower will be 42 meters.

### Solution 16

Since Δ ABC ~ ΔPQR

So, their respective sides will be in proportion

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R (2)

Since, AD and PM are medians so they will divide their opposite sides in equal halves.

From equation (1) and (3)

So, we had observed that two respective sides are in same proportion in both triangles and also angle included between them is respectively equal

Hence, Δ ABD ~ Δ PQM (by SAS rule)