Request a call back

Join NOW to get access to exclusive study material for best results

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.

CBSE Class 10 NCERT solutions Triangles-Ex6_1_2-1

Two squares with sides 1 cm and 2 cm

CBSE Class 10 NCERT solutions Triangles-Ex6_1_2-2

(ii) Trapezium and Square

CBSE Class 10 NCERT solutions Triangles-Ex6_1_2-3

Triangle and Parallelogram

CBSE Class 10 NCERT solutions Triangles-Ex6_1_2-4

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)

CBSE Class 10 NCERT solutions Triangles-Ex6_2_1-1

Let EC = x

Since DE || BC.

Therefore, by basic proportionality theorem,

CBSE Class 10 NCERT solutions Triangles-Ex6_2_1-2

(ii)

CBSE Class 10 NCERT solutions Triangles-Ex6_2_1-3

Let AD = x

Since DE || BC,

Therefore by basic proportionality theorem,

CBSE Class 10 NCERT solutions Triangles-Ex6_2_1-4

Solution 2

(i)

CBSE Class 10 NCERT solutions Triangles-Ex6_2_2-1

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

Now,

CBSE Class 10 NCERT solutions Triangles-Ex6_2_2-2

(ii)

CBSE Class 10 NCERT solutions Triangles-Ex6_2_2-3

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

CBSE Class 10 NCERT solutions Triangles-Ex6_2_2-4

(iii)

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

CBSE Class 10 NCERT solutions Triangles-Ex6_2_2-5

Solution 3

In the given figure

Since LM || CB,

Therefore by basic proportionality theorem,

 

Solution 4

CBSE Class 10 NCERT solutions Triangles-Ex6_2_4-1

In Δ ABC,

Since DE || AC

CBSE Class 10 NCERT solutions Triangles-Ex6_2_4-2

CBSE Class 10 NCERT solutions Triangles-Ex6_2_4-3

Solution 5

CBSE Class 10 NCERT solutions Triangles-Ex6_2_5-1

In Δ POQ

Since DE || OQ

CBSE Class 10 NCERT solutions Triangles-Ex6_2_5-2

CBSE Class 10 NCERT solutions Triangles-Ex6_2_5-3

CBSE Class 10 NCERT solutions Triangles-Ex6_2_5-4

Solution 6

CBSE Class 10 NCERT solutions Triangles-Ex6_2_6

Solution 7

CBSE Class 10 NCERT solutions Triangles-Ex6_2_7-1

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

CBSE Class 10 NCERT solutions Triangles-Ex6_2_7-2

i.e. AQ = QC

Or, Q is midpoint of AC.

Solution 8

CBSE Class 10 NCERT solutions Triangles-Ex6_2_8-1

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

CBSE Class 10 NCERT solutions Triangles-Ex6_2_8-2

And hence basic proportionality theorem is verified

So, PQ||BC

Solution 9

CBSE Class 10 NCERT solutions Triangles-Ex6_2_9

Solution 10

CBSE Class 10 NCERT solutions Triangles-Ex6_2_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]

CBSE Class 10 NCERT solutions Triangles-Ex6_3_1

(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

CBSE Class 10 NCERT solutions Triangles-Ex6_3_3-1

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)

CBSE Class 10 NCERT solutions Triangles-Ex6_3_3-2

Solution 4

In Δ PQR

PQR = PRQ

Therefore PQ = PR    (i)

Given,

CBSE Class 10 NCERT solutions Triangles-Ex6_3_4

Solution 5

CBSE Class 10 NCERT solutions Triangles-Ex6_3_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)

CBSE Class 10 NCERT solutions Triangles-Ex6_3_6

Solution 7

(i)

CBSE Class 10 NCERT solutions Triangles-Ex6_3_7-1

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)

 CBSE Class 10 NCERT solutions Triangles-Ex6_3_7-2

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)

CBSE Class 10 NCERT solutions Triangles-Ex6_3_7-3

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)

CBSE Class 10 NCERT solutions Triangles-Ex6_3_7-4

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

CBSE Class 10 NCERT solutions Triangles-Ex6_3_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)

CBSE Class 10 NCERT solutions Triangles-Ex6_3_9

Solution 10

CBSE Class 10 NCERT solutions Triangles-Ex6_3_10-1

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)

CBSE Class 10 NCERT solutions Triangles-Ex6_3_10-2

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

CBSE Class 10 NCERT solutions Triangles-Ex6_3_12-1 

Median divides opposite side.

CBSE Class 10 NCERT solutions Triangles-Ex6_3_12-2

CBSE Class 10 NCERT solutions Triangles-Ex6_3_12-3

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

CBSE Class 10 NCERT solutions Triangles-Ex6_3_13-1

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

CBSE Class 10 NCERT solutions Triangles-Ex6_3_13-2

Solution 14

CBSE Class 10 NCERT solutions Triangles-Ex6_3_14

Solution 15

m

CBSE Class 10 NCERT solutions Triangles-Ex6_3_15-1

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

CBSE Class 10 NCERT solutions Triangles-Ex6_3_15-2

So, height of tower will be 42 meters.

Solution 16

CBSE Class 10 NCERT solutions Triangles-Ex6_3_16-1

Since Δ ABC ~ ΔPQR

So, their respective sides will be in proportion

CBSE Class 10 NCERT solutions Triangles-Ex6_3_16-2

Also, A = P, B = Q, C = R                     (2)

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

CBSE Class 10 NCERT solutions Triangles-Ex6_3_16-3

From equation (1) and (3)

CBSE Class 10 NCERT solutions Triangles-Ex6_3_16-4

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)

CBSE Class 10 NCERT solutions Triangles-Ex6_3_16-5