# NCERT Solutions for Class 10 Maths Chapter 6 - Triangles

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## Chapter 6 - Triangles Exercise Ex. 6.4

Let us assume two similar triangles as ABC ~ PQR. Let AD and PS be the medians of these triangles.

A = P, B = Q, C = R

Since, AD and PS are medians

Since AB || CD

OAB = OCD (Alternate interior angles)

OBA = ODC (Alternate interior angles)

AOB = COD (Vertically opposite angles)

Therefore AOB ~ COD (By AAA rule)

Since ABC and DBC are one same base,

Therefore ratio between their areas will be as ratio of their heights.

Let us draw two perpendiculars AP and DM on line BC.

In APO and DMO,

APO = DMO = 90

AOP = DOM (vertically opposite angles)

OAP = ODM (remaining angle)

Therefore APO ~ DMO (By AAA rule)

Since D and E are mid points of ABC

Let us assume two similar triangles as ABC ~ PQR. Let AD and PS be the medians of these triangles.

A = P, B = Q, C = R

Since, AD and PS are medians

Let ABCD be a square of side a.

Therefore its diagonal

Two desired equilateral triangles are formed as ABE and DBF

Side of an equilateral triangle ABE described on one of its side = a

Side of an equilateral triangle DBF described on one of its diagonal

We know that equilateral triangles are having all its angles as 60º and all its sides of same length. So, all equilateral triangles are similar to each other. So, ratio between areas of these triangles will be equal to the square of the ratio between sides of these triangles.

We know that equilateral triangles are having all its angles as 60º and all its sides of same length. So, all equilateral triangles are similar to each other. So, ratio between areas of these triangles will be equal to the square of the ratio between sides of these triangles.

Let side of ABC = x

Hence, (c)

Given that sides are in the ratio 4:9.

Hence, (d).

## Chapter 6 - Triangles Exercise Ex. 6.1

(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.

Two squares with sides 1 cm and 2 cm

(ii) Trapezium and Square

Triangle and Parallelogram

## Chapter 6 - Triangles Exercise Ex. 6.2

Let EC = x

Since DE || BC.

Therefore, by basic proportionality theorem,

(ii)

Let AD = x

Since DE || BC,

Therefore by basic proportionality theorem,

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

In the given figure

Since LM || CB,

Therefore by basic proportionality theorem,

In ABC,

Since DE || AC

In POQ

Since DE || OQ

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.

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

## Chapter 6 - Triangles Exercise Ex. 6.3

(i) A = P = 60°

B = Q = 80°

C = R = 40°

Therefore ABC ~ 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 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 DEF and PQR

D = P = 70°

E = Q = 80°

F = R = 30°

Therefore DEF ~ PQR [by AAA rule]

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 ODC ~ OBA,

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

Therefore OAB = 55°

In DOC and BOA

AB || CD

Therefore CDO = ABO [Alternate interior angles]

DCO = BAO [Alternate interior angles]

DOC = BOA [Vertically opposite angles]

Therefore DOC ~ BOA [AAA rule)

PQR = PRQ

Therefore PQ = PR (i)

Given,

In RPQ and RST

RTS = QPS [given]

R = R [common angle]

RST = RQP [ Remaining angles]

Therefore RPQ ~ RTS [by AAA rule]

Since ABE ACD

Therefore AB = AC (1)

AD = AE (2)

Now, in ADE and ABC,

Dividing equation (2) by (1)

(i)

In AEP and CDP

Since CDP = AEP = 90°

CPD = APE (vertically opposite angles)

PCD = PAE (remaining angle)

Therefore by AAA rule,

AEP ~ CDP

(ii)

In ABD and CBE

ADB = CEB = 90°

ABD = CBE (common angle)

DAB = ECB (remaining angle)

Therefore by AAA rule

ABD ~ CBE

(iii)

In AEP and ADB

AEP = ADB = 90°

PAE = DAB (common angle)

APE = ABD (remaining angle)

Therefore by AAA rule

AEP ~ ADB

(iv)

In PDC and BEC

PDC = BEC = 90°

PCD = BCE (common angle)

CPD = CBE

Therefore by AAA rule

PDC ~ BEC

ABE and CFB

A = C (opposite angles of a parallelogram)

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

ABE = CFB (remaining angle)

Therefore ABE ~ CFB (by AAA rule)

In ABC and AMP

ABC = AMP = 90

A = A (common angle)

ACB = APM (remaining angle)

Therefore ABC ~ AMP (by AAA rule)

Since ABC ~ FEG

Therefore A = F

B = E

As, ACB = FGE

Therefore ACD = FGH (angle bisector)

And DCB = HGE (angle bisector)

Therefore ACD ~ FGH (by AAA rule)

And DCB ~ HGE (by AAA rule)

and ACD = FGH

Therefore DCA ~ HGF (by SAS rule)

In ABD and ECF,

Given that AB = AC (isosceles triangles)

So, ABD = ECF

ADB = EFC = 90

BAD = CEF

Therefore ABD ~ ECF (by AAA rule)

Median divides opposite side.

Therefore ABD ~ PQM (by SSS rule)

Therefore ABD = PQM (corresponding angles of similar triangles)

Therefore ABC ~ PQR (by SAS rule)

In ADC and BAC

Given that ADC = BAC

ACD = BCA (common angle)

CAD = CBA (remaining angle)

Hence, ADC ~ BAC [by AAA rule]

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

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 ABE ~ CDF

So, height of tower will be 42 meters.

Since ABC ~ PQR

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, ABD ~ PQM (by SAS rule)

## Chapter 6 - Triangles Exercise Ex. 6.5

i.Given that sides are 7 cm, 24 cm, and 25 cm. Squaring the lengths of these sides we get 49, 576, and 625.

Clearly, 49 + 576 = 625 or 7^{2} + 24^{2} = 25^{2} .

Therefore, given triangle is satisfying Pythagoras theorem. So, it is a right triangle. The longest side in a right angled triangle is the hypotenuse.

Therefore length of hypotenuse of this triangle = 25 cm.

ii.Given that sides are 3 cm, 8 cm, and 6 cm. Squaring the lengths of these sides we may get 9, 64, and 36. Clearly, sum of squares of lengths of two sides is not equal to square of length of third side. Therefore given triangle is not satisfying Pythagoras theorem. So, it is not a right triangle

iii.Given that sides are 50 cm, 80 cm, and 100 cm. Squaring the lengths of these sides we may get 2500, 6400, and 10000. Clearly, sum of squares of lengths of two sides is not equal to square of length of third side. Therefore given triangle is not satisfying Pythagoras theorem. So, it is not a right triangle.

iv.Given that sides are 13 cm, 12 cm, and 5 cm. Squaring the lengths of these sides we may get 169, 144, and 25. Clearly, 144 +25 = 169 Or, 12^{2} + 5^{2} = 13^{2}.

Therefore given triangle is satisfying Pythagoras theorem. So, it is a right triangle.

The longest side in a right angled triangle is the hypotenuse.

Therefore length of hypotenuse of this triangle = 13 cm.

iii. In DCA & DAB

DCA = DAB = 90Âº

CDA = ADB (common angle)

DAC = DBA (remaining angle)

Given that ABC is an isosceles triangle.

Therefore AC = CB

Applying Pythagoras theorem in ABC (i.e. right angled at point C)

Let AD be the altitude in given equilateral triangle ABC.

We know that altitude bisects the opposite side.

So, BD = DC = a

Since in an equilateral triangle, all the altitudes are equal in length.

So, length of each altitude will be

In AOB, BOC, COD, AOD

Applying Pythagoras theorem

Let OB be the pole and AB be the wire.

Therefore by Pythagoras theorem,

Let these distances are represented by OA and OB respectively.

Now applying Pythagoras theorem

Let CD and AB be the poles of height 11 and 6 m.

Therefore CP = 11 - 6 = 5 m

From the figure we may observe that AP = 12m

In APC, by applying Pythagoras theorem

Therefore distance between their tops = 13 m.

In ACE,

Let side of equilateral triangle be a. And AE be the altitude of ABC

Now, in ABE by applying Pythagoras theorem

AB

^{2}= AE

^{2}+ BE

^{2}

Given that AB = cm, AC = 12 cm and BC = 6 cm

We may observe that

AB

^{2}= 108

AC

^{2}= 144

And BC

^{2}= 36

AB

^{2}+BC

^{2}= AC

^{2}

Thus the given triangle ABC is satisfying Pythagoras theorem

Therefore triangle is a right angled triangle right angled at B

Therefore B = 90°.

Hence, (c).

## Chapter 6 - Triangles Exercise Ex. 6.6

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given that PS is angle bisector of QPR.

QPS = SPR (1)

SPR = PRT (As PS || TR) (2)

QPS = QTR (As PS || TR) (3)

Using these equations we may find

PRT = QTR from (2) and (3)

So, PT = PR (Since PTR is isosceles triangle)

Now in QPS and QTR

QSP = QRT (As PS || TR)

QPS = QTR (As PS || TR)

Q is common

QPS ~ QTR

(i). Let us join DB.

DN || CB

DM || AB

So, DN = MB

DM = NB

The condition to be proved is the case when DNBM is a square or D is the midpoint of side AC.

Then CDB = ADB = 90°

2 + 3 = 90° (1)

In CDM

1 + 2 + DMC = 180°

1 + 2 = 90° (2)

In DMB

3 + DMB + 4 = 180°

3 + 4 = 90° (3)

From equation (1) and (2)

1 = 3

From equation (1) and (3)

2 = 4

BDM ~ DCM

(ii). Similarly in DBN

4 + 3 = 90° (4)

In DAN

5 + 6 = 90° (5)

In DAB

4 + 5 = 90° (6)

From equation (4) and (6)

3 = 5

From equation (5) and (6)

4 = 6

DNA ~ BND

AB

^{2}= AD

^{2}+ DB

^{2}(1)

In ACD applying Pythagoras theorem

AC

^{2}= AD

^{2}+ DC

^{2}

AC

^{2}= AD

^{2}+ (DB + BC)

^{2}

AC

^{2}= AD

^{2}+ DB

^{2}+ BC

^{2}+ 2DB x BC

Now using equation (1)

AC

^{2}= AB

^{2}+ BC

^{2}+ 2BC . BD

AD

^{2}+ DB

^{2}= AB

^{2}

AD

^{2}= AB

^{2}- DB

^{2}(1)

In ADC applying Pythagoras theorem

AD

^{2}+ DC

^{2}= AC

^{2}(2)

Now using equation (1)

AB

^{2}- BD

^{2}+ DC

^{2}= AC

^{2}

AB

^{2}- BD

^{2}+ (BC - BD)

^{2}= AC

^{2}

AC

^{2}= AB

^{2}- BD

^{2}+ BC

^{2}+ BD

^{2}- 2BC. BD

= AB

^{2}+ BC

^{2}- 2BC. BD

AM

^{2}+ MD

^{2}= AD

^{2 }(1)

In AMC

AM

^{2}+ MC

^{2}= AC

^{2}(2)

AM

^{2}+ (MD + DC)

^{2}= AC

^{2}

(AM

^{2}+ MD

^{2}) + DC

^{2}+ 2MD.DC = AC

^{2}

Using equation (1) we may get

AD

^{2}+ DC

^{2}+ 2MD.DC = AC

^{2}

(ii). In ABM applying Pythagoras theorem

AB

^{2}= AM

^{2}+ MB

^{2}

= (AD

^{2}- DM

^{2}) + MB

^{2}

= (AD

^{2}- DM

^{2}) + (BD - MD)

^{2}

= AD

^{2}- DM

^{2}+ BD

^{2}+ MD

^{2 - }2BD.MD

= AD

^{2}+ BD

^{2 - }2BD.MD

(iii). InAMB

AM

^{2}+ MB

^{2}= AB

^{2}(1)

In AMC

AM

^{2}+ MC

^{2}= AC

^{2}(2)

Adding equation (1) and (2)

2AM

^{2}+ MB

^{2}+ MC

^{2}= AB

^{2}+ AC

^{2}

2AM

^{2}+ (BD - DM)

^{2}+ (MD + DC)

^{2}= AB

^{2}+ AC

^{2}

2AM

^{2}+BD

^{2}+ DM

^{2}- 2BD.DM + MD

^{2}+ DC

^{2}+ 2MD.DC = AB

^{2}+ AC

^{2}

2AM

^{2}+ 2MD

^{2}+ BD

^{2}+ DC

^{2}+ 2MD (-BD + DC) = AB

^{2}+ AC

^{2}

Let ABCD be a parallelogram

Let us draw perpendicular DE on extended side AB and AF on side DC.

In DEA

DE

^{2}+ EA

^{2}= DA

^{2}(i)

In DEB

DE

^{2}+ EB

^{2}= DB

^{2}

DE

^{2}+ (EA + AB)

^{2}= DB

^{2}

(DE

^{2}+ EA

^{2}) + AB

^{2}+ 2EA. AB = DB

^{2}

DA

^{2}+ AB

^{2}+ 2EA.AB = DB

^{2}(ii)

In ADF

AD

^{2}= AF

^{2}+ FD

^{2}

In AFC

AC

^{2}= AF

^{2}+ FC

^{2}

= AF

^{2}+ (DC - FD)

^{2}

= AF

^{2}+ DC

^{2}+ FD

^{2}- 2DC - FD

= (AF

^{2}+ FD

^{2}) + DC

^{2}- 2DC . FD

AC

^{2}= AD

^{2}+ DC

^{2}- 2DC FD (iii)

Since ABCD is a parallelogram

AB = CD (iii)

And BC = AD (iv)

In DEA and ADF

DEA = AFD

EAD = FDA (EA || DF)

EDA = FAD (AF || ED)

AD is common in both triangles.

Since respective angles are same and respective sides are same

DEA AFD

So EA = DF (v)

Adding equation (ii) and (iii)

DA

^{2}+ AB

^{2}+ 2EA.AB + AD

^{2}+ DC

^{2}- 2DC.FD = DB

^{2}+ AC

^{2}

DA

^{2}+ AB

^{2}+ AD

^{2 }+ DC

^{2}+ 2EA.AB - 2DC.FD = DB

^{2}+ AC

^{2}

BC

^{2}+ AB

^{2}+ AD

^{2}+ DC

^{2}+ 2EA.AB-2AB.EA = DB

^{2}+ AC

^{2}

AB

^{2}+ BC

^{2}+ CD

^{2}+ DA

^{2}= AC

^{2}+ BD

^{2}

Let us join CB

(i) In APC and DPB

APC = DPB {Vertically opposite angles}

CAP = BDP {Angles in same segment for chord CB}

APC ~ DPB {BY AA similarly criterion}

(ii) We know that corresponding sides of similar triangles are proportional

(i) In PAC and PDB

P = P (common)

PAC = PDB (exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

PCA = PBD

PAC ~ PDB

(ii) We know that corresponding sides of similar triangles are proportional.

AD = AD (common)

So, DBA ~ DCA (By SSS)

Now, corresponding angles of similar triangles will be equal.

BAD = CAD

AD is angle bisector of BAC

Let AB be the height of tip of fishing rod from water surface. Let BC be the horizontal distance of fly from the tip of fishing rod.

Then, AC is the length of string.

AC can be found by applying Pythagoras theorem in ABC

AC^{2} = AB^{2} + BC^{2}

AC^{2} = (1.8)^{2} + (2.4)^{2}

AC^{2} = 3.24 + 5.76

AC^{2} = 9.00

AC = = 3

Now, she pulls string at rate of 5 cm per second.

So, string pulled in 12 seconds = 12 x 5 = 60 cm = 0.6 m

Let after 12 second Fly be at point D.

Length of string out after 12 second is AD

AD = AC - string pulled by Nazima in 12 seconds

= 3.00 - 0.6

= 2.4

In ADB

AB

^{2}+ BD

^{2}= AD

^{2}

(1.8)

^{2}+ BD

^{2}= (2.4)

^{2}

BD

^{2}= 5.76 - 3.24 = 2.52

BD = 1.587

Horizontal distance of fly = BD + 1.2

= 1.587 + 1.2

= 2.787

= 2.79 m

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### Other Chapters for CBSE Class 10 Mathematics

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