Chapter 6 Triangles - Ncert Solutions for Class 10 Mathematics CBSE

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Chapter 6 - Triangles Excercise 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.

Chapter 6 - Triangles Excercise 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 ABC,
Since DE || AC


Solution 5


In 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

Chapter 6 - Triangles Excercise Ex. 6.3

Solution 1

(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]

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 ODC ~ OBA,
Therefore  OCD = OAB    [corresponding angles equal in similar triangles]

Therefore  OAB = 55

Solution 3


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)
Solution 4
In PQR
PQR = PRQ
Therefore PQ = PR    (i)
Given,
Solution 5


In RPQ and RST
RTS = QPS              [given]
R = R            [common angle]
RST = RQP                      [ Remaining angles]
Therefore RPQ ~ RTS    [by AAA rule]
Solution 6
Since ABE ~ ACD
Therefore     AB = AC               (1)
AD = AE                (2)
Now, in ADE and ABC,
Dividing equation (2) by (1)
Solution 7

(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

Solution 8
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)

Solution 9

In ABC and AMP
ABC = AMP = 90
A = A                 (common angle)
ACB = APM            (remaining angle)
Therefore ABC ~ AMP        (by AAA rule)

Solution 10


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)

Solution 11

In ABD and ECF,
Given that AB = AC        (isosceles triangles)
So, ABD = ECF    
ADB = EFC = 90
BAD = CEF
Therefore ABD ~ ECF     (by AAA rule)

Solution 12
 

Median divides opposite side.


Therefore ABD ~ PQM    (by SSS rule)
Therefore ABD = PQM    (corresponding angles of similar triangles)
Therefore ABC ~ PQR     (by SAS rule)

Solution 13


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
Solution 14
Solution 15


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.

Solution 16


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 Excercise Ex. 6.4

Solution 1
Solution 2


Since AB || CD
OAB = OCD            (Alternate interior angles)
OBA = ODC            (Alternate interior angles)
AOB = COD            (Vertically opposite angles)
Therefore AOB ~ COD        (By AAA rule)
Solution 3


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)

Solution 4
Let us assume two similar triangles as ABC ~ PQR
Solution 5


Since D and E are mid points of ABC

Solution 6


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
Solution 7


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.
Solution 8


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)
Solution 9
If, two triangles are similar to each other, ratio between areas of these triangles will be equal to the square of the ratio between sides of these triangles.
Given that sides are in the ratio 4:9.



Hence, (d).

Chapter 6 - Triangles Excercise Ex. 6.5

Solution 1

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 72 + 242 = 252 .

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, 122 + 52 = 132.

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.

Solution 2



Solution 3
 

iii.     In DCA & DAB

DCA = DAB = 90º
         CDA = ADB            (common angle)
         DAC = DBA            (remaining angle)
       
Solution 4


Given that ABC is an isosceles triangle.
Therefore AC = CB
Applying Pythagoras theorem in ABC (i.e. right angled at point C)
Solution 5


Solution 6


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
Solution 7
 

In AOB, BOC, COD, AOD
Applying Pythagoras theorem
Solution 8


Solution 9


Solution 10
 

Let OB be the pole and AB be the wire.
Therefore by Pythagoras theorem,

Solution 11
 


Let these distances are represented by OA and OB respectively.
Now applying Pythagoras theorem

Solution 12

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.
Solution 13


          In ACE,


Solution 14
   
 
Solution 15

 


Solution 16


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

Now, in ABE by applying Pythagoras theorem
AB2 = AE2 + BE2
Solution 17


Given that AB = cm, AC = 12 cm and BC = 6 cm
We may observe that
AB2 = 108
AC2 = 144
And BC2 = 36
AB2 +BC2 = AC2

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 Excercise Ex. 6.6

Solution 1


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
Solution 2
(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
Solution 3
In ADB applying Pythagoras theorem
AB2 = AD2 + DB2        (1)
In ACD applying Pythagoras theorem
AC2 = AD2 + DC2
AC2 = AD2 + (DB + BC)2
AC2 = AD2 + DB2 + BC2 + 2DB  x  BC
Now using equation (1)
AC2 = AB2 + BC2 + 2BC . BD

Solution 4
In ADB applying Pythagoras theorem
AD2 + DB2 = AB2
AD2 = AB2 - DB2                (1)
In ADC applying Pythagoras theorem
AD2 + DC2 = AC2                (2)
Now using equation (1)
AB2 -  BD2 + DC2 = AC2
AB2 -  BD2 + (BC -  BD)2 = AC2
AC2 = AB2 -  BD2 + BC2 + BD2 - 2BC. BD
      = AB2 + BC2 - 2BC. BD

Solution 5
(i).  In AMD
AM2 + MD2 = AD2            (1)
In AMC
AM2 + MC2 = AC2            (2)
AM2 + (MD + DC)2 = AC2
(AM2 + MD2) + DC2 + 2MD.DC = AC2
Using equation (1) we may get
AD2 + DC2 + 2MD.DC = AC2


(ii).  In ABM applying Pythagoras theorem
AB2 = AM2 + MB2
= (AD2 - DM2) + MB2
= (AD2 -  DM2) + (BD -  MD)2
= AD2 -  DM2 + BD2 + MD2 - 2BD.MD
= AD2 + BD2 - 2BD.MD


(iii).  InAMB
AM2 + MB2 = AB2        (1)
In  AMC    
AM2 + MC2 = AC2        (2)
Adding equation (1) and (2)
2AM2 + MB2 + MC2 = AB2 + AC2
2AM2 + (BD - DM)2 + (MD + DC)2 = AB2 + AC2
2AM2+BD2 + DM2 - 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2
2AM2 + 2MD2 + BD2 + DC2 + 2MD (-BD + DC) = AB2 + AC2
Solution 6


Let  ABCD be a parallelogram
Let us draw perpendicular DE on extended side AB and AF on side DC.
In DEA
DE2 + EA2 = DA2                      (i)
In DEB
DE2 + EB2 = DB2
DE2 + (EA + AB)2 = DB2
(DE2 + EA2) + AB2 + 2EA. AB = DB2
DA2 + AB2 + 2EA.AB = DB2        (ii)
In ADF
AD2 = AF2 + FD2
In  AFC
AC2 = AF2 + FC2
= AF2 + (DC - FD)2
= AF2 + DC2 + FD2 - 2DC - FD
= (AF2 + FD2) + DC2 - 2DC . FD
AC2 = AD2 + DC2 - 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)
DA2 + AB2 + 2EA.AB + AD2 + DC2 - 2DC.FD = DB2 + AC2
DA2 + AB2 + AD2 + DC2 + 2EA.AB - 2DC.FD = DB2 + AC2
BC2 + AB2 + AD2 + DC2 + 2EA.AB-2AB.EA = DB2 + AC2
AB2 + BC2 + CD2 + DA2 = AC2 + BD2


Solution 7


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
Solution 8

(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.

Solution 9

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

Solution 10



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
AC2 = AB2 + BC2
AC2 = (1.8)2 + (2.4)2  
AC2 = 3.24 + 5.76  
AC2 = 9.00

AC =  = 3

Thus, length of string out is 3 m. 
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
AB2 + BD2 = AD2
(1.8)2 + BD2 = (2.4)2
BD2 = 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