NCERT Solution for Class 10 Mathematics Chapter 6 - Triangles RR ACADEMY MEERUT

NCERT Solution for Class 10 Mathematics Chapter 6 - Triangles 

Excercise 6.1

Question 1

Fill in the blanks using correct word given in the brackets:-
(i)     All circles are __________. (congruent, similar)
(ii)     All squares are __________. (similar, congruent)
(iii)     All __________ triangles are similar. (isosceles, equilateral)
(iv)    Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

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.

Question 2

Give two different examples of pair of
 
(i) Similar figures
(ii) Non-similar figures

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 

Question 3

State whether the following quadrilaterals are similar or not: 

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.

NCERT Solution for Class 10 Mathematics Chapter 6 - Triangles Page/Excercise 6.2

Question 1

In figure.  (i) and (ii) below, DE || BC. Find EC in (i) and AD in (ii). 

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,

Question 2

E and F are points on the sides PQ and PR respectively of a PQR. For each of the following cases, state whether EF || QR. 
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

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 

Question 3

In figure, if LM || CB and LN || CD, prove that 

Solution 3

 
In the given figure
Since LM || CB,
Therefore by basic proportionality theorem,
 

Question 4

In figure , DE || AC and DF || AE. Prove that 

Solution 4

 

In ABC,
Since DE || AC 
 

Question 5

In figure , DE || OQ and DF || OR, show that EF || QR. 

Solution 5

 

In POQ
Since DE || OQ
 
 

Question 6

Solution 6

Question 7

Using Basic proportionality Theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side.

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.

Question 8

Using converse of Basic Proportionality Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

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

Question 9

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that 

Solution 9

Question 10

The diagonals of a quadrilateral ABCD intersect each other at the point O such that  Show that ABCD is a trapezium.

Solution 10

NCERT Solution for Class 10 Mathematics Chapter 6 - Triangles Page/Excercise 6.3

Question 1

State which pairs of triangles in figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:    

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]

Question 2

In figure , ODC ~ OBA, BOC = 125� and CDO = 70�. Find DOC, DCO and OAB 

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�

Question 3

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that  

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)

Question 4

 

Solution 4

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

Question 5

S and T are point on sides PR and QR of PQR such that P = RTS. Show that RPQ ~ RTS.

Solution 5

 

In RPQ and RST
RTS = QPS              [given]
R = R            [common angle]
RST = RQP                      [ Remaining angles]
Therefore RPQ ~ RTS    [by AAA rule]

Question 6

In figure, If ABE ~ ACD, show that ADE ~ ABC. 

Solution 6

Since ABE ~ ACD
Therefore     AB = AC               (1)
AD = AE                (2)
Now, in ADE and ABC,
Dividing equation (2) by (1)

Question 7

In figure , altitudes AD and CE of ABC intersect each other at the point P. Show that: 

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

Question 8

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ABE ~ CFB

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)

Question 9

In figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that: 

 

Solution 9

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

Question 10

CD and GH are respectively the bisectors of ACB and EGF such that D and H lie on sides AB and FE of ABC and EFG respectively. If ABC ~ FEG, Show that: 

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) 

Question 11

In figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD  BC and EF  AC, prove that ABD ~ ECF 

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)

Question 12

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of PQR. 
Show that ABC ~ PQR.

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)

Question 13

D is a point on the side BC of a triangle ABC such that ADC = BAC. Show that 

CA2 = CB.CD

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

Question 14

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that 
ABC ~ PQR

Solution 14

Question 15

A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

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.

Question 16

If AD and PM are medians of triangles ABC and PQR, respectively where ABC ~ PQR prove that 

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)

NCERT Solution for Class 10 Mathematics Chapter 6 - Triangles Page/Excercise 6.4

Question 1

Let ABC ~ DEF their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

Solution 1

Question 2

Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

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)

Question 3

In figure 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that  

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)

Question 4

If the areas of two similar triangles are equal, prove that they are congruent.

Solution 4

Let us assume two similar triangles as ABC ~ PQR 

Question 5

D, E and F are respectively the mid-points of sides AB, BC and CA of ABC. Find the ratio of the area of DEF and ABC.

Solution 5

 

Since D and E are mid points of ABC

Question 6

Prove that the ratio of the areas of two similar triangles is equal to the square        
of the ratio of their corresponding medians.

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 

Question 7

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

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.

Question 8

ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
 
(A)    2 : 1
(B)    1 : 2
(C)    4 : 1
(D)    1 : 4

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)

Question 9

Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
 
(A)    2 : 3
(B)    4 : 9
(C)    81 : 16
(D)    16 : 81

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).

NCERT Solution for Class 10 Mathematics Chapter 6 - Triangles Page/Excercise 6.5

Question 1

Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i)  7 cm, 24 cm, 25 cm
(ii)  3 cm, 8 cm, 6 cm
(iii)  50 cm, 80 cm, 100 cm
(iv)  13 cm, 12 cm, 5 cm

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.

Question 2

PQR is a triangle right angled at P and M is a point on QR such that PM  QR. Show that PM2 = QM x MR.

Solution 2

 

Question 3

In figure 6.53, ABD is a triangle right angled at A and AC  BD. Show that
(i)    AB2 = BC x BD
(ii)    AC2 = BC x DC
(iii)    AD2 = BD x CD

Solution 3

 
 
iii.     In DCA & DAB 

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

Question 4

ABC is an isosceles triangle right angled at C. prove that AB2 = 2 AC2.

Solution 4

 

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

Question 5

ABC is an isosceles triangle with AC = BC. If AB2 = 2 AC2, prove that ABC is a right triangle.

Solution 5

 

Question 6

ABC is an equilateral triangle of side 2a. Find each of its altitudes.

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 

Question 7

Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

Solution 7

 

In AOB, BOC, COD, AOD
Applying Pythagoras theorem

Question 8

In figure 6.54, O is a point in the interior of a triangle ABC, OD  BC, OE  AC and OF  AB. Show that 
 

(i)    OA2 + OB2 + OC2 - OD2 - OE2 - OF2 = AF2 + BD2 + CE2
(ii)    AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Solution 8

 

Question 9

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Solution 9

 

Question 10

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Solution 10

 

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

Question 11

An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after  hours?

Solution 11

 

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

Question 12

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

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.

Question 13

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2

Solution 13

 

          In ACE, 

Question 14

      The perpendicular from A on side BC of a ABC intersect BC at D such that DB = 3 CD 

      Prove that 2 AB2 = 2 AC2 + BC2

Solution 14

   
 

Question 15

In an equilateral triangle ABC, D is a point on side BC such that BD = BC . Prove that 9 AD2 = 7 AB2

Solution 15

 

Question 16

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

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

Question 17

Tick the correct answer and justify: In ABC, AB =  cm, AC = 12 cm and BC = 6 cm.
The angle B is:

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).

NCERT Solution for Class 10 Mathematics Chapter 6 - Triangles Page/Excercise 6.6

Question 1

In the given figure, PS is the bisector of  QPR of   PQR. Prove that 

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

Question 2

In the given figure, D is a point on hypotenuse AC of ABC, DM  BC and DN  AB, Prove that:
(i).    DM2 = DN.MC    (ii).    DN2 = DM.AN

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

Question 3

In the given figure, ABC is a triangle in which  ABC> 90° and AD  CB produced.

Prove that AC2 = AB2 + BC2 + 2BC . BD. 

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

Question 4

In the given figure, ABC is a triangle in which  ABC < 90° and AD  BC. Prove that 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

Question 5

In the given figure, AD is a median of a triangle ABC and AM  BC. Prove that: 

 

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

Question 6

Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

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 

Question 7

In the given figure, two chords AB and CD intersect each other at the point P. prove that:
(i)   APC ~ DPB    (ii)    AP.PB = CP.DP

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

Question 8

In the given figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

(i)    PAC ~ PDB    (ii) PA.PB = PC.PD

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.

Question 9

in the given figure, D is a point on side BC of  ABC such that .

Prove that AD is the bisector of  BAC. 

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

Question 10

Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds? 

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