NCERT Class 10 Maths Solutions Chapter - 6 Triangles, Ex - 6.6

Ex - 6.6

Question 1.  In the given figure, PS is the bisector of  QPR of   PQR. Prove that


Solution


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

(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

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

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

(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 = AC<sup>2

(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.</sup>  Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Solution



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


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

(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

 
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

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