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Class 7 Maths R S Aggarwal solutions

Contents

13. Lines and Angles

14.  PROPERTIES OF PARALLEL LINES

15. PROPERTIES OF TRIANGLES 15A, 15B, 15C, 15D

16. CONGRUENCE

 

EXERCISE-13,  LINES AND ANGLES

 

Q1.         I]       Given angle = 35o. Let the measure of its complement be xo. Then,

                X + 35o = 90     => x = 90 – 35 = 55o . Hence, the complement of the given angle= 55o .

Q2.         i]             Let the given angle measure 80o. Then, x + 80 = 180.

π  X = 180 – 80 = 100o .   www.rsmaths99.com

Q3.         We know, LA + LB = 180o .   Let LA = xo . Then LB = X + 36O. Now,

                X + (X +36O)= 180O   => 2X = 144 =>  X = 72O   =>  72 + 36 = 108O.

Q4.         X + X = 180O .   =>  2X = 180  => X = 90O .

Q5.         I] No      II] No       III]  Yes.

Q6.         Since, LAOB is a st. line , then, LAOC + LBOC = 180o .  

=> 64 + x = 180o.    x = 116o.

Q7.         Since, AOB is a st. line , then,

                LAOC + LBOC = 180O.  => 2X – 10 + 3X + 20 = 180O.   5X + 10 = 180O.

π  5x= 170o.  =>  x = 34o.   Hence, LAOC = 2 X 34 -10 = 58O.

π  LBOC = 3X +20 = 3 X 34 + 20 = 122O.

Q8.         Since, AOB is st. line . Then, LAOC + LCOD + LBOD = 180O. => 65 + X + 70 = 180O.

π  X = 180 – 135 = 45O.   www.rsmaths99.com

Q9.         Since, AOB is a st. line . Then, LAOC + LCOB = 180O.

π  45O + LCOB = 180O. => LCOB = 180 – 42 = 138O. Now,

π  LAOC = LCOB = 42O  (vertically opposite Ls.).And,

π  LBOC = LAOD = 138O  (vertically opposite Ls.).

 LAOD = 138O. LBOD = 42O, LBOC = 138O.

Q10.       Since, PQ is a straight line and RS insert at O, we have, LPOS +LPOR = 180O.

                114O +LPOR =180O => LPOR = 180 – 114 = 66O.  Now,

                LPOR = LQOS = 66O.  ( vertically opposite Ls.).

And LPOS = LROQ = 114 O. ( vertically opposite Ls.).

Q11.       ATQ,      LAOB + LBOC + LCOD +LDOA = 360O.        rsmaths99.com

                56O +  100O + XO + 74O = 360O.     HENCE, XO = 130O.

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EXERCISE – 14, PROPERTIES OF PARALLEL LINES

 

                [i]     {L1  =L5,    = L4 = L8,  L2   = L6,   L3  = L7 }are corresponding angles.

                [ii]    {  L3 = L5,  L4 = L6 } are alternative interior Ls.  www.rsmaths99.com

                [iii]    L3 + L6= 180O.    L4 + L5 = 180O    

                [iv]   {L1 , L7, L4, L8} alternative interior Ls or co-interior angles.   

                --------------------------------------------------------------------------

Q1.                         L1  =L5, = 70o.     (corresponding Ls)

L

 
  L3 = L5,  = 70o .  (alternate interior Ls.)                 

L4 + L5 = 180o.   (consecutive interior Ls)

M

 
L4 + L70 = 180o.   => L4 = 110o.

L8 = L4 = 110o.  (corresponding Ls).

Q2.         Since, line L is a st. line , then,

                7x + 5x = 180o . => 12x = 180o => x = 15o.

                 L1 = 5 x 15 = 75o.            L2 = 7 x 15 = 105o.

Now, L1 = L5 = 75o. (corresponding angle)

L5 = L3= 75o  (alternative interior angle)

And,      L3 = L1 = 75o.  (vertically opp. angles)  rsmaths99.com

L8 = L4= (corresponding Ls) = L2  (vertically opp. angles)=105.

Q3.         2x – 8 + 3x – 7 = 180o.    => 5x – 15 = 180o.    => x = 39o.

 =>  2x – 8 = 2 x 39 – 8 = 78 – 8 = 70o.

=> 3x – 7 = 3 x 39 – 7 – 117 -7 = 110o.

Q4.         Since, L is  a st. line . Then, L1 + L2 = 180o.  => 50 + L2 = 180o.  => L2 = 130o .

                Since, M is a st. line .Then, LC + LD = 180o.  => LC + 65o = 180o. => LC = 180o -65o= 115o.

Q5.         LBCA = LCAE = 45O.   (alternative interior angle)

LCBA = LBAD = 45O.   (alternative interior angle)

Q6.         Exterior l = sum of interior opposite L.

                Now,  LBAC = LACE = 80O. (  Alternate angle)

                Since, BD is a st. line LBCA + LACE + LECD = 180O.

π  LBCA + 80o + 35o = 180O.    =>  LBCA = 65O.

 Now, LABC + LACB LBAC = 180O.  =>  LABC + 65O + 80O = 180O.

                LABC = 180O – 145O = 35o.   rsmaths99.com

Q7.         Given, AO ||CD and OB || CE  , LAOB = 50. To find LECD.

                LCDB = LAOB  (corresponding angle) = 50O.

Q8.        

               

Given, AB||CD, LABO = 50o,  LCDO = 40o.  To find LBOD.

                CONST.: Draw EF ||AB||CD and name the angles as L1 & L2 as shown in figure.

                Now,  since, AB ||EF    L1 = 50 (alt Ls)

                                EF||CD   L2 = 40  (alt.  Ls).    LBOD =  L1 + L2 = 50o + 40o = 90o.

Q9.

               

 

Given,  AB||CD         

GL and HM are bisectors of  LAGH and LGHD

                 L1= L2 ------(1) and  L3 = L4 ------(2)      TO prove GL||HM.

                Now,  LAGH = LGHD  (alt Ls).       L1+ L2 = L3 + L4.

                 2L2 = 2L3  [from (1)& (2)].  rsmaths99.com

                But there are alt Ls of lines GL and HM .    GL||HM.

Q10.       Given, AB ||CD ,  LECD = 100o,  LABE = 120O,   LBEC = XO. To find value of x.

                Construct : Draw FG||AB||CD.

                Now, CD||FG  => L1 + 100 = 180o     [co-interior angles]

π  L1 = 80o.       Also, AB||FG   => LABE = LBEG    [ALT. Ls]

π  120o   = x + L1    => x = 120 – L1.    =>  120o – 80o = 40o.  x = 40o.

 

Q11.       Given,  ABCD is a quadrilateral in which AB||DC and AD||BC.

To prove LADC = LABC.

Proof : since, DC||AB.    LADC + LABC = 180o  -----(1)   [co-interior angles]

Also, AD||BC    =>  LDAB + LABC = 180o  -------(2)   [co-interior angles]

From (1) & (2) ,  =>     LADC + LABC = LDAB + LABC   => LADC = LABC. Proved.

 

Q12.       Given,   l||m, p||q.  To find measure of La, Lb,  Lc , Ld.

                Soln.:  a = 65o      (VOA).

                Since,    l||m     =>   a + d = 180o    [co-interior angles]

π  65 + d = 180o.    =>   d = 180o – 65 = 115o.

Since,  p||q  => c + d = 180o    => c + 115 = 180o    => c = 180 – 115 = 65o.

                Now,  l||m   => c + b = 180o.   => b = 115o.

                 a = 65o, b = 115o,  c = 65o, d = 115o.  

               

Q13.       Given, AB||DC,  AD||BC,   LBAC = 35O,,  LCAD = 40O, LACB = XO, LACD = YO.

                To find x and y.      rsmaths99.com

                Soln.:    AD||BC  => X = 40O  (alt. Ls).          AB||DC  => 35o  (alt. Ls).  

 

Q14.       Given,  AB||CD,  LBAE = 125O,    LBAC = XO,   LABD =XO , LBDC =Y O,  LACD = ZO.

                To find   x, y , z

                Soln. x + 125o = 180o     (linear pair)

π  X = 180 – 125 = 55o.   AB||CD  =>  x + z = 180O   (Co-interior angles)

π  55 + z = 180o.  => z = 180 – 55 = 125o.

Also, x + y = 180o   => 55 = y = 180o    => y = 125o.

                 x = 55o,   y = 125o,    z = 125o.

Q15.       I]       L1 + L130o = 180o  => L1 = 50o.  For l to be parallel to m.

                L1 should be equal to 40o.    l  m.

Ii]      L1 = 35o     (VOA’s). For l to be parallel to m L1 + 145o should be 180o.

Now, L1 + 145o = 1800.     l||m.    rsmaths99.com

Iii]      L1 + 60o = 180o   => L1 = 120o. for l to be parallel to m L1 = 125o.  (corres. Ls)

Which is not so.     l  m.

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EXERCISE – 15A,PROPERTIES OF TRIANGLES

 

Q1.         In a ∆ABC,  LA = 72o,, LB = 63o.  To find LC = 180o -  (LA + LB)

                = 180o – (72 + 63)  = 180o – 135o = 55o.

Q2.         LD = 180o – (105 + 40)  = 180 – 145 = 35o.

Q3.         LY = 180o – (LX + LZ)    = 180 – 138 = 42o.

Q4.         Let the ratio of common multiple be x .

Then, the three Ls are 4xo, 3xo and 2xo respectively.

 By L sum property of a triangle.   => 4x + 3x + 2x = 180o.

π  9x = 180 => x = 20o.    the three Ls are 4x = 4 X 20 = 80o.

π  3x = 3 X 20 = 60o.   => 2x = 2 X 20 = 40o

Q5.         Let the other L be x. Then, x + 36 = 180o. 

π  X = 180o – 36o= 144o.   rsmaths99.com

Q6.         Let the ratio of common multiple be x. then, the angles are 2x and x .

                Clearly 2x + x = 90o. => 3x = 90o.   => x = 30o..

                 the angles are 2x = 2 X 30 = 60o. x = 30o.

Q7.         Let the two equal angles be xo each . Then, by angle sum property  of a ∆

                X + x + 100o = 180o.   =>  2x = 80o.  => x = 40o.

                 The two equal angles are 40o each.

 

Q8.         Let the equal angles be xo each. Then, ATQ, third angle = 2xo.

                By angle sum prop. Of a ∆

                X +x + 2x = 180o.   => x = 60o.

                 angles are x = 60o, x = 60o, 2x = 120o.

Q9.         Let ABC be the given ∆.   Let LA = LB + LC  ----(1)

                Now, LA + LB + LC = 180o.  =>  LA + LA = 180O.    FROM  (1)

π   2 LA = 180O.  =>  2X = 90O.   rsmaths99.com

 ABC is a right angle triangle.

Q10.       Given , in a ∆ABC,  2 LA = 3 LB = 6 LC. To, find LA, LB and LC.

                2 LA = 3 LB = 6 LC = x.  => LA =   LB  = .

                By angle sum prop. of triangle ;  LA + LB + LC = 180o.

π   = 180o.  =>    => x = 180o.

 

   LA =   LB =   LC =

Q11.       Since all the angle of an equilateral ∆ are equal.

                Let the measure of each equal angle be xo. Then, by angle sum prop. of a ∆;

                X + x + x = 180o.  => x = 60o.     measure of each angle of an equilateral  ∆ = 60o.

Q12.       Given, DE||BC, LA = 65O, LB = 55O. To find  I]  LADE, II] LAED  III] LC

                Soln.  : I]    DE||BC      LADE = LB = 55O.   

                II]    In ∆ADE ,    LA + LD + LE = 1800.    rsmaths99.com

                  65o + 55o + LE = 180o.  =>  LE     = 180o – 120o = 60o.  LAED = 60o.

                Iii]    In ∆ ABC,     LA + LB + LC = 180o.   =>  65 + 55 + LC = 180o.

π  LC = 180o – 120o = 60o.

OR  

 

DE||BC    => LC = LE     [CORRESPONDING Ls]    => LC = 60o.

Q13.       I] No, ii]   no,  iii] yes,  iv]  no, v]  no, vi]   yes.

Q14.       i]  yes,    ii]  yes   iii] no   iv]  yes.

Q15.       I]    Obtuse   ii]  complementary   iii]  45o   iv]   60o    v]  hypotenuse  vi]  perimeter.

 

EXERCISE – 15B EXTERIOR AND INTERIOR OPPOSITE ANGLES

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Q1.        For ∆ ABC,     Exterior  L ACD = LA + LB =  75o + 45o = 120o.

Q2.        For ∆ ABC,   Exterior  L ACD = LA + LB   => 130o = x + 68o.

                X = 130o – 68o = 62o.    now,   y + 130 = 180   [linear pair]

                Y = 180o – 130o = 50o.

Q3.        For ∆ ABC,   Exterior  L ACD = LA + LB     => x + 32 = 65o.   =>  x = 65 – 32 = 33o.

                Y + 65 = 180o   (linear pair).    => y = 180o – 65o = 115o .  rsmaths99.com

 

Q4.         Let the ratio of common multiple be x. then, interior opp. Angles are 2xoand 3xo resp.

                By exterior angle prop. of a ∆;   2x + 3x = 110o   =>  x =

 Angle of a   are 2x = 2 x 22 = 44o.   3x = 3 x 22 = 66o.

                                3rd  angle = 180o – 1100 = 70o.

 

Q5.         Let each interior equal angle be xo. then, by exterior angle prop. of a ∆;

                X + x = 100. => 2x = 100o.     => x = 50o    rsmaths99.com

                   Each angle of a ∆ are x = 50o,    x = 50o,   180 – 100 = 80o.

 

Q6.         For ∆ ABC,   Exterior  L ACD =45o + 25o = 70o.

                For  ECD exterior   L AED = L ECD + L CDE = 70o + 40o = 110o.

 

Q7.         I]   In ∆ ABC,   L A = 180 – (LB + LC)  =  180 – (40 + 100)  = 40o.

                   for ∆ ABC  exterior  L ACD = LA + LB = 40o + 40o = 80o.

                Ii]   In ∆ ACD,    LA + LC + LD = 180o.   =>  50o + 80o + LADC = 180o.  

π   LADC  = 180o – 130o = 50o.   rsmaths99.com

Iii]    for ∆ ABD,     exterior  LDAE = LB + LD =  40o + 50o = 90o.

   LACD  = 80o,     LADC  = 50o,  LDAE = 90o.

Q8.         X : y = 2 : 3,    Let the ratio of common multiple be t. Then,

X = 2t, & y = 3t.      for ∆ ABC , x + y = 130o.    =>  2t + 3t = 130o.

π  t =      x = 2t = 2 x 26 = 52o.  & y = 3t = 3 x 26 = 78o.

now, z + 130o = 180o   (linear pair).

 

  z = 180o – 130 = 50o.

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EXERCISE – 15C  TRIANGLE INEQUALITY

Q1.         I]     Sides are 1cm, 1cm, 1cm.   since, 1 + 1 > 1.  is possible.

                Ii]    Sides are 2cm, 3cm, 4cm.  since,  2cm + 3cm > 4cm.   3cm +4cm > 2 cm.

                       4cm + 2cm > 3cm.   is possible.

Q2.         Two sides of the given  are 5cm,& 9cm.   rsmaths99.com

  third side must be greater than 5 + 9 = 14cm.

Q3.         I]   PA + PB > AB.    II]  PB + PC > BC   III]  AC < PA + PC

Q4.         Given, AM is median of ABC.   To prove AB + BC + CA > 2AM.

                Proof : since, AM is median.    BM = CM ------(1).

                Now, in ∆ABM, AB + BM > AM  [inequality prop. of a ∆ ]

                                In ∆ACM, AC + BM > AM.  [inequality prop. of a ∆ ]

                Adding there two inequalities, we get,

                AB + (BM + CM) + AC > 2AM.  => AB + BC + CA > 2AM. Proved.

Q5.         To prove AB + BC + CA > 2AP.

                Proof : in ABP,  AB + BP > AP -----(1)     [inequality prop. of a ∆ ]

                In ∆ACP,   AC + PC >AP  ------(2)       [inequality prop. of a ∆ ]

                Adding  (1)& (2),   AB + (BP + PC) + AC > 2AP

π  AB + BC + CA > 2AP.  Proved.   rsmaths99.com

Q6.         Given, ABCD is a quadrilateral.  To prove; AB + BC + CD + DA > AC + BD.

                Proof:  ABD,     AB + AD > BD.    [inequality prop. of a ∆ ]

  In ∆BCD,  BC + CD > BD               [inequality prop. of a ∆ ]

  In ∆ABC,    AB + BC > AC.            [inequality prop. of a ∆ ]

  In ∆ACD,      AD + CD > AC.         [inequality prop. of a ∆ ]

Adding we get,    2(AB + BC + CD + DA) > 2 (AC + BD)

π   AB + BC + CD + DA > AC + BD. Proved.

Q7.         Given, O is a point in the exterior of ABC.

To prove 2(OA + OB + OC)> AB + BC + CA.

Const.: join OA, OB,& OC.   rsmaths99.com

Proof :      In ∆AOB,      OA + OB > AB         [inequality prop. of a ∆ ]

                  In ∆BOC,        OB + OC > BC         [inequality prop. of a ∆ ]

                  In ∆AOC,       OA + OC > CA         [inequality prop. of a ∆ ]

Adding  all we get,   2(OA + OB + OC) > AB + BC + CA. proved.

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EXERCISE – 15D   PYTHAGORAS THEOREM

Q1.         Let the length of hypotenuse be ‘h’ cm. Then, by Pythagoras theorem.

h2 = (9)2+ (12)2     =  81 + 144 = 225.   => h =    = 15 cm.

Q2.         Let the third side be x cm. Then, by Pythagoras Th.  ;

                X2 + 102 = 262.     =>  X2 = 262 – 102 = (26 - 10)(26 - 10)

                = 16 x 36 = 576. X =   = 24 cm.   rsmaths99.com

Q3.         Let the third side be x cm. Then, by Pythagoras Th .  ;

                X2 + (4.5)2 = (7.5)2   => x2 =     

                =    x 6 x 24 = 36.    x = 6 cm.

Q4.         Given, h2 = 50, length of other two sides are equal. To find length of each leg.

                Soln. By Pythagoras Th.  X2 + x2 = 50. 

 => 2x2 = 50.    => x2 = 25.        => x = 5.

                 Length of each equal legs = 5 units.

Q5.         15cm, 36cm, 39cm.   392 = 1521.  rsmaths99.com

                Also, 362 + 152 = 1296 + 225  = 1521.

                 392 = 362 + 152.    

   the triangle is rt. Angled triangle.

Q6.         Given, length of two sides of triangles are a = 6 cm. b = 4.5 cm.

To find – the length of hypotenuse.

Soln. :    (hypotenuse)2 = a2 + b2  

=  62 + (4.5)2  = 36 +     [we can write 4.5 =   ]

= 36 +   .      

    (hypotenuse)2 =   .

π  Hypotenuse =       www.rsmaths99.com

Q7.        [i)   C2 =  252 = 625.   a2 + b2 = 152 + 202         = 225 + 400 = 625.

                       C2 = a2 + b2.        triangle is right – angled triangle.

               

[ii]     C2 = 162 =  256.         a2 + b2 = 92 + 122.   = 81 + 144 = 225.

                      C2   a2 + b2.        triangle is not a right angled triangle.

               

[iii]   a = 10cm, b = 24 cm, c = 26 cm.   c2 = 262= 676.

π    a2 + b2 = 102 + 242  = 100 + 576 = 676.

      c2 =  a2 + b2.                  triangle is right – angled triangle.

 

Q8.         Clearly LA = 180 – (LB + LC ) = 180 – (35 + 55) = 180 – 90 = 90o.

                             BC2 = AB2 + AC2.

     Option (iii ) is correct.  www.rsmaths99.com

 

Q9.         By Pythagoras Th.  X2 + 122 = 152. => x2 = 225 – 144 = 81.  => x = 9 cm.

                 Distance of the foot of ladder from wall = 9 cm.

Q10.       Lat the required distance be x cm.

Then, by Pythagoras Th.  X2 +( 4.8) 2 = 52

π  X2 = 52 - ( 4.8) 2 =  (5 – 4.8)( 5 + 4.8) = 0.2 x 9.8 = 1.96m.

π         =   =>   x =  

 

Q11.      Let AB be the tree, let P be the point where the tree breaks down and it top touches A1.

                Given, PB = 9m, A1B = 12m. Let A1P = x m. To find height of tree = (x + 9)m.

Soln.:   clearly LB = 90o.   In right angle-triangle PBA1,

π  X2 =  92 + 122           [By Pythagoras Th. ]   www.rsmaths99.com

π       =  81 + 144 = 225 = (15)2     =>  x = 15m.

  height of tree = x + 9  = 15 + 9 = 24 m.

Q12.       Let AB and CD be the two given poles.

                Given, AB = 18m, CD = 13m, BD = 12m.  To find AC.

                Construct : Draw CP||DB.

Soln. :   In right angle  APC ,    AP = 18 – 13 = 5m, PC = 12m.

                 By Pythagoras Th. ,  AC2 = AP2 + PC2 = (5)2  + (12)2 = 25 + 144 = 169 = (13)2

π  AC = 13.        Distance between their tops = 13m.

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Q13.       Let A be the initial position.  Given, AB = 35m,  BC = 12m.  To find AC.

                Soln. :   By Pythagoras Th. ,  AC2 = AB2 + BC2   =  (35)2 + (12)2   = 1225 + 144.

                = > AC2 = 1369 = (37)2                        AC = 37 m.

 

Q14.       Let P be the initial position. Given, PQ = 3 km, QR = 4 km. To find RP.

                Soln. : By Pythagoras Th.         PR2 = RQ2 + QP2.   =  42 + 32.

                =  16 + 9 = 25 = (5)2.           PR = 5 km.

 

Q15.       Let ABCD be the rectangle.   Let AB = 16 cm,  BC = 12 cm. to find AC and BD.

                Soln. :  In right angle  ABC .   AC2 = AB2 + BC2   =  (16)2 + (12)2

π  AC2 = 256 + 144  = 400 = (20)2.        AC = 20 cm = BD.

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Q16.       In right angle  ABC,    x2 = (41)2 – (40)2  = (41 - 40)(41 + 40) = 1 x 81 = 81 = (9)2.

                  x =  9 cm.      Breadth  = 9 cm.

                   Perimeter = 2 (length + breadth)   = 2 (40 + 9) = 2 x 49 = 98 cm.

 

Q17.       Let ABCD be the rhombus, Given, AC = 16 cm.

                  AO =  = 8 cm.     BD = 30 cm,      BO =   

                       In right angle  AOB,     AB2 = AO2 + BO2  =  82 + 152  = 64 + 225  =  289.

π   AB2 =  (17)2          AB = 17 cm.  [Fig. in book]  www.rsmaths99.com

EXERCISE – 16 CONGRUENCE

 

Q1.         [i]  ABC   ∆ EFD.    A E, B  C D.      AB = EF, BC = FD, AC = ED

                LA = LE, LB = LF,  LC = LD.

                [ii]    ∆CAB   ∆ QRP,    C               Q, A R,  BP.

                  CA = QR, AB = RP, CB = QP

                  LC = LQ,  LA = LR,  LB = LP.

                [iii]  XYZ   ∆ QPR,

                X  Q, Y  P,   Z  R  www.rsmaths99.com

                 XY = QP, YZ = PR,  XZ = QR

                LX = LQ, LY = LP,  LZ = LR

                [iv]  MPN    SQR

                M S,  P Q, N R

                 MP = SQ, PN = SR, MN = SR

                LM = LS, LP = LQ,  LN = LR.

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Q2.         [i]  SAS  [ii]  RHS  [iii]  SSS   [iV]  ASA  [ V] AAS

 

Q3.         Yes, PLO    PMO  by RHS congruence as in PLO  and PMO 

                L L = L M (90)o    (R)

                PO = PO (common)  (H)

                PL = PM    (given)  (s)

Q4.         Given,  AD = BC,  AD||BC. To find AB = DC.

                Soln. : in ∆ADC and ∆ CBA

                AD = BC   (Given)  (S)

                L1 = L2  (alternate angle)  (A)

                AC = CA   (Common)  (S)  www.rsmaths99.com

                 By SAS congruence rule ADC   ∆CBA            AB = DC.

Q5.         Given, AB = AC, BD = DC.   To prove   ADB    ∆ADC

                I]   LADB = LADC = 90O    II]    LBAD = LCAD

                Proof: In    ∆ ABD and  ACD

                AB = AC   (given)  (S)   

                AD = DA    (common)  (S)

                BD = CD                       (given)   (S)

                   by SSS cong. Rule;    ABD  ∆ACD.

                I] Hence, LADB = LADC     (By Corresponding  Part of Congruent Triangle)

                But  LADB +  LADC   = 180o   (linear pare) 

                  LADB + LADB = 180O    =>  2 LADB = 180O    => LADB = 90O =  LADC.

                II]     Also L BAD = LCAD   (BY CPCT)  www.rsmaths99.com

 

Q6.         Given,   AD bisector of LA.          L1 = L2.     Also, AD |BC.   L3 = L4 = 90o.

                To prove :  ∆ABC is isosceles.

                Proof :     in ∆ABD  and  ∆ ACD

                                L1 = L2.     (given) (A)

                                AD = DA   (common)  (S)

                                 L3 = L4       (each 90o)  (A)

                 By ASA cong. rule    ∆ ABD    ∆ACD          AB = AC     (BY CPCT)

                    ∆ABC is isosceles. 

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Q7.         Given, AB = AD, CB = CD.  To prove - ∆ ABC    ADC  . 

                Proof :                  In  ∆ ABC  and  ∆ADC 

π      AB = AD  (given)  (S)     

π      CB = CD   (given)   (S)

π      AC = AC    (common)    (S)

 By SSS cong. rule  ABC    ∆ADC  . Proved.

 

 

Q8.         Given,    PA |  AB,     QB   |   AB,   PA = QB. To prove : ∆ OAP    ∆OBQ

                and to check whether OA = OB.

                Proof :      In ∆ OAP  and  ∆OBQ. 

                LAOP = LBOQ    (V.O. Angles)  (A)

                LPAO = LQBO     (90O)  (A)

                PA = QB    (Given)   (S)   www.rsmaths99.com

                     By ASA cong. rule    ∆ OAP    ∆OBQ.  Also,   OA = OB  (CPCT).

 

Q9.         Given, Two right –triangles    ∆ABC and  DCB.

                Such that  LA = LD = 90O.  AC = DB.  To prove   ∆ABC    ∆DCB.

                Proof : In  ∆ABC and  ∆DCB.

                 LA = LD   (90o)   (given)  (R)

                BC = BC  (Common)   (H)  www.rsmaths99.com

                AC = DB  (given)   (S)

                  By RHS cong. rule   ∆ABC    ∆DCB.

 

Q10.       Given, ABC  is an isosceles   ∆.   In which AB = AC.   E and F are mid. Points of AC and AB

                To prove : BE = CF.

                Proof :  In   ∆ ABE and  ∆ ACF

π    AB = AC   (Given)  (S)   

π    LA = LA     (common)  (A)

π     AE = AF   (         )   (S)  www.rsmaths99.com

  By  SAS cong. rule    ∆ ABE   ∆ ACF.     BE = CF  (By CPCT).

 

Q11.       ABC is an isosceles  in which  AB = AC,   AP = AQ.   To prove BQ = CP.

                Proof :  In  ABQ  and   ∆ ACP. 

AB = AC    (given)  (S)

LA  = LA     (common)  (A)

AQ = AP  (Given)  (S)

  By SAS  cong. rule   ABQ  and   ∆ ACP.      BQ = CA. Proved.

 

Q12.       Given,     ABC is an isosceles   ∆ in which   AB = AC, BD = CE. To prove BE = CD.

                Proof :  In ∆ ACD  and  ∆ ABE

                AD  = AE     (AB + BD  = AC + CE )  (S)

                LA  = LA      (common)   (A)  www.rsmaths99.com

                AC = AC       (given)  (S)

                 By SAS cong. rule    ACD    ∆ ABE.     CD = BE.    Or  BE = CD. Proved.

 

Q13.    Given,  ABC is an isosceles  ∆ in which AB = AC, BD = CD.

 To prove :  AD bisects  LA and LD.

Proof :   Let us name the angles as  L1, L2,  L3, L4 as shown in figure.

In    ABD  and    ∆ACD

π   AB = AC    (given)   (S)

π  BD = CD    (Given)   (S)

π  AD  =  AD    (Common)  (S)

  By SSS   cong. rule  ABD      ∆ACD.

  L1 = L2   & L3 = L4      (By CPCT).    www.rsmaths99.com

 AD bisects  LA  and LD.  Proved.

Q14.    If two triangles have their corresponding angles equal, they are not necessarily congruent.

            In    ABC  and  ∆PQR.   Their corresponding angles are equal 

LA = L P, LB = L Q, LC = LR , but they are not congruent, since BC QR.

 

             

Q15.    No, two triangles are not congruent if their two corresponding sides and one angle are equal. They will be congruent only if the said angle is the included angle between the sides.                                                                                     www.rsmaths99.com

 

Q16.   

           

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