Class IX Math NCERT Solution for Quadrilaterals 8.1

NCERT TEXTBOOK QUESTIONS SOLVED
EXERCISE: 8.1 (Page 146)
Q1.   The angles of quadrilateral are in the ratio 3 : 5 : 9 : quadrilateral.
Sol:   Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
           ∵ Sum of all the angles of quadrilateral = 360°
           ∴ 3x + 5x + 9x + 13x = 360°
           ⇒ 30x = 360°
           
           ∴ 3x = 3 × 12° = 36°
                5x = 5 × 12°= 60°
                9x = 9 × 12° = 108°
                13x = 13 × 12° = 156°
           ⇒ The required angles of the quadrilateral are 36°, 60°, 108° and 156°.
Q2.   If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Sol:   A parallelogram ABCD such that AC = BD In ΔABC and ΔDCB,
                AC = DB
[Given]
                AB = DC
[Opposite sides of a parallelogram]
                BC = CB
[Common]
                ΔABC = ΔDCB
[SSS criteria]
        ∴Their corresponding parts are equal.
        ⇒∠ABC = ∠DCB
...(1)
        ∵AB || DC and BC is a transversal.
[∵ ABCD is a parallelogram]
        ∴∠ABC + ∠DCB = 180°
...(2)
        From (1) and (2), we have
                ∠ABC = ∠DCB = 90°
        i.e. ABCD is parallelogram having an angle equal to 90°.
        ∴ABCD is a rectangle.
Q3.   Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Sol:   We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O.
           ∴ In ΔAOB and ΔAOD, we have
                AO = AO
[Common]
                OB = OD
[Given that O in the mid-point of BD]
                ∠AOB = ∠AOD
[Each = 90°]
                ΔAOB ≌ ΔAOD
[SAS criteria]
           Their corresponding parts are equal.
 
AB = AD
...(1)
Similarly,
AB = BC
...(2)
 
BC = CD
...(3)
 
CD = AD
...(4)
           ∴ From (1), (2), (3) and (4), we have AB = BC CD = DA
           Thus, the quadrilateral ABCD is a rhombus.
Q4.   Show that the diagonals of a square are equal and bisect each other at right angles.
Sol:   We have a square ABCD such that its diagonals AC and BD intersect at O.
           (i) To prove that the diagonals are equal, i.e. AC = BD
                 In ΔABC and ΔBAD, we have
                 AB = BA
                 [Common]
                 BC = AD
[Opposite sides of the square ABCD]
                 ∠ABC = ∠BAD
[All angles of a square are equal to 90°]
                 ∴ΔABC ≌ ΔBAD
[SAS criteria]
                 ⇒Their corresponding parts are equal.
                 ⇒AC = BD
...(1)
           (ii) To prove that 'O' is the mid-point of AC and BD.
                 ∵ AD || BC and AC is a transversal.
[∵ Opposite sides of a square are parallel]
                 ∴∠1 = ∠3
[Interior alternate angles]
                 Similarly, ∠2 = ∠4
[Interior alternate angles]
                 Now, in ΔOAD and ΔOCB, we have
                 AD = CB
[Opposite sides of the square ABCD]
                 ∠1 = ∠3
[Proved]
                 ∠2 = ∠4
[Proved
                 ΔOAD ≌ ΔOCB
[ASA criteria]
                 ∴Their corresponding parts are equal.
                 ⇒OA = OC and OD = OB
                 ⇒O is the mid-point of AC and BD, i.e. the diagonals AC and BD bisect each other at O.
...(2)
           (iii) To prove that AC 1 BD.
                   In ΔOBA and ΔODA, we have
 
OB = OD
[Proved]
 
BA =DA
[Opposite sides of the square]
 
OA = OA
[Common]
                   ∴
ΔOBA ≌ ΔODA
[SSS criteria]
                 ⇒Their corresponding parts are equal.
                 ⇒ ∠AOB = ∠AOD
                 But ∠AOB and ∠AOD form a linear pair.
 
∠AOB + ∠AOD = 180°
 
∠AOB = ∠AOD = 90°
                 ⇒AC ⊥ BD
...(3)
                 From (1), (2) and (3), we get AC and BD are equal and bisect each other at right angles.
Q5.   Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Sol:   We have a quadrilateral ABCD such that ��O�� is the mid-point of AC and BD. Also AC ⊥ BD.
           Now, in ΔAOD and ΔAOB, we have
                 AO = AO
[Common]
                 OD = OB
[∵ O is the mid-point of BD]
                 ∠AOD = ∠AOB
[Each = 90°]
                 ∴ ΔAOD ≌ ∠AOB
[SAS criteria]
                 ∴Their corresponding parts are equal.
                 ⇒ AD = AB
...(1)
           Similarly, we have
AB = BC
...(2)
 
BC = CD
...(3)
 
CD = DA
...(4)
           From (1), (2), (3) and (4) we have: AB = BC = CD = DA
           ∴Quadrilateral ABCD is having all sides equal.
           In ΔAOD and ΔCOB, we have
 
AO = CO
[Given]
 
OD = OB
[Given]
 
∠AOD = ∠COB
[Vertically opposite angles]
           ∴
ΔAOD ≌ ΔCOB
 
           ⇒Their corresponding pacts are equal.
           ⇒                       ∠1 = ∠2
           But, they form a pair of interior alternate angles.
           ∴AD || BC
           Similarly, AB || DC
           ∴ ABCD is' a parallelogram.
           ∵ Parallelogram having all of its sides equal is a rhombus.
           ∴ ABCD is a rhombus.
           Now, in ΔABC and ΔBAD, we have
 
AC = BD
[Given]
 
BC = AD
[Proved]
 
AB = BA
[Common]
 
ΔABC ≌ ΔBAD
[SSS criteria]
 
           Their corresponding angles are equal.
 
∠ABC = ∠BAD
 
           Since, AD || BC and AB is a transversal.
           ∴∠ABC + ∠BAD = 180°
[Interior opposite angles are supplementary]
           i.e. The rhombus ABCD is having one angle equal to 90°.
           Thus, ABCD is a square.
Q6.   Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that
           (i) it bisects ∠C also,
(ii) ABCD is a rhombus.
Sol:   We have a parallelogram ABCD in which diagonal AC bisects ∠A.
           ⇒ ∠DAC = ∠BAC
           (i) To prove that AC bisects ∠C.
                ∵ABCD is a parallelogram.
                ∴AB || DC and AC is a transversal.
                ∴∠l = ∠3
[Alternate interior angles] ...(1)
                Also, BC || AD and AC is a transversal.
                ∴∠2 = ∠4
[Alternate interior angles] ...(2)
                But AC bisects ∠A.
[Given]
                ∴∠1 = ∠2
...(3)
                From (1), (2) and (3), we have
                                ∠3 = ∠4
                ⇒AC bisects ∠C.
           (ii) To prove ABCD is a rhombus.
                In ΔABC, we have                ∠1 = ∠4
[∵∠1 = ∠2 = ∠4]
                ⇒
BC = AB
Sides opposite to equal angles are equal] ...(4)
                Similarly,
AD = DC
...(5)
                But ABCD is a parallelogram
 
[Given]
 
AB = DC
[Opposite sides of a parallelogram] ...(6)
                From (4), (5) and (6), we have
                                                         AB = BC = CD = DA
                 Thus, ABCD is a rhombus.
Q7.   ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Sol:   ABCD is a rhombus.
          ∴ AB = BC = CD = AD
          Also, AR || CD an AD || BC
          Now,                    AD = CD ⇒∠1 = ∠2 ...(1)
[Angles opposite to equal sides-are equal]
         Also, CD II AB
[Opposite sides of the parallelogram]
          ∴ ∠1 = ∠3
          and AC is AC is transversal.
          ∴ ∠1 = ∠3
          From (1) and (2), we have
                              ∠2 = ∠3 and ∠1 = ∠4
          ⇒AC bisects ∠C as well as ∠A.
          Similarly, we prove that BD bisects ∠B as well as ∠D.
Q8.   ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
           (i) ABCD is a square
(ii) diagonal BD bisects ∠B as well as ∠D.
Sol:   We have a rectangle ABCD such that AC bisects ∠A as ∠C.
           i.e.                                            ∠1 = ∠4 and ∠2 = ∠3 ...(1)
           (i) Since, rectangle is a parallelogram.
                ∴ ABCD is a parallelogram.
                ⇒AB || CD and AC is a transversal.
                ∴∠2 = ∠4
[Alternate interior angles] ...(2)
                From (1) and (2), we have
                                            ∠3 = ∠4
                ⇒AB = BC
[ ∵ Sides opposite to equal angles in ΔABC are equal.]
                ∴AB = BC = CD = AD
                ⇒ABCD is a rectangle having all of its sides equal.
                ∴ABCD is a square.
           (ii) Since, ABCD is a square, and diagonals of a square bisect the opposite angles.
                ∴ BD bisects ∠B as well as ∠D.
Q9.   In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that:
           (i) ΔAPD ≌ ΔCQB
           (ii) AP = CQ
           (iii) ΔAQB ≌ ΔCPD
           (iv) AQ = CP
           (v) APCQ is a parallelogram.
Sol:   We have parallelogram ABCD. BD is a diagonal and ‘P’ and ‘Q’ are such that
                PD = QB
[Given]
           (i) To prove that ΔAPD ≌ ΔCQB
                ∵ AD || BC and BD is a transversal.
[∵ABCD is a parallelogram.]
                ∴∠ADB = ∠CBD
[Interior alternate angles]
                ⇒∠ADP = ∠CBQ
                Now, in ΔAPD and ΔCQB, we have
                AD =CB
[Opposite side of the parallelogram]
                PD = QB
[Given]
                ∠CBQ = ∠ADP
[Proved]
                ∴ Using SAS criteria, we have
                ΔAPD ≌ ΔCQB
           (ii) To prove that AP = CQ
                Since ΔAPD ≌ ΔCQB
[Proved]
                ∴ Their corresponding parts are equal.
                ⇒AP = CQ
           (iii) To prove that ΔAQB ≌ ΔCPD.
                AB || CD and BD is a transversal.
[ ∵ ABCD is a parallelogram.]
                ∴∠ABD = ∠CDB
                ⇒∠ABQ = ∠CDP
                Now, in ΔAQB and ΔCPD, we have
 
QB = PD
[Given]
 
∠ABQ = ∠CDP
[Proved]
 
AB = CD
[Opposite sides of parallelogram ABCD]
 
∴ΔAQB ≌ ΔCPD
[SAS criteria]
           (iv) To prove that AQ = CP.
                Since                ΔAQB ≌ ΔCPD
[Proved]
                ∴Their corresponding parts are equal.
                ⇒                AQ = CP.
           (v) To prove that APCQ is a parallelogram.
                Let us join AC.
                Since, the diagonals of a || gm bisect each other
                ∴AO = CO
                and                                BO = DO
                ⇒(BO – BQ) = (DO – DP)
[∵ BQ = DP (Given)]
                ⇒QO = PO
...(2)
                Now, in quadrilateral APCQ, we have
                                AO = CO and QO = PO
                i.e. AC and QP bisect each other at O.
                ⇒APCQ is a parallelogram.
Q10.   ABCD is a parallelogram and AP and perpendiculars from vertices A and C on diagonal BD (see Show that
           (i) ΔAPB ≌ ΔCQD
(ii) AP = CQ
Sol:   (i) In ΔAPB and ΔCQD, we have
                ∠APB = ∠CQD [90° each]
                AB = CD
[Opposite sides of parallelogram ABCD]
                          ∠ABP = ∠CDQ
                ⇒Using AAS criteria, we have
                ∠APB ≌ ΔCQD
           (ii) Since, ΔAPB ≌ ΔCQD
                ∴ Their corresponding parts are equal.
                ⇒AP = CQ
Q11.   In ΔABC and ΔDEF, AB = DE, AB || DE, and BC || EF. vertices A, B and C are joined to vertices F respectively (see figure). Show that
              (i) quadrilateral ABED is a parallelogram.
              (ii) quadrilateral BEFC is a parallelogram.
              (iii) AD || CF and AD = CF.
              (iv) quadrilateral ACFD is a parallelogram.
              (v) AC = DF
              (vi) ΔABC ≌ ΔDEF:
Sol:      (i) To prove that ABED is a parallelogram.
                   Since “A quadrilateral is a parallelogram if a pair of equal length.”
                   Now
AB = DE
[Given]
 
AB || DE
[Given]
                   i.e. ABED is a quadrilateral in which a pair of opposite and of equal length.
                   ∴ ABED is a parallelogram.
           (ii) To prove that BECF is a parallelogram.
                   ∵
BC = EF
[Given]
                   and
BC || EF
[Given]
                   i.e. BECF is a quadrilateral in which a pair of opposite sides (BC and EF) is parallel and of equal length.
                   ∴ BECF is a parallelogram.
           (iii) To prove that AD || CF and AD = CF
                ∵ABED is a parallelogram.
[Proved]
                ∴ Its opposite sides are parallel and equal.
                ⇒ AD || BE and AD = BE
...(1)
                Also BEFC is a parallelogram.
[Proved]
                ∴ BE || CF and BE = CF
[∵Opposite sides of a parallelogram are parallel and equal] ...(2)
                From (1) and (2), we have AD || CF and AD = CF
           (iv) To prove that ACFD is a parallelogram.
                ∵ AD || CF
[Proved]
                and AD = CF
[Proved]
                i.e. In quadrilateral ACFD, one pair of opposite sides (AD and CF) is parallel and of equal length.
                ∴ Quadrilateral ACFD is a parallelogram.
           (v) To prove that AC = DE.
                ∵ACFD is a parallelogram.
[Proved]
                ∴ Its opposite sides are parallel and of equal length.
                i.e.                            AC = DF
           (vi) To prove that ΔABC ≌ ΔDEF
                In ΔABC and ΔDEF, we have:
                AB = DE
[Opposite sides of a parallelogram]
                BC = EF
[Opposite sides of a parallelogram]
                AC = DF
[Proved]
                ∴ Using SSS criteria, we have ΔABC ≌ ΔDEF.
Q12.   ABCD is a trapezium in which AB II CD and AD = BC (see figure).
                Show that
                (i) ∠A = ∠B
                (ii) ∠C = ∠D
                (iii) ∠ABC ≌ ∠BAD
                (iv) Diagonal AC = Diagonal BD
                Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.
Sol:   We have AB || CD and AD = BC
           (i) To prove that ∠A = ∠B.
                Produce AB to E and draw CE || AD.
                ∴
AB || DC
                ⇒ AE || DC
[Given]
                Also
AD || CE
                ∴AECD is a parallelogram.
                ⇒
AD = CE
[opposite sides of the parallelogram AECD]
                But
AD = BC
[Given]
                ∴
BC = CE
 
 
BC = CE
 
                Now, in ΔBCE, we have
 
BC = CE
 
                ⇒
∠CBE = ∠CEB
...(1)
[∵ Angles opposite to equal sides of a triangle are equal]
                Also, ∠ABC + ∠CBE = 180°
[Linear pair] ...(2)
                and ∠A + ∠CEB = 180°
[∵Adjacent angles of a parallelogram are supplementary]
...(3)
                From (2) and (3), we get
                ∠ABC + ∠CBE = ∠A + ∠CEB
                But
∠CBE = ∠CEB
                ∴
∠ABC = ∠A
                or
∠B = ∠A
                or
∠A = ∠B
           (ii) To prove that ∠C = ∠D.
                AB || CD and AD is a transversal.
                ∠A + ∠D = 180°
[Sum of interior opposite angles]
                Similarly, ∠B + ∠C = 180°
                ⇒∠A + ∠D = ∠B + ∠C
                But ∠A = ∠B
[Proved]
                ∴∠C = ∠D
           (iii) To prove ΔABC ≌ ΔBAD
           In ΔABC and ΔBAD, we have
                AB = BA
[Common]
                BC = AD
[Given]
                ∠ABC = ∠BAD
[Proved]
                ∴ΔABC ≌ ΔBAD
[Using SAS criteria]
           (iv) To prove that diagonal AC = diagonal BD
                ΔABC ≌ ΔBAD
[Proved]
                ∴ Their corresponding parts are equal.
                ⇒the diagonal AC = the diagonal BD.

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