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TEXTBOOK QUESTIONS SOLVED

Exercise 8.1

1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Sol. Let the angles be 3x, 5x, 9x and 13x.

Then 3x + 5x + 9x + 13x = 360°       [Sum of angles of a quadrilateral is 360]

⇒       30x = 360°   ⇒     x  =  12°

∴ Anlges are 36°, 60°, 108° and 156°.

2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Sol. Consider triangles DAB and CBA,

AD = BC                         [Opposite sides of a parallelogram]

AB is common.                                [Given]

∴ ΔDAB ≅  ΔCBA                            [SSS]

⇒  ∠ DAB =  ∠ CBA                                 …(i)              [CPCT]

As ABCD is a parallelogram. $AD\parallel BC$  and AB is transversal.

∴   ∠ DAB +  ∠ CBA = 180° [Sum of interior angles on the same side of transversal is 180.]

⇒   2∠DAB = 180°                                             [From (i)]

⇒   ∠DAB = 90°

As in a parallelogram, ∠DAB = 90°. Hence, the parallelogram is a rectangle.

3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Sol. Consider triangles AOB and COD,

AO = OC                                          [Given]

OB = OD                                          [Given]

∠A0B = ∠COD                               [90° each]

∴ ∠AOB  ≅ ∠COD                          [SAS]

⇒      AB = CD                                               ……(i)

Similarly, we can show that BC = DA           ……(ii)

Consider triangles AOB and BOC,

AO = OC                                                            ………[Given]

BO is common.

and   ∠AOB  =  ∠BOC                                          [90°  each ]

∴   ΔAOB = ΔBOC                                                 [ SAS]

⇒         AB = BC                                                ……(iii)

From (i), (ii) and (iii), we get(iii)

AB = BC = CD = DA

Hence, ABCD is a rhombus.

4. Show that the diagonals of a square are equal and bisect each other at right angles.

Sol. Consider triangles DAB and CBA,

AD = BC                                         [Sides of a square]

AB is common.

∠DAB = ∠CBA                             [90° each]

∴ ΔDAB ≅ ΔCBA                          [SAS]

⇒       BD = AC                                       [CPCT]

and ∠1 = ∠2 [CPCT]

Proving as above we can show ∠3 = ∠4.                      …(ii)

Also, ∠2 = ∠3                                       [‘: AB = BC, angles opposite to equal sides are equal.]

and    ∠1 =  ∠4                                        [From (i), (ii),(iii)]

Consider triangles AOD and COD,

AD = DC                                                [Sides of a square]

OD is common.

∠1 = ∠4                                                    [Proved above]

∴   ΔAOD   ≅   ΔCOD                              [SAS]

∴     OA = OC                                          …(iv) [CPCT]

Similarly, we can show that

OB = OD

and      ∠AOD = ∠COD                          [CPCT]

Also,    ∠AOD + ∠COD = 180°             [Linear pair]

⇒      2∠AOD = 1800                            [Using (v)]

⇒        ∠AOD = 90°                                    …(vi)

Hence, diagonals are equal and bisect each other at right angles.

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Sol.
Consider triangles AOB and COD.

AO = DC                                                          [Given]

OB = OD                                                          [Given]

∠AOB = ∠COD                                                [90° each]

∴   ΔAOB  ≅    ΔCOD                                      [SAS]

⇒     AB = CD                                                         …(i)

Similarly, we can show that BC = DA              …(ii)

Consider triangles AOB and BOC,

AO = OC                                                          [ Given ]

BO is common.

and   ∠AOB = ∠BOC

∴        ΔAOB  ≅   ΔCOB

⇒       AB = BC                                                  …..(iii)

From (i), (ii) and (iii), we get

AB = BC = CD = DA

Hence, ΔBCD is a rhombus.

Further, consider ΔDAB and ΔCBA

AB is common

BD = AC                                                          [Given]

∴  ΔDAB  ≅   ΔCBA                                               [SSS]

⇒  ∠DAB = ∠CBA                                                       ….(i)

Also,   as $AD\parallel BC$      (opposite sides of a rhombus) and AB is transversal.

∴ ∠DAB + ∠CBA = 180°                    (Sum of interior angles on  the same side of transversal is 180°.)

⇒   2∠DAB = 180°                                            [From (i)]

⇒     ∠DAB = 90°.

As in a rhombus one angle is 90°. Hence rhombus is a square.

6. Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that

(i) it bisects ∠C also,
(ii) ABCD is a rhombus.

Sol. (i) Consider triangles ABC and ADC,
AB = CD [Opposite sides of parallelogram]

AC is common.
AD = BC                                                     [Opposite sides of parallelogram]

∴  ΔDAC   ≅  ΔBCA                                            [SSS]

⇒      ∠DAC = ∠BCA                                     ….(i)[CPCT]

⇒     ∠DCA = ∠BAC                                      …(ii) (CPCT)

Also,  ∠DAC = ∠BAC                                   …(iii) [(Given]

⇒        ∠DCA = ∠BCA                                      [From (i), (ii), (iii)]

∴  AC bisects ∠C also,

(ii) In parallelogram ∠DAB = ∠DCB,                 [Opposite angles of a parallelogram are equal.]

⇒   ${1 \over 2}\angle DAB = {1 \over 2}\angle DCB$

⇒       ∠DAC   = ∠DCA                                                 [ AC is bisector of ∠A and ∠C.]

∴ CD = AD                                                    [Sides opposite to equal angles are equal.]

In parallelogram, as adjacent sides are equal, hence ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Sol. Consider triangles ADC and ABC,

AD = AB                                                     [Sides of a rhombus]

AC is common.

CD = CB                                                     [Sides of a rhombus]

⇒       ∠DAC = ∠BAC                                ……(i)[CPCT]

and   ∠DCA = ∠BCA                                …(ii) [CPCT]

Hence AC bisects ∠A and ∠C.
Similarly, by taking triangles BAD and BCD, we can show that BD bisects ∠B and ∠D.

8. 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.
(i) Consider triangles ADC and ABC,

∠DAC = ∠BAC                                   [AC is bisector of ∠A]

∠DCA =   ∠BCA                                 [AC is bisector of ∠C]

AC is common.

As in rectangle ABCD, adjacent sides are equal. Hence ABCD is a square.

(ii) Consider triangles DAB and BCD,

AB = BC = CD = DA                                                [Sides of a square]

BD is common.

∴  ΔDAB   ≅  ΔDCB                                                [SSS]

and ∠ABD = ∠CBD                                               [CPCT]

∴  BD bisects ∠B and ∠D.                                    [Using above results ]

9. 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  ≅ CPD
(iv) AQ = CP
(v) APCQ is a parallelogram.

Sol.
(i) Consider triangles APD and CQB,   $AD\parallel BC$  and BD is transversal.

∴    ∠1 = ∠2                                             [Alternate angled ]

AD = BC                                                 [Opposite Aides of a parallelogram]

DP = BQ                                                 [Given]

∴    ΔAPD ≅  ΔCQB.                             [SAS]

(ii) AP = CQ                                                     [CPCT] [From result (1)]

(iii) Consider triangles AQB and CPD,

AB = CD                                               [Opposite sides of a parallelogram ]

∠ABQ  =   ∠CDP                                 [Alternate interior angles as $AB\parallel CD$ and BD is
transversal]

BQ = DP                                              [Given]

∴  ΔAQB  ≅  ΔCPD                           [SAS]

(iv) From result (iii),

ΔAQB   ≅  ΔCPD

∴              AQ = CP                             [Corresponding sides]
AP = CQ                                             [From result (ii)]

AQ = CP                                             [From result (iv)]

Thus, opposite sides of quadrilateral APCQ are equal. Hence, AFCQ is a parallelogram.

10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that

(i)  ΔAPB  ≅  ΔCQD.
(ii)  AP = CQ

Sol. Consider triangles APB and CQD,
∠1 = ∠2                                                          [Alternate angles, $AB\parallel CD$ , BD is transversa ]

∠APB = ∠DQC                                             [90° each]

AB = CD                                                         [ Opposite sides of a parallelogram]

(i) ∴ ΔAPB   ≅   ΔCQD                                 [AAS]

(ii)      AP = CQ.                                            [CPCT]

11. In ΔABC and ΔDEF, AB = DE, $AB\parallel DE$, BC = EF and $BC\parallel EF$  Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that

(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) $AD\parallel CF$   and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC =  DF
(vi) ΔABC ≅  ΔDEE

Sol.

AB = DE  and   $AB\parallel DE$                                                                     [Given]

⇒  ABED is a parallelogram.
[In a quadrilateral if a pair of opposite sides is equal and parallel, then it is a parallelogram.]

BC = EF and $BC\parallel EF$                                                                      [Given]

⇒     BEFC is a parallelogram.                                                                              [Reason same as above]
(iii) From result (i), ABED is a parallelogram.

$AD\parallel BE$     and AD = BE

From result (ii), BEFC is a parallelogram.

$BE\parallel CF$     and BE = CF

⇒  $AD\parallel CF$    and AD = CF.                                                          [From results (i) and (ii)]

(iv) As $AD\parallel CF$  and AD = CF                                                    [From result (iii)]

∴  Quadrilateral ACFD is a parallelogram.
(v) AC = DF.                                                                                 [ ACFD is a parallelogram, result (iv)]
(vi) Consider triangles ABC and DEF,
AB = DE                                                                                        [Given]

BC = EF                                                                                         [Given]

AC = DF                                                                                        [From result (v)]

∴  ΔABC  ≅   ΔDEF.                                                                             [SSS]

12. ABCD is a trapezium in which  $AB\parallel CD$   and AD = BC (see figure). Show that

(i) ∠A  = ∠B
(ii) ∠C = ∠D
(iv) diagonal AC = diagonal BD
[Hint: Extend AB and draw a line  through C parallel to DA intersecting AB produced at E.]

Sol.

(i) Construction: Draw $CE\parallel AD$ , meeting AB produced at E
Proof :   $AB\parallel CD$                                                               [Given ]
and         $AD\parallel CE$                              [Construction]

∴   AECD is a parallelogram.

⇒   AD = CE                                                              [ Opposite sides of arallelogram ]

∴        CE = BC.

⇒    ∠1 = ∠2                            …(i)                     [Angles opposite to equal sides are equal]

Also, ∠D = ∠2                       …(ii)                     [Opposite angles of a parallelogram]

and ∠1 = ∠3                          …(iii)                     [Alternate angles]

⇒  ∠D = ∠3                                                         [From equations (i), (iii)]

⇒    ∠D  =   ∠C

As $AB\parallel CD$    and AD, BC are transversals.

∴   ∠A  + ∠D= 180°                                  …(iv)

[Sum of interior angles on the same side of transversal is 180°.]

∠B  +  ∠C= 180°                                       …(v) [Reason same as above]

Also,  ∠C = ∠D                                         …(vi) [Proved above]

∴    ∠A  = ∠B                                                       [From equations (iv), (v), (vi)]
(iii) Construction: Draw AC and BD.
Proof: Consider triangles DAB and CBA.

AB is common.

∠DAB = ∠CBA                                                [Proved above]

(iv) AC = BD.                                                  [CPCT]