Quadrilaterals (Exercise 8.1)

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Quadrilaterals


Chapter 8


Exercise 8.1


EX 8.1 QUESTION 1.


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

Solution:

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
∴ 3x + 5x + 9x + 13x = 360°           [Angle sum property of a quadrilateral]
⇒ 30x = 360°
⇒ x = 360°/30 = 12°
∴First angle =  3x = 3 x 12° = 36°
Second angle = 5x = 5 x 12° = 60°
Third angle = 9x = 9 x 12° = 108°
Forth angle = 13x = 13 x 12° = 156°


EX 8.1 QUESTION 2.


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

Solution:

In ∆ABC and ∆DCB,
AC = DB [Given]
AB = DC [Opposite sides of a parallelogram are equal]
BC = CB [Common]
∴ ∆ABC ≅ ∆DCB [By SSS congruency]
⇒ ∠ABC = ∠DCB [By C.P.C.T.] …(1)
Now, AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram]
∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles]
From (1) and (2), we have
∠ABC = ∠DCB = 90°
i.e., ABCD is a parallelogram having an angle equal to 90°.
∴ ABCD is a rectangle.


EX 8.1 QUESTION 3.


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

Solution:

∴ In ∆AOB and ∆AOD, we have
AO = AO [Common]
OB = OD [O is the mid-point of BD]
∠AOB = ∠AOD [Each 90]
∴ ∆AQB ≅ ∆AOD [By,SAS congruency
∴ AB = AD [By C.P.C.T.] ……..(1)
Similarly, AB = BC .. .(2)
BC = CD …..(3)
CD = DA ……(4)
∴ From (1), (2), (3) and (4), we have
AB = BC = CD = DA
Thus, the quadrilateral ABCD is a rhombus.
Alternatively: ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus.


EX 8.1 QUESTION 4.


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

Solution:

In ∆ABC and ∆BAD,
AB = BA [Common]
BC = AD [Sides of a square ABCD]
∠ABC = ∠BAD [Each angle is 90°]
∴ ∆ABC ≅ ∆BAD [By SAS congruency]
AC = BD [ C.P.C.T.]

(ii)In ∆AOB and ∆COD
∴ ∠OAB = ∠OCD           [Alternate interior angles are equal]
AB = CD                         [Opposite side of a square]
∠OAB = ∠OCD               [Alternate interior angles are equal]
∆ABC ≅ ∆BAD                [ASA congruency]
AO = OC , BO =  OD      [C.P.C.T.]

(iii) In ∆AOB and ∆AOD,
OB = OD                       [Proved]
AB = AD                       [Sides of a square ABCD]
OA = OA                      [Common]
∴ ∆BAD ≅ ∆ABC         [By SSS congruency]
⇒ ∠AOB = ∠AOD       [By C.P.C.T.]
∠AOB + ∠AOD = 180° [ linear pair.]
⇒2∠AOB = 180°
⇒ ∠AOB = 180°/2 = 90°
The diagonals of a square are equal and bisect each other at right angles.


EX 8.1 QUESTION 5.


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

Solution:

Now, in ∆BAD and ∆ABC,
AD = BC [Each 90°]
BD = AC [Common]
∴ ∆BAD ≅ ∆ABC [SAS congruency]
⇒ ∠BAD = ∠ABC [ C.P.C.T.] …(1)
∠BAD + ∠ABC = 180°   [Co – interior angles]
∠ABC = 180°/2 = 90°


EX 8.1 QUESTION 6.


Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

 

Solution:

(i) ABCD is a parallelogram.
∠DAC = ∠BAC ———(i)
∴ ∠DAC = ∠BCA ——(ii)       [Alternate interior angles are equal]
∠BAC = ∠ACD ——–(iii)       [Alternate interior angles are equal]
From equation (I),(ii) and (iii)
∠ACD = ∠BCA———(iv)
⇒ AC bisects ∠C.

(ii) From equation (ii) and (iv)
∠ACD = ∠DAC
In ∆ADC,
∠ACD = ∠DAC
AD = DC  [Angles opposite to equal sides of a triangle are equal]

Parallelogram whose adjacent sides are equal is a rhombus. Hence, ABCD is a rhombus.


EX 8.1 QUESTION 7.


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

Solution:

ABCD is a rhombus.
⇒ AB = BC = CD = DA
Also, AB || CD and AD || BC
Now, CD = AD ⇒ ∠1 = ∠2 …….(1) [ ∵ Angles opposite to equal sides of a triangle are equal]
Also, AD || BC and AC is the transversal.       [ ∵ Every rhombus is a parallelogram]
⇒ ∠1 = ∠3 …(2)                         [ ∵ Alternate interior angles are equal]
From (1) and (2),
∠2 = ∠3 …(3)
Since AB || DC and AC is transversal.
∴ ∠2 = ∠4 …(4)                                  [ ∵ Alternate interior angles are equal] From (1) and (4),
∠1 = ∠4
∴ AC bisects ∠C as well as ∠A.
Similarly, we can prove that BD bisects ∠B as well as ∠D.


EX 8.1 QUESTION 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.

Solution:


EX 8.1 QUESTION 9.


In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that:

(i) ΔAPD ≅ ΔCQB

(ii) AP = CQ

(iii) ΔAQB ≅ ΔCPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Solution:

(i) In ∆APD and ∆CQB
DP = BQ
∴ ∠ADB = ∠CBD [Alternate interior angles are equal]
AD = BC     [Opposite sides of a parallelogram are equal]
∴ ∆APD ≅ ∆CQB [SAS congruency]

(ii) ∆APD ≅ ∆CQB      [Proved]
⇒ AP = CQ               [C.P.C.T.]

(iii) In ∆AQB and ∆CPD,
QB = DP         [Given]
∠ABQ = ∠CDP [Alternate interior angles are equal]
AB = CD [ Opposite sides of a parallelogram are equal]
∴ ∆AQB = ∆CPD [SAS congruency]

(iv) ∆AQB = ∆CPD  [Proved]
⇒ AQ = CP          [ C.P.C.T.]

(v) In a quadrilateral ∆PCQ,
Opposite sides are equal. [Proved]
∴ ∆PCQ is a parallelogram.


EX 8.1 QUESTION 10.


. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that

(i) ΔAPB ≅ ΔCQD

(ii) AP = CQ

Solution:

(i) In ∆APB and ∆CQD,
∠APB = ∠CQD      [Each 90°]
AB = CD [ ∵ Opposite sides of a parallelogram are equal]
∠ABP = ∠CDQ  [Alternate angles are equal ]
∴ ∆APB = ∆CQD [AAS congruency]

(ii) ∆APB ≅ ∆CQD [Proved]
⇒ AP = CQ [ C.P.C.T.]


EX 8.1 QUESTION 11.


 In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22).

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.

Solution:

(i) In ABED
AB = DE [Given]
AB || DE [Given]
∴ ABED is a parallelogram.

(ii)In BEFC BC = EF [Given]
BC = EF [Given]
BC || EF [Given]
∴ BEFC is a parallelogram.

(iii) In ABED
∴ AD || BE and AD = BE …(i)[Opposite sides of a parallelogram are equal and parallel]
In BEFC
BE || CF and BE = CF …(ii) [Opposite sides of a parallelogram are equal and parallel]
From (i) and (ii), we have
AD || CF and AD = CF

(iv) In ACFD
AD || CF and AD = CF [Proved]
∴Quadrilateral ACFD is a parallelogram.

(v)In ACFD [Proved]
AC =DF     [Opposite sides of a parallelogram are equal]

(vi) In ∆ABC and ∆DFF, we have
AB = DE [Given]
BC = EF [Given]
AC = DE [Proved]
∆ABC ≅ ∆DFF [SSS congruency]


EX 8.1 QUESTION 12.


ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ΔABC ≅ ΔBAD

diagonal AC = diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Solution:

(i) Produce AB to E and draw CF || AD.. .(1)
∵ AB || DC
⇒ AE || DC Also AD || CE
∴ AECD is a parallelogram.
⇒ AD = CE …(1)      [Opposite sides of the parallelogram are equal]
AD = BC …(2)   [Given]
BC = CE     [From equation (1) and (2)]
Now, in ∆BCE, we have BC = CF
⇒ ∠CEB = ∠CBE …(3)    [Angles opposite to equal sides of a triangle are equal]
∠ABC + ∠CBE = 180° … (4)     [Linear pair]
∠A + ∠CEB = 180° …(5)     [Co-interior angles of a parallelogram ADCE]
From (4) and (5),
∠ABC + ∠CBE = ∠A + ∠CEB
⇒ ∠ABC = ∠A [From (3)]
⇒ ∠B = ∠A …(6)

(ii) ABCD is a trapezium in which AB || CD
∴ ∠A + ∠D = 180° …(7) [Co-interior angles]
∠B + ∠C = 180° … (8)
From (7) and (8), we get
∠A + ∠D = ∠B + ∠C
⇒ ∠C = ∠D [From (6)]

(iii) In ∆ABC and ∆BAD, we have
AB = BA [Common]
BC = AD [Given]
∠ABC = ∠BAD [Proved]
∴ ∆ABC = ∆BAD [By SAS congruency]

(iv) Since, ∆ABC = ∆BAD [Proved]
⇒ AC = BD [By C.P.C.T.]