- NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1
- NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2
NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1
Ex 8.1 Class 9 Maths Question 1.
If the diagonals of a parallelogram
are equal, then show that it is a rectangle.
Solution:
Let ABCD is a parallelogram such
that AC = BD.
AC = DB [Given]
AB = DC [Opposite sides of a parallelogram]
BC = CB [Common]
∴ ∆ABC ≅ ∆DCB [By SSS congruency rule]
⇒ ∠ABC = ∠DCB [By CPCT] …(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 get
∠ABC = ∠DCB = 90°
i.e., ABCD is a parallelogram having an angle equal to 90°.
∴ ABCD is a rectangle.
Ex 8.1 Class 9 Maths Question 2.
Show that the diagonals of a square
are equal and bisect each other at right angles.
Solution:
Let ABCD be a square such that its
diagonals AC and BD intersect at O.
(i) To prove that the diagonals of a square are
equal, we need to prove AC = BD.
In ∆ABC and ∆BAD, we have
AB = BA
[Common]
BC = AD [Sides of a
square ABCD]
∠ABC = ∠BAD [Each angle is 90°]
∴ ∆ABC ≅ ∆BAD [By SAS congruency rule]
AC = BD [By CPCT] …..…(1)
(ii) AD || BC and AC is a transversal. [∵ A square is a parallelogram]
∴ ∠DAC = ∠ACB
[Alternate interior angles are equal]
Similarly, ∠ADB = ∠DBC
Now, in ∆OAD and ∆OCB, we have
AD = CB [Sides
of a square ABCD]
∠DAO = ∠OCB
[Proved above]
∠ADO = ∠OBC
[Proved above]
∴ ∆OAD ≅ ∆OCB [By
ASA congruency rule]
⇒ OA = OC and OD = OB [By CPCT]
i.e., the diagonals AC and BD bisect each other at O. …….(2)
(iii)
In ∆OBA and ∆ODA, we have
OB = OD [Proved above]
BA = DA [Sides of
a square ABCD]
OA = OA [Common]
∴ ∆OBA ≅ ∆ODA [By SSS congruency rule]
⇒ ∠AOB = ∠AOD [By CPCT] ….…(3)
∵ ∠AOB and ∠AOD form a linear pair.
∴ ∠AOB + ∠AOD = 180°
∴ ∠AOB = ∠AOD = 90° [By(3)]
⇒ AC ⊥ BD ……(4)
From (1), (2) and (4), we get diagonals of a
square are equal and bisect each other at right angles.
Ex 8.1 Class 9 Maths Question 3.
Diagonal AC of a parallelogram ABCD
bisects ∠A (see figure). Show that
(i)
it bisects ∠C also,
(ii)
ABCD is a rhombus.
Solution:
We have a parallelogram ABCD in which diagonal AC bisects ∠A.
⇒ ∠DAC = ∠BAC
∴ AB || DC and AC is a transversal.
∴ ∠1 = ∠3 …(1) [ ∵ Alternate interior angles are equal]
Also, BC || AD and AC is a transversal.
∴ ∠2 = ∠4 …(2) [ ∵ Alternate interior angles are equal]
Also, ∠1 = ∠2 …(3) [ ∵ AC bisects ∠A]
From (1), (2) and (3), we have
∠3 = ∠4
⇒ AC bisects ∠C.
(ii) In
∆ABC, we have
∠1 = ∠4 [From (2) and (3)]
⇒ BC = AB …(4) [ ∵ Sides opposite to equal angles of a ∆ are equal]
Similarly, AD = DC ……..(5)
But, ABCD is a parallelogram. [Given]
∴ AB = DC …..…(6)
From (4), (5) and (6), we have
AB = BC = CD = DA
Hence, ABCD is a rhombus.
Ex 8.1 Class 9 Maths Question 4.
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:
We have a rectangle ABCD such that AC bisects ∠A as well as ∠C.
i.e., ∠1 = ∠4 and ∠2 = ∠3 ……..(1)
(i)
Since every rectangle is a parallelogram.
∴ ABCD is a parallelogram.
⇒ AB || DC and AC is a transversal.
∴ ∠2 = ∠4 …..…(2) [ ∵ Alternate interior
angles are equal]
From (1) and (2), we have
∠3 = ∠4
In ∆ABC, ∠3 = ∠4
⇒ AB = BC [ ∵ Sides opposite to
equal angles of a A are equal]
Similarly, CD = DA
So, ABCD is a rectangle having adjacent sides equal.
⇒ ABCD is a square.
(ii)
Since ABCD is a square and diagonals of a square bisect the opposite angles.
So, BD bisects ∠B as well as ∠D.
Ex 8.1 Class 9 Maths Question 5.
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.
Solution:
We have a parallelogram ABCD, BD is the diagonal and points P and Q are such that PD = QB
(i)
Since AD || BC and BD is a transversal.
∴ ∠ADB = ∠CBD [ ∵ Alternate interior
angles are equal]
⇒ ∠ADP = ∠CBQ
Now, in ∆APD and ∆CQB, we have
AD = CB [Opposite
sides of a parallelogram ABCD are equal]
PD = QB [Given]
∠ADP = ∠CBQ [Proved
above]
∴ ∆APD ≅ ∆CQB [By
SAS congruency rule]
(ii)
Since ∆APD ≅ ∆CQB [Proved above]
⇒ AP = CQ [By CPCT]
(iii)
Since AB || CD and BD is a transversal.
∴ ∠ABD = ∠CDB
⇒ ∠ABQ = ∠CDP
Now, in ∆AQB and ∆CPD, we have
QB = PD [Given]
∠ABQ = ∠CDP [Proved above]
AB = CD [Opposite
sides of a parallelogram ABCD are equal]
∴ ∆AQB ≅ ∆CPD [By
SAS congruency rule]
(iv) Since
∆AQB ≅ ∆CPD [Proved above]
⇒ AQ = CP
[By CPCT]
(v)
In a quadrilateral APCQ,
Opposite sides are equal. [Proved
above]
∴ APCQ is a parallelogram.
Ex 8.1 Class 9 Maths Question 6.
ABCD is a parallelogram and AP and CQ are
perpendiculars from vertices A and C on diagonal BD (see figure). Show that
(i)
∆APB ≅ ∆CQD
(ii)
AP = CQ
Solution:
(i) In ∆APB and ∆CQD, we have
∠APB = ∠CQD [Each
90°]
AB = CD [ ∵ Opposite sides of a
parallelogram ABCD are equal]
∠ABP = ∠CDQ
[ ∵ Alternate angles are equal as AB || CD and BD is a transversal]
∴ ∆APB ≅ ∆CQD [By
AAS congruency rule]
(ii)
Since ∆APB ≅ ∆CQD [Proved
above]
⇒ AP = CQ [By CPCT]
Ex 8.1 Class 9 Maths Question 7.
ABCD is a trapezium in which AB || 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].
We have given a trapezium ABCD in which AB || CD and
AD = BC.
(i)
Produce AB to E and draw CE || AD
⇒ AE || DC Also AD || CE
∴ AECD is a parallelogram.
⇒ AD = CE …..…(1)
[ ∵ Opposite sides of the parallelogram are equal]
But AD = BC …..…(2) [Given]
From (1) and (2), we get BC = CE
Now, in ∆BCE, we have BC = CE
⇒ ∠CEB = ∠CBE ….…(3)
[∵ Angles opposite to equal sides of a triangle are equal]
Also, ∠ABC + ∠CBE = 180° …..…(4) [Linear pair]
and ∠A + ∠CEB = 180° …..…(5) [Co-interior angles of a
parallelogram ADCE]
From (4) and (5), we get
∠ABC + ∠CBE = ∠A + ∠CEB
⇒ ∠ABC = ∠A [From
(3)]
⇒ ∠B = ∠A …..…(6)
(ii)
AB || CD and AD is a transversal.
∴ ∠A + ∠D = 180° ….…(7)
[Co-interior angles]
Similarly, ∠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
above]
∴ ∆ABC ≅ ∆BAD [By
SAS congruency rule]
(iv)
Since ∆ABC ≅ ∆BAD [Proved above]
⇒ AC = BD [By CPCT]
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