NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1
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9 Maths Chapter 8 Quadrilaterals Ex 8.1 are the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 8 Quadrilaterals Ex 8.1.
- NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1
- NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.2
Ex 8.1 Class 9 Maths 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°
[Using angle sum property of a
quadrilateral]
⇒ 30x = 360°
⇒ x = 360°/30 = 12°
∴ 3x = 3 × 12° = 36°
5x = 5 × 12° = 60°
9x = 9 × 12° = 108°
13x = 13 × 12° = 156°
Thus,
the required angles of the quadrilateral are 36°, 60°, 108° and 156°.
Ex 8.1 Class 9 Maths Question 2.
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 3.
Show that if the diagonals of a
quadrilateral bisect each other at right angles, then it is a rhombus.
Solution:
Let ABCD be a quadrilateral such
that the diagonals AC and BD bisect each other at right angles at O.
AO = AO [Common]
OB = OD [O is the mid-point of BD]
∠AOB = ∠AOD [Each 90°]
∴ ∆AQB ≅ ∆AOD [By SAS congruency rule]
∴ AB = AD [By CPCT] ……..(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 Class 9 Maths Question 4.
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 5.
Show that if the diagonals of a
quadrilateral are equal and bisect each other at right angles, then it is a
square.
Solution:
Let ABCD be a quadrilateral such
that diagonals AC and BD are equal and bisect each other at right angles. So,
ABCD is a parallelogram.
BO = OD (Given)
AO = OA (Common)
∠AOB = ∠AOD (Given 90°)
∴ ΔABO ≅ ΔADO (By SAS congruency rule)
AB = AD (By CPCT)
Similarly, AB = BC, BC = CD and CD = AD
i.e.,
AB = BC = CD = DA ……..(i)
Again,
in ΔABC and ΔBAD, we have
AB =
BA (Common)
AC
= BD
(Given)
BC =
AD
[From (i)]
∴ ΔABC ≅ ΔBAD
∴ ∠ABC = ∠BAD ……..(ii)
But ∠ABC + ∠BAD = 180° (Sum of co-interior angles of a
parallelogram is 180°)
∴ ∠ABC = ∠BAD = 90° [From eq. (ii)]
Thus, AB = BC = CD = DA and ∠A = 90°
∴ ABCD is a square.
Ex 8.1 Class 9 Maths Question 6.
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 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:
Since, ABCD is a rhombus.
⇒ AB = BC = CD = DA
Also, AB || DC 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 a transversal.
⇒ ∠1 = ∠3 ….…(2) [ ∵ Alternate interior
angles are equal]
From (1) and (2), we have
∠2 = ∠3 ….…(3)
Since, AB || DC and AC is a transversal.
∴ ∠2 = ∠4 ……(4) [ ∵ Alternate interior
angles are equal]
From (1) and (4), we have ∠1 = ∠4
∴ AC bisects ∠C as well as ∠A.
Similarly, we can prove that BD bisects ∠B as well as ∠D.
Ex 8.1 Class 9 Maths 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:
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 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 ≅ 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 10.
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 11.
In ∆ABC and ∆DEF, AB = DE, AB || DE, BC = EF and BC
|| 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 || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ∆ABC ≅ ∆DEF
Solution:
(i) We have AB = DE [Given]
and AB || DE [Given]
i.e., ABED is a quadrilateral in which a pair of opposite sides (AB and DE) are
parallel and of equal length.
∴ ABED is a parallelogram.
(ii)
BC = EF [Given]
and BC || EF [Given]
i.e. BEFC is a quadrilateral in which a pair of opposite sides (BC and EF) are
parallel and of equal length.
∴ BEFC is a parallelogram.
(iii)
ABED is a parallelogram [Proved in (i)]
∴ AD || BE and AD = BE ….…(1)
[ ∵ Opposite sides of a parallelogram are equal and parallel]
Also, BEFC is a parallelogram. [Proved in (ii)]
BE || CF and BE = CF ….…(2)
[ ∵ Opposite sides of a parallelogram are equal and parallel]
From (1) and (2), we have
AD || CF and AD = CF
(iv)
Since AD || CF and AD = CF [Proved
above]
i.e., In quadrilateral ACFD, one pair of opposite sides (AD and CF) are
parallel and of equal length.
∴ Quadrilateral ACFD is a parallelogram.
(v)
Since ACFD is a parallelogram. [Proved
above]
So, AC = DF [∵ Opposite sides of a
parallelogram are equal]
(vi)
In ∆ABC and ∆DFF, we have
AB = DE [Given]
BC = EF [Given]
AC = DF [Proved
in (v)]
∆ABC ≅ ∆DFF [By
SSS congruency rule]
Ex 8.1 Class 9 Maths Question 12.
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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