NCERT Solutions for Class 9 Maths 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 8.1

NCERT Solutions for Class 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.

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.

In ∆ABC and ∆DCB,
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.

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 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,
Now, in ∆OAD and ∆OCB, we have
AD = CB                                                              [Sides of a square ABCD]
DAO = OCB                                                   [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.

In ΔABO and ΔADO, we have
BO = OD                                    (Given)
AO = OA                                    (Common)
AOB = AOD                         (Given 90°)
ΔABO ΔADO                     (By SAS congruency rule)
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)

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
(i) Since, ABCD is a parallelogram.
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]
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]

Now, in ∆APD and ∆CQB, we have
AD = CB                               [Opposite sides of a parallelogram ABCD are equal]
PD = QB                               [Given]
∆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]

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
(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)]
[
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

i.e., In quadrilateral ACFD, one pair of opposite sides (AD and CF) are parallel and of equal length.

(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

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

Solution:

We have given a trapezium ABCD in which AB || CD and AD = BC.

(i) Produce AB to E and draw CE || AD

AB || DC
AE || DC Also AD || CE
AECD is a parallelogram.
[
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]