NCERT Solutions For Class 9 Maths Chapter 8

NCERT Solutions Class 9 Maths Quadrilaterals

NCERT solutions for class 9 Maths chapter 8 Quadrilaterals include solutions for all the questions present in class 9 NCERT maths textbooks. This is the best study material for students preparing for their finals. NCERT solutions for class 9 Maths chapter 8 are available in a free PDF format to help students in practising Quadrilaterals on a regular basis. All the solutions present in this study materials are solved by the subject experts and are explained in a lucid manner along with the more examples, formulas, and other important questions. Students can download these NCERT solutions for class 9 Maths chapter 8 pdf files and prepare for their exams.

NCERT Solutions For Class 8 Maths Chapter 8 Exercises

Q1. The angles of quadrilateral are in the ratio 3:5:9:13. Find the angles of the quadrilateral.

Solution:

Let the common ratio between the angles be x.
We know that the ‘Sum of the interior angles of the quadrilateral’ = 360
Now,
3x+5x+9x+13x=360
30x=360
x=12
Therefore the Angles of the quadrilateral are:
3x=3×12=36

5x=5×12=60 9x=9×12=108 13x=13×12=156

 

Q2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution:

1

Given,
PQ = RS
To show,

PQRS is a rectangle we have to prove that one of its interior angle is right angled.
Proof,
In PQRandQPS,
QR=QP (Common side)
PR=PS (Opposite sides of a parallelogram are equal)
PR=QS (Given)
Therefore, PQRQPSby SSS congruence condition.
∠P = ∠Q (by CPCT)
also,
∠P + ∠Q = 180° (Sum of the angles on the same side of the transversal)
⇒ 2∠P = 180°
⇒ ∠P = 90°
Thus PQRS is a rectangle.

 

Q3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution:

2

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.

Given,
OA=OC,OB=ODandAOB=BOC=OCD=ODA=90

To show,
ABCD is parallelogram and AB=BC=CD=DA
Proof,
In AOBandCOB,
OA=OC (Given)
AOB=COB (Opposite sides of a parallelogram are equal)
OB=BO (Common)
Therefore, AOBbigtriangleupCOB (by SAS congruence condition).
Thus, AB=BC (by CPCT-Corresponding parts of Congruent)
Similarly we can prove,
AB=BC=CD=DA
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

 

Q4. Show that the diagonals of a square are equal and bisect each other at right angles.

Solution:

3

Let PQRS be a square and its diagonals PR and QS intersect each other at O.
To show,
PR=QS,PO=ORandPOQ=90

Proof,
In PQRandQRS,
QR=QP (Common)
PQR=QPS=90
PR=PS (Given)
Therefore, PQRQRS by SAS congruence condition.
Thus, PR=PS (by CPCT).

Therefore, diagonals are equal.
Now,
In POQandROS,
QPO=SRO (Alternate interior angles)
POQ=ROS (Vertically opposite)
PQ=RS (Given)
Therefore, ΔAOB ≅ ΔCOD (by AAS congruence condition).
Thus, PO=RO by CPCT. (Diagonal bisect each other.)
Now,
In POQandROQ,
OQ=QO (Given)
PO=RO (diagonals are bisected)
PQ=RQ (Sides of the square)
Therefore, POQROQ by SSS congruence condition.
also, POQ=ROQ

POQ+ROQ=180 (Linear pair)
Thus, POQ=ROQ=90 (Diagonals bisect each other at right angles)

 

Q5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Solution:

4
Given,
Let ABCD be a quadrilateral in which diagonals ACandBD bisect each other at right angle at O.
To prove,
Quadrilateral ABCD is a square.
Proof,
In AOBandCOD,
AO=CO (Diagonals bisect each other)
AOB=COD (Vertically opposite)
OB=OD (Diagonals bisect each other)
Therefore, AOBCODby SAS congruence condition.
Thus, AB=CD by CPCT. …………… (i)
also,
OAB=OCD (Alternate interior angles)
ABCD

Now,
In AODandCOD,
AO=CO (Diagonals bisect each other)
AOD=COD (Vertically opposite)
OD=OD (Common)
Therefore, AODCOD (by SAS congruence condition).
Thus, AD=CD (by CPCT).  ………………… (ii)
also,
AD=BCandAD=CD
AD=BC=CD=AB  ……………….(iii)
also,  ADC=BCD (by CPCT).
and ADC+BCD=180 (co-interior angles)
2ADC=180
ADC=90   ……………. (iv)
One of the interior angle is right angle.
Thus, from (i), (ii) , (iii) and (iv) the given quadrilateral ABCD is a square.

 

Q6. Diagonal AC of a parallelogram ABCD bisects A. Show that
(i) it bisects C also,
(ii) ABCD is a rhombus.

Solution:

5

(i)In ADCandCBA,
AD=CB (Opposite sides of a parallelogram)
DC=BA (Opposite sides of a parallelogram)
AC=CA (Common)
Therefore, ADCCBA by SSS congruence condition.
Thus,
ACD=CAB (by CPCT)
and CAB=CAD (Given)
ACD=BCA
Thus, AC bisects C also.

(ii) ACD=CAD (Proved)
AD=CD (Opposite sides of equal angles of a triangle are equal)
Also, AB=BC=CD=DA (Opposite sides of a parallelogram)
Thus, ABCD is a rhombus.

 

Q7. ABCD is a rhombus. Show that diagonal AC bisects A as well as C and diagonal BD bisects B as well as D.
Solution:

6

Let ABCD is a rhombus and ACandBD are its diagonals.
Proof,
AD=CD (Sides of a rhombus)
DAC=DCA (Angles opposite of equal sides of a triangle are equal.)
also, ABCD
DAC=BCA (Alternate interior angles)
DCA=BCA
Therefore, ACbisectC.
Similarly, we can prove that diagonal ACbisectA.

Also, by preceding above method we can prove that diagonal BDbisectBaswellasD.

 

Q8. PQRS is a rectangle in which diagonal PR bisects P as well as R. Show that:
(i) PQRS is a square
(ii) diagonal QS bisects Q as well as S.

Solution:

7

(i) SPR=SRP (PR bisects P as well as R)
PS=RS (Sides opposite to equal angles of a triangle are equal)
also, RS=PQ (Opposite sides of a rectangle)
Therefore, PQ=QR=RS=SP
Thus, PQRS  is a square.

(ii) In QRS,
QR=RS
RSQ=RQS (Angles opposite to equal sides are equal)
also, RSQ=PQS (Alternate interior angles)
RQS=PQS
Thus, QS bisects Q.
Now,
RQS=PSQ
RSQ=PSQ
Thus, BD bisects D

 

Q9. In parallelogram ABCD, two points PandQ are taken on diagonal BD such that DP=BQ. Show that:
(i) APDCQB
(ii) AP=CQ
(iii) AQBCPD
(iv) AQ=CP
(v) APCQ is a parallelogram

Solution:

8

(i) In APDandCQB,
DP=BQ (Given)
ADP=CBQ (Alternate interior angles)
AD=BC (Opposite sides of a ||gm)
Thus, APDCQB (by SAS congruence condition).

(ii) AP=CQ by CPCT as APDCQB.

(iii) In AQBandCPD,
BQ=DP (Given)
ABQ=CDP (Alternate interior angles)
AB=BC=CD (Opposite sides of a parallelogram)
Thus, AQBCPD (by SAS congruence condition).

(iv) AQ=CP by CPCT as AQBCPD.

(v) From (ii)  and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus, APCQ is a parallelogram.

 

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) APBCQD
(ii) AP = CQ
 

Solution:

9

(i) In APBandCQD,
ABP=CDQ (Alternate interior angles)
APB=CQD (equal to right angles as AP and CQ are perpendiculars)
AB = CD (ABCD is a parallelogram)
Thus, APBCQD (by AAS congruence condition).

(ii) AP = CQ by CPCT as APBCQD.

Q11. In ABCandDEF, AB=DE,ABDE,BC=EFandBCEF. Vertices A, B and C are joined to vertices D, E and F respectively.
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) ADCF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC=DF

(vi) ABCDEF
Solution:

10

(i) AB=DEandABDE (Given)

Thus, quadrilateral ABED is a parallelogram because two opposite sides of a quadrilateral are equal and parallel to each other.

(ii) Again BC=EFandBCEF.

Thus, quadrilateral BEFC is a parallelogram.

(iii)  Since ABED and BEFC are parallelograms.
AD=BEandBE=CF (Opposite sides of a parallelogram are equal)

Thus, AD=CF.

Also, ADBEandBECF (Opposite sides of a parallelogram are parallel)

Thus, ADCF.

(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other. Thus, it is a parallelogram.
(v) ACDFandAC=DF because ACFD is a parallelogram.

(vi) ABCandDEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

Thus, ABCDEF (by SSS congruence condition).

 

Q12. ABCD is a trapezium in which ABCD and AD=BC .Show that
(i) A=B

(ii) C=D
(iii) ABCBAD
(iv) diagonal AC = diagonal BD

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

11

Construction: Draw a line through C parallel to DA intersecting AB produced at E.
(i) CE = AD (Opposite sides of a parallelogram)
AD = BC (Given)
Therefor, BC = CE
CBE=CEBalso,
A+CBE=180 (Angles on the same side of transversal and CBE=CEB)
B+CBE=180 (Linear pair)
A=B

(ii) A+D=B+C=180 (Angles on the same side of transversal)
A+D=A+C   (as, A=B)

D=C

(iii) In ABCandBAD,
AB = AB (Common)
DBA=CBA
AD = BC (Given)
Thus, ABCBAD (by SAS congruence condition).

(iv)  Diagonal AC = diagonal BD (by CPCT as ABCBAD.)

 

Exercise 2:

Q1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB,BC,CDandDA. AC is a diagonal. Show that :
(i) SRAC and SR=12AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

12

Solution:

(i) In DAC,
R is the mid point of DC and S is the mid point of DA.
Thus by mid point theorem, SRAC and SR=12AC

(ii) In BAC,
P is the mid point of AB and Q is the mid point of BC.
Thus by mid point theorem, PQAC and PQ=12AC
also, SR=12AC
Thus, PQ = SR

(iii) SRAC– from (i)
and, PQAC– from (ii)
SRPQ– from (i) and (ii)
also, PQ = SR
Thus, PQRS is a parallelogram.

 

Q2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Solution:

13

Given,
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.
To Prove,
PQRS is a rectangle.
Construction,
Join AC and BD

Proof,
In DRSandBPQ,
DS = BQ (Halves of the opposite sides of the rhombus)
SDR=QBP (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
Thus, DRSBPQ (by SAS congruence condition).
RS = PQ (by CPCT)    ………… (i)
In QCRandSAP,
RC = PA (Halves of the opposite sides of the rhombus)
RCQ=PAS (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
Thus, QCRSAP (by SAS congruence condition).
RQ = SP (by CPCT)    ……(ii)
Now,
In CBD,
R and Q are the mid points of CD and BC respectively.
QRBD
also,
P and S are the mid points of AD and AB respectively.
PSBD
QRPS
Thus, PQRS is a parallelogram.
also, PQR=90
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
Q=90
Thus, PQRS is a rectangle.

 

Q3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

14

Solution:

Given,
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.
Construction,
AC and BD are joined.
To Prove,
PQRS is a rhombus.
Proof,
In ABC
P and Q are the mid-points of AB and BC respectively
Thus, PQAC and PQ=12AC (Mid point theorem) ……….(i)
In ADC,
SRAC and SR=12AC (Mid point theorem)   ……(ii)
So, PQSR and PQ=SR
As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.
PSQR and PS=QR  (Opposite sides of parallelogram)  ….(iii)
Now,
In BCD,
Q and R are mid points of side BC and CD respectively.
Thus, QRBDandQR=12BD (Mid point theorem)   ……(iv)
AC = BD (Diagonals of a rectangle are equal)   ……… (v)
From equations (i), (ii), (iii), (iv) and (v),
PQ = QR = SR = PS
So, PQRS is a rhombus.

 

Q4. ABCD is a trapezium in which ABDC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F. Show that F is the mid-point of BC.

 15
Solution:

Given,
ABCD is a trapezium in which ABDC, BD is a diagonal and E is the mid-point of AD.
To prove,
F is the mid-point of BC.
Proof,
BD intersected EF at G.
In BAD,
E is the mid point of AD and also EGAB.
Thus, G is the mid point of BD (Converse of mid point theorem)
Now,
In BDC,
G is the mid point of BD and also GFABDC.
Thus, F is the mid point of BC (Converse of mid point theorem)


Q5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.

16 

Solution:

Given
ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively.
To show,
AF and EC trisect the diagonal BD.
Proof,

ABCD is a parallelogram
Therefore, ABCD

also, AEFC

Now,
AB=CD  (Opposite sides of parallelogram ABCD)
12AB=12CD
AE=FC (E and F are midpoints of side AB and CD)
AECF is a parallelogram (AE and CF are parallel and equal to each other)
AFEC (Opposite sides of a parallelogram)
Now,
In DQC,
F is mid point of side DC and FPCQ (as AFEC).
P is the mid-point of DQ (Converse of mid-point theorem)
DP=PQ   ………(i)
Similarly,
In APB,
E is mid point of side AB and EQAP (as AFEC).
Q is the mid-point of PB (Converse of mid-point theorem)
PQ=QB    …….(ii)
From equations (i) and (i),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD.

 

Q6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Solution:

17

Let ABCD be a quadrilateral and P, Q, R and S are the mid points of AB, BC, CD and DA respectively.
Now,
In ACD,
R and S are the mid points of CD and DA respectively.
Thus, SRAC.
Similarly we can show that,
PQAC
PSBD
QRBD
Thus, PQRS is parallelogram.
PR and QS are the diagonals of the parallelogram PQRS. So, they will bisect each other.

 

Q7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MDAC
(iii) CM=MA=12AB

18

 

Solution:
(i) In ACB,
M is the mid point of AB and MDBC
Thus, D is the mid point of AC (Converse of mid point theorem)

(ii) ACB=ADM (Corresponding angles)
also, ACB=90
Thus, ADM=90and MDAC

(iii)  In AMDandCMD,
AD = CD (D is the midpoint of side AC)
ADM=CDM (Each 90°)
DM = DM (common)
Thus, AMDCMD (by SAS congruence condition).
AM = CM (by CPCT)
also, AM=12AB (as M is mid point of AB)
Hence, CM=AM=12AB.