Ncert Solutions For Class 9 Maths Ex 8.1

Ncert Solutions For Class 9 Maths Chapter 8 Ex 8.1

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.)