Ncert Solutions For Class 9 Maths Ex 7.1

Ncert Solutions For Class 9 Maths Chapter 7 Ex 7.1

Question 1:

In quadrilateral PQRS, PR = PQ and PQ bisects P (look at the fig.). Show that the PQR = PQS. What can you say about QR and QS?

1

Solution:

In PQR and PQS , we have

PR = PS

RPQ=SPQ (PQ bisects P)

PQ = PQ (common)

PQR=PQS (By SAS congruence)

Hence Proved.

Therefore, QR = QS (CPCT)

 

Question 2:

ABCD is a quadrilateral in which AD = BC and DAB= CBA (see Fig.). Prove that

(i) ABD=BAC

(ii) BD = AC

(iii) ABD=BAC

2

Solution:

In the given figure, ABCD is a quadrilateral in which AD = BC and DAB= CBA.

In DAB and BAC, we have

AD = BC [Given]

DAB=CBA [Given]

AB = AB [Common]

ABD=BAC [By SAS congruence]

BD = AC [CPCT]

and ABD=BAC [CPCT]

Hence Proved.

 

Question 3:

AD and BC are equal perpendiculars to a line segment AB (see Fig.). Show that CD bisects AB.

3

Solution:

In AOD=BOC, we have,

AOD=BOC [Vertically opposite angles)

CBO=DAO(Each=90o)

AD = BC [Given]

AOD=BOC [By AAS congruence]

Also, AO = BO [CPCT]

Hence, CD bisects AB Proved.

 

Question 4:

L and m are two para;;e; lines intersected by another pair of parallel lines p and q (see fig.). Show that ABC+CDA .

4

Solution:

In the given figure, ABCD is a parallelogram in which AC is a diagonal, i.e., AB || DC and BC || AD.

In ABCandCDA, we have,

BAC=DCA (Alternate angles)

BCA=DAC (Alternate angles)

AC = AC (Common)

ABC=CDA (By SAS congruence)

Hence Proved.

 

Question 5:

Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendicular from B to the arms of A (See fig.). Show that:

(i)  APB and AQB

(ii) BP = BQ or B is equidistant from the arms of A

5

Solution:

In APB and AQB , we have

PAB=AQB (l is the bisector of A)

APB=AQB (Each=90o)

AB = AB

APB=AQB (By AAS congruence)

Also, BP = BQ (By CPCT)

i.e, B is equidistant from the arms of A

Hence proved.

 

Question 6:

In the fig, AC = AE, AB = AD and BAD=EAC . Show that BC = DE.

6

Solution:

BAD=EAC (Given)

BAD+DAC=EAC+DAC(AddingDAC to both sides) BAC=EAC

Now, in ABCandADE , We have

AB = AD

AC = AE

BAC=DAE

ABC=ADE (By SAS congruence)

BC = DE (CPCT)

Hence Proved.

 

Question 7:

AB is a line segment and p is its mid-point. D and E are points on the same side of AB such that BAD=ABE and EPA=DPB (see fig.). Show that

            (i) DAP=EBP

            (ii) AD = BE

7

Solution:

In DAP and EBP, we have

AP = BP (P is the midpoint of the line segment AB)

BAD=ABE (Given)

EPB=DPA (EPA=DPBEPA+DPE=DPB+DPE) DPA=EPB

AD = BE (ASA)

AD = BE

Hence Proved.

 

Question 8:

In right triangle ABC, right angles at C, M is the mid-point of hypotenuse AB, C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that :

            (i) AMC=BMD

            (II) DBC is a right angle.

            (iii) DBC=ACB

            (iv) CM=12AB

8

Solution:

In BMBandDMC , we have

(i) DM = CM (given)

BM = AM (M is the midpoint of AB)

DMB=AMC (Vertically opposite angles)

AMC=BMD (By SAS)

Hence Proved.

(ii) AC || BD (DBM and CAM are .alternate angles)

DBC+ACB=180o (Sum of co-interior angles)

DBC=90o

Hence proved.

(iii) In DBC amd ACB, We have

DB = AC (CPCT)

BC = BC (Common)

DBC=ACB(Each=90o)

DBC=ACB (By SAS)

Hence proved.

(iv) AB = CD

12AB=12CD (CPCT)

Hence, 12AB=CM Proved.