NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 7 Triangles Ex 7.1. 



Ex 7.1 Class 9 Maths Question 1.
In a quadrilateral ACBD, AC = AD, and AB bisects A (see figure). Show that ΔABC ΔABD. What can you say about and BD?

Solution:

Given: In quadrilateral ACBD, AC = AD and BAC = DAB

To prove: ΔABC ΔABD
Proof: In ΔABC and ΔABD, AC = AD       [given]
BAC = DAB                                           [given]
and AB = AB                                              [common side]
Hence, ΔABC
ΔABD                             [by SAS congruence rule]
Then, BC = BD
Thus, BC and BD are equal.

 

Ex 7.1 Class 9 Maths Question 2.
ABCD is a quadrilateral in which AD = BC and DAB = CBA (see figure). Prove that
(i) ΔABD 
 ΔBAC
(ii) BD = AC
(iii)
ABD = BAC

Solution:


Given: In quadrilateral ABCD, AD = BC and
DAB = CBA.

To prove:
(i) ΔABD  ΔBAC
(ii) BD = AC
(iii)
ABD = BAC
Proof: (i) In ΔABD and ΔBAC,
AD = BC                                     [given]
DAB = CBA                          [given]
and AB = AB                             [common side]
ΔABD  ΔBAC                      [by SAS congruence rule]

(ii) From part (i), ΔABD  ΔBAC
Then, BD = AC                         [by CPCT]

(iii) From part (i), ΔABD  ΔBAC
Then,
ABD = BAC              [by CPCT]
Hence proved.

 

Ex 7.1 Class 9 Maths Question 3.
AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.


Solution:
Given:
 AD = BC, AD

AB and BC AB
To prove: CD bisects AB.

Proof: In ΔAOD and ΔBOC.
AD = BC                              [given]
OAD = OBC                  [each 90°]
and
AOD = BOC          [vertically opposite angles]
ΔAOD
ΔBOC                  [by AAS congruence rule]
Then, OA = OB                 [by CPCT]
Thus, CD bisects AB.

 

Ex 7.1 Class 9 Maths Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ΔABC ΔCDA.


Solution:
Given:
l || m and p || q
To prove: ΔABC
ΔCDA
Proof: In ΔABC and ΔADC,
BAC = ACD                       [alternate interior angles as p || q]
ACB = DAC                       [alternative interior angles as l || m]
AC = AC                                 [common side]
ΔABC
 ΔCDA                      [by AAS congruence rule]

 

Ex 7.1 Class 9 Maths Question 5.
Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of A (see figure). Show that
(i) ΔAPB
ΔAQB
(ii) BP = BQ or B is equidistant from the arms of 
A


Solution:
(i)
Consider ΔAPB and ΔAQB, we have
APB = AQB = 90°
PAB = QAB                     [ AB bisects PAQ]
AB = AB                                [common]
ΔAPB   ΔAQB                [by AAS congruence rule]
(ii) From part (i), ΔAPB 
  ΔAQB.

Then, BP = BQ                     [by CPCT]

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

 

Ex 7.1 Class 9 Maths Question 6.
In figure, AC = AE, AB = AD and BAD = EAC. Show that BC = DE.

Solution:

In ΔABC and ΔADE,
AB = AD                             [Given]
BAD = EAC                   [Given]               …(i)
On adding
DAC on both sides in Eq. (i), we get
BAD + DAC = EAC + DAC
BAC = DAE
and AC = AE                      [given]
ΔABC 
ΔADE                   [by SAS congruence rule]
BC = DE                          [by CPCT]

 

Ex 7.1 Class 9 Maths 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 figure). Show that
(i) ΔDAP
ΔEBP
(ii) AD = BE


Solution:

(i) EPA = DPB                     [Given]         …. (i)
BAD = ABE                         [Given]         .…(ii)
On adding 
EPD on both sides in Eq. (i), we get
EPA + EPD = DPB + EPD
DPA = EPB                                                …(iii)
Now, in ΔDAP and ΔEBP, we have
DPA = EPB                         [from eq.(iii)]
DAP = EBP                         [from eq.(ii)]

AP = BP                                    [ P is the mid-point of AB (Given)]
ΔDAP
 ΔEBP                         [by ASA congruence rule]
(ii) Since ΔDAP
 ΔEBP

AD = BE                                    [by CPCT]

 

Ex 7.1 Class 9 Maths Question 8.
In right triangle ABC, right angled 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 figure). Show that
(i) ∆AMC
∆BMD
(ii)
DBC is a right angle
(iii) ∆DBC
∆ACB
(iv) CM = ½ AB


Solution:
(i) In ∆AMC and ∆BMD,
CM = DM                                 [Given]
AMC = BMD                      [Vertically opposite angles]
AM = BM                                 [Since M is the mid-point of AB]
∆AMC ∆BMD                  [By SAS congruence rule]

(ii) Since ∆AMC ∆BMD
MAC = MBD                  [By CPCT]
But they form a pair of alternate interior angles.
AC || DB
Now, BC is a transversal which intersects parallel lines AC and DB,
BCA + DBC = 180°          [Co-interior angles]
But
BCA = 90°                       [∆ABC is right angled at C]
90° + DBC = 180°
DBC = 90°

(iii) Again, since ∆AMC ∆BMD
AC = BD                                 [By CPCT]
Now, in ∆DBC and ∆ACB, we have
BD = CA                                    [Proved above]
DBC = ACB                          [Each 90°]
BC = CB                                     [Common]
∆DBC ∆ACB                      [By SAS congruence rule]

(iv) As ∆DBC ∆ACB
DC = AB                                     [By CPCT]
But DM = CM                           [Given]
CM = ½ DC = ½ AB
CM = ½ AB


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