NCERT Solutions For Class 9 Math Chapter – 7 Exercise – 7.1

NCERT Solutions For Class 9 Math Chapter – 7 Exercise – 7.1

NCERT Solutions For Class 9 Math Chapter – 7  Triangles helps students in learning basic concepts of probability which is include in the second term’s CBSE syllabus 2021-2022  NCERT Solutions For Class 9 Maths provide answers to all the questions in exercise present at the end of the chapter These solutions are prepared by our mathematics experts who are highy experienced in the field of education.

 Experts
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Q1.   In quadrilateral ABCD  , AC = AD and AB bisects A  Show that ABC   ̴ ∆ABD.  What can you say about BC and BD ?

Solution:-

Ncert solutions class 9 chapter 7-1

AC = AD (Given)

AB = AB [Common]

∟CAB = ∟DAB [ AB bisects ∟CAD]

∆ABC ̴ ∆ABD [SAS Rule]

Therefore BC = BD [CPCT]

 

Q2. ABCD is a quadrilateral in which AD = BC and DAB = CBA (see figure) Prove that

  • ABD BAC
  • BD = AC
  • ABD =

Solution:-

Ncert solutions class 9 chapter 7-2

  • Consider triangles ABD and ABC.

We have AD = BC [Given]

AB = BA [Common]

And

∟DAB = ∟CBA [Given]

∆ABD  ∆BAC

BD = AC

∟ABD = ∟BAC [CPCT]

 

Q3. AD and BC are equal perpendicular to a line segment AB Show that CD bisects AB.

Solution:-

Ncert solutions class 9 chapter 7-3

 

 

 

 

 

 

AD and BC are perpendicular to AB.

AD // BC and CD is transversal.

BCD =  ∟ADC ∟BO = ∟ADO  [Alternate angle]

Triangles BCO and ADO,

BC = AD

OBC = ∟OAD

BCO = ∟ADO

OBC  ∆OAD

OB = OA.

Hence , CD bisects AB.

 

Q4. l and M are two parallel lines intersected by another pair of parallel lines p and q Show thta ∆ABC  ∆CDA.

Solution:-

 

Ncert solutions class 9 chapter 7-4

 

 

 

∟1 = ∟2 (I) [Alternate angles]

P // Q and AC is transversal

∟3 = ∟4  (ii) (Alternate angles)

 

In ∆ADC and ∆ABC,

AC = CA

∟1 = ∟2 and ∟3 = ∟4

∆ABC  ∆ ABC

AC = CA

∟1 = ∟2  and ∟3 = ∟4

∆ABC  ∆CDA

 

 

Q5. Line l is the bisector of an angle ∟A and B is any point on  l BP and BQ  are perpendicular from  to the arms of ∟A Show that:

  • ∆APB AQB
  • BP = BQ or B is equidistant from the arma of A

Solution:-

Ncert solutions class 9 chapter 7-5

Line l is the bisector of ∟QAP.

QAB = ∟PAB.

AB = PA

BQA = ∟BPA

 

  • In triangles AQB and APB.

AB = BA and ∟BQA= ∟BPA

APB  ∆AQB.

  • BP = BQ

 

Q6. In figures AC = AE , AB = AD and ∟BAD = ∟EAC Show that BC = DE.

Solution:-

 

Ncert solutions class 9 chapter 7-6

 

 

 

 

 

∟BAD + ∟DAC = ∟DAC + ∟CAE   [ ∟DAC is added to both sides]

∟BAC = ∟DAE    (i)

 

In triangles ∆BAC and ∆DAE ,

AB = AD

AC = AE

∟BAC = ∟DAE

∆BAC   ∆DAC

BC = DE.

 

Q7. AB is a line segment and P is its mid- point D and E are points on the same sides of AB such that BAQ = ABE and EPA = DPB Show that

  • ∆DAP EBP
  • AD = BE.

Solution:-

Ncert solutions class 9 chapter 7-7 

∟EPA + ∟EPD = ∟EPD + ∟DPB

∟APD = ∟BPE

AP = BP

∟DAP = ∟EBP

  • ∟APD = ∟BPE , AP = BP and ∟DAP = ∟EBP.

∆DAP ∆EBP.

  • AD = BE.

 

Q8. 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

Show that:

  • AMC BMD
  • ∟DBC is a right angle.
  • ∆DBC ACB
  • CM = 1/2AB.

Solution:-

Ncert solutions class 9 chapter 7-8

  • In ∆AMC and DMB ,

∟AMC = ∟BMD (VOA)

∆AMC  ∆BMD [ SAS rule]

 

  • ∟BAC = ∟DBA [CPCT]

AB is the transversal

∟DBC is a right angle

 

  • In ∆ABC and ∆DCB.

AC = DB [Since ∆AMC  ∆BMD ]

BC = CB [Common]

∟ACB = ∟DBC [Each 900]

∆DBC  ∆ACB [SAS rule]

 

  • DC = AB [CPCT from part (iii]

2CM = AB

CM = 1/2AB.

CM  = 1/2AB.