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 at Sunstarup  have created the NCERT Solutions after extensive search on each topic Students can refer to this study study material to boost their confidence and attempt the second term exam smartly. The concepts are explained with steps, shortcuts to remember formula , tips and tricks to solve the numerical problems wisely and quickly.

 

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

Solution:-

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:-

• 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:-

 

 

 

 

 

 

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:-

 

 

 

 

∟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:-

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:-

 

 

 

 

 

 

∟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:-

 

∟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:-

• 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.