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NCERT Solutions for Class 9 Maths Exercise 7.1
Last Updated: June 18, 2018 by myCBSEguide
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NCERT Solutions for Class 9 Maths Exercise 7.1 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.
NCERT solutions for Class 9 Maths Triangles Download as PDF
NCERT Solutions for Class 9 Maths Exercise 7.1
NCERT Solutions for Class 9 Mathematics Triangles
1. In quadrilateral ABCD (See figure). AC = AD and AB bisects A. Show that ABC ABD. What can you say about BC and BD?
Ans. Given: In quadrilateral ABCD, AC = AD and AB bisects A.
To prove: ABC ABD
Proof: In ABC and ABD,
AC = AD [Given]
BAC = BAD [ AB bisects A]
AB = AB [Common]
ABC ABD [By SAS congruency]
Thus BC = BD [By C.P.C.T.]
2. ABCD is a quadrilateral in which AD = BC and DAB = CBA. (See figure). Prove that:
(i) ABD BAC
(ii) BD = AC
(iii) ABD = BAC
Ans. (i) In ABC and ABD,
BC = AD [Given]
DAB = CBA [Given]
AB = AB [Common]
ABC ABD [By SAS congruency]
Thus AC = BD [By C.P.C.T.]
(ii) Since ABC ABD
AC = BD [By C.P.C.T.]
(iii) Since ABC ABD
ABD = BAC [By C.P.C.T.]
3. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB (See figure)
Ans. In BOC and AOD,
OBC = OAD = [Given]
BOC = AOD [Vertically Opposite angles]
BC = AD [Given]
BOC AOD [By ASA congruency]
OB = OA and OC = OD [By C.P.C.T.]
NCERT Solutions for Class 9 Maths Exercise 7.1
4. and are two parallel lines intersected by another pair of parallel lines and (See figure). Show that ABC CDA.
Ans. AC being a transversal. [Given]
Therefore DAC = ACB [Alternate angles]
Now [Given]
And AC being a transversal. [Given]
Therefore BAC = ACD [Alternate angles]
Now In ABC and ADC,
ACB = DAC [Proved above]
BAC = ACD [Proved above]
AC = AC [Common]
ABC CDA [By ASA congruency]
NCERT Solutions for Class 9 Maths Exercise 7.1
5. Line is the bisector of the angle A and B is any point on BP and BQ are perpendiculars from B to the arms of A. Show that:
(i) APB AQB
(ii) BP = BQ or P is equidistant from the arms of A (See figure).
Ans. Given: Line bisects A.
BAP = BAQ
(i) In ABP and ABQ,
BAP = BAQ [Given]
BPA = BQA = [Given]
AB = AB [Common]
APB AQB [By ASA congruency]
(ii) Since APB AQB
BP = BQ [By C.P.C.T.]
B is equidistant from the arms of A.
NCERT Solutions for Class 9 Maths Exercise 7.1
6. In figure, AC = AB, AB = AD and BAD = EAC. Show that BC = DE.^
Ans. Given that BAD = EAC
Adding DAC on both sides, we get
BAD + DAC = EAC + DAC
BAC = EAD ……….(i)
Now in ABC and AED,
AB = AD [Given]
AC = AE [Given]
BAC = DAE [From eq. (i)]
ABC ADE [By SAS congruency]
BC = DE [By C.P.C.T.]
NCERT Solutions for Class 9 Maths Exercise 7.1
7. AB is a line segment and P is the mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB. Show that:
(i) DAF FBP
(ii) AD = BE (See figure)
Ans. Given that EPA = DPB
Adding EPD on both sides, we get
EPA + EPD = DPB + EPD
APD = BPE ……….(i)
Now in APD and BPE,
PAD = PBE [ BAD = ABE (given),
PAD = PBE]
AP = PB [P is the mid-point of AB]
APD = BPE [From eq. (i)]
DPA EBP [By ASA congruency]
AD = BE [ By C.P.C.T.]
NCERT Solutions for Class 9 Maths Exercise 7.1
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 = AB
Ans. (i) In AMC and BMD,
AM = BM [AB is the mid-point of AB]
AMC = BMD [Vertically opposite angles]
CM = DM [Given]
AMC BMD [By SAS congruency]
ACM = BDM ……….(i)
CAM = DBM and AC = BD [By C.P.C.T.]
(ii) For two lines AC and DB and transversal DC, we have,
ACD = BDC [Alternate angles]
AC DB
Now for parallel lines AC and DB and for transversal BC.
DBC = ACB [Alternate angles] ……….(ii)
But ABC is a right angled triangle, right angled at C.
ACB = ……….(iii)
Therefore DBC = [Using eq. (ii) and (iii)]
DBC is a right angle.
(iii) Now in DBC and ABC,
DB = AC [Proved in part (i)]
DBC = ACB = [Proved in part (ii)]
BC = BC [Common]
DBC ACB [By SAS congruency]
(iv) Since DBC ACB [Proved above]
DC = AB
AM + CM = AB
CM + CM = AB [ DM = CM]
2CM = AB
CM = AB
Answer:
i) △AMC≅△BMD
Proof: As 'M' is the midpoint
BM=AM
And also it is the mid point of DC then
DM=MC
And AC=DB (same length)
∴Therefore we can say that
∴△AMC≅△BMD
ii) ∠DBC is a right angle
As △DBC is a right angle triangle and
(Pythagoras)
So, ∠B=90°
∴∠DBC is 90°
iii) △DBC≅△ACB
As M is the midpoint of AB and DC. So, DM=MC and AB=BM
∴DC=AB (As they are in same length)
And also, AC=DB
and ∠B=∠C=90°
By SAS Axiom
∴△DBC≅△ACB
iv) CM=
As △DBC≅△ACB
CM=
2
DC
∴DC=AB(△DBC≅△ACB)
So, CM=
∴CM=
