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NCERT Solutions of class 9 Maths exercise 7.1 – Triangles
NCERT Solutions of class 9 Maths exercise 7.1 of chapter 7- Triangles are the solutions of unsolved questions of class 9 NCERT maths textbook, these NCERT solutions are solved by an expert of maths as per the CBSE norms. The NCERT solutions of exercise 7.1 of chapter 7-Triangles are based on the congruency of the triangle. Every year one or two questions of 3 to 4 marks are always asked from this exercise 7.1, so it is very important for class 9 students to study NCERT solutions of exercise 7.1.
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NCERT Solutions of Class 9 Science : Chapter 1 to Chapter 15
Q1.In quadrilateral ACBD, AC = AD, and AB bisect ∠A (see the given figure). Show that ΔABC≅ΔABD. What can you say about BC and BD?
Ans.
GIVEN:In quadrilateral ACBD, AC = AD
AB is the bisector of ∠A
So, ∠BAC = ∠DAB
TO PROVE: ΔABC≅ΔABD
PROOF: In ΔABC and ΔABD
We have
AC = AD (given)
AB = AB (common)
∠BAC = ∠DAB (given)
ΔABC≅ΔABD (SAS rule)
BC = BD (by CPCT)
Q2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see the given figure ).Prove that
(i) ΔABD ≅ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC
Ans.
(i) GIVEN: ABCD is a quadrilateral
AD = BC and ∠DAB = ∠CBA
TO PROVE: ΔABD ≅ΔBAC
PROOF: In ΔABD and ΔBAC
AD = BC (given)
AB = AB (common)
∠DAB = ∠CBA
ΔABD ≅ΔBAC (SAS)
Hence proved
(ii) BD = AC (By CPCT)
(iii) ∠ABD = ∠BAC(By CPCT)
Q3. AD and BC are equal perpendiculars to a line segment AB (see the given figure).Show that CD bisects AB.
Ans.
GIVEN: AD = BC
AB ⊥ BC and AB ⊥ BC
∴∠OBC = ∠OAD = 90°
TO PROVE: CD bisects AB i.e AO = BO
PROOF: In ΔAOD and ΔBOD
Since BOA and COD is a line
∠DOA = ∠BOC (vertically opposite angle)
AD = BC (Given)
∠OBC = ∠OAD = 90°(given)
ΔBOC ≅ ΔAOD(AAS)
AO = BO (By CPCT)
Hence Proved
See the video for the solutions of Q1 TO Q3
Q4. l and m are two parallel lines intersected by another pair of parallel lines p and q(see the given figure). Show that ΔABC ≅ CDA.
Ans.
GIVEN: p ∥ q and l∥ m
TO PROVE: ΔABC ≅ ΔCDA
PROOF:In the figure
l∥ m and AC is the transversal(given)
So, ∠DAC = ∠ACB (alternate angles)
p ∥ q and AC is the transversal(given)
∠BAC = ∠ACD(alternate angles)
AC = AC(given)
ΔABC ≅ Δ CDA(ASA rule)
Hence proved
Q5.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 the given figure). Show that
(i) ΔAPB ≅ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A
Ans.
(i) GIVEN: ,BP and BQ are perpendiculars from B to the arms of ∠A
∴∠AQB = ∠APB = 90°
Line l is the bisector of an angle ∠A and
∴∠QAB = ∠PAB
TO PROVE: ΔAPB ≅ΔAQB
PROOF:∠AQB = ∠APB = 90°(given)
AB = AB (common)
∠QAB = ∠PAB (given)
ΔAPB ≅ΔAQB (AAS rule)
(ii) BP = BQ(by CPCT)
Hence proved
Q6. In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC =DE.
Ans.
GIVEN: AC = AE
AB = AD
∠BAD = ∠EAC
TO PROVE: BC =DE
PROOF: We have to prove BC =DE, which are sides of the triangle ΔABC and ΔADE
In triangles ΔABC and ΔADE
AB = AD(given)
AC = AE(given)
∠BAD = ∠EAC (given)
Adding both sides ∠DAC
∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE(proved)
ΔABC ≅ ΔADE (SAS rule)
BC = DE (by CPCT )
Hence proved
See the video for the solutions of Q4 TO Q6
Q7. AB is a line segment and p is its midpoint. D and E are points on the same sides of AB such that ∠BAD = ∠ABE and ∠EPA=∠DPB(see the given figure). Show that
(i) ΔDAP≅ ΔEBP
(ii) AD = BE
Ans.(i)
GIVEN: ∠DAP = ∠EBP
∠EPA = ∠DPB
P is the mid point of AB
∴AP = BP
TO PROVE:
(i) ΔDAP≅ ΔEBP
(ii) AD = BE
PROOF: In ΔDAP and ΔEBP
∠BAD = ∠ABE(given)
∠EPA=∠DPB(given)
Adding ∠EPD both sides
∠EPA +∠EPD =∠DPB+ ∠EPD
∠APD = ∠EPB
AP = BP(given)
ΔDAP≅ ΔEBP(ASA rule)
Q8.In right triangle ABC ,right angled at C , M is th 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 the given figure). Show that
(i) ΔAMC ≅ BMD
(ii) ∠DBC is a right angle
(iii) ΔDBC ≅ ΔACB
Ans.
(i) GIVEN:ΔABC is a right triagle ,∠C = 90°
M is the midpoint of AB
∴BM = AM
AM = AB/2
DM = CM
TO PROVE: ΔAMC ≅ BMD
PROOF:DM = CM (given)
BM = AM (given)
∠BMD = ∠AMC (Vertically opposite angle)
ΔAMC ≅ BMD(SAS rule)
Hence proved
(ii) Since ΔAMC ≅ BMD(proved)
∠DBM = ∠CAM (By CPCT)
∠DBM and ∠CAM are alternate angles
∴DB ∥ AC
∠DBC + ∠ACB = 180° (co-interior angles)
∠ACB = 90°
∠DBC + 90° = 180°
∠DBC = 90°
Hence proved
(iii) Since ΔAMC ≅ BMD(proved)
BD = AC (By CPCT)
BC = BC(common)
∠DBC = ∠ACB =90°(Proved above)
ΔDBC ≅ ΔACB (SAS rule)
proved above
(iv) CM = DM(given)
CM + DM = DC
CM + CM = DC
2CM = DC
CM = 1/2 DC
Since we have proved
ΔDBC ≅ ΔACB
So, DC = AB (by CPCT)
Therefore
CM = 1/2 AB
Hence proved
See the video of the solutions for question 7 and question 8
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