NCERT solutions for class 9 Maths Chapter 7
Looking for NCERT solutions for chapter 7 of class 9 mathematics? In this post we are sharing the solutions to all the exercises of Chapter 7.
NCERT solutions for class 9 Maths Chapter 7 Exercise 7.1
Given: AC = AD and AB bisects ∠A, therefore, ∠CAB = ∠DAB
To prove: Triangle ABC ≅ Triangle ABD
Proof: In triangles ABC & ABD
AC = AD (Given)
AB = AB (Common)
∠CAB = ∠DAB (Given)
Triangle ABC ≅ Triangle ABD by SAS (Side-Angle-Side) criteria
Given: BC = AD and, ∠CBA = ∠DAB
To prove: Triangle ABC ≅ Triangle DAB, BD =AC and ∠ABD = ∠BAC
Proof:
In triangles ABC & DAB
BC = AD (Given)
∠CBA = ∠DAB (Given)
AB = AB (Common)
Therefore Triangle ABC ≅ Triangle DAB by SAS (Side-Angle-Side) criteria
BD = AC (by CPCT)
∠ABD = ∠BAC (by CPCT)
Given: BC = AD and ∠CBO = ∠OAD = 900
To prove: Triangle OBC ≅ Triangle 0AD
Proof: In triangles OBC & OAD
BC = AD (Given)
∠CBO = ∠OAD = 900
(Given)
∠BOC = ∠AOD (Vertically opposite angles)
Therefore Triangle OBC ≅ Triangle OAD by AAS
OA = OB (by CPCT)
Therefore CD bisects AB
Given: l//m, p//q and they all intersect each other
To prove: Triangle ABC ≅ Triangle CDA
Proof: l//m and p//q, therefore, ABCD is a //gm
In Triangles ABC & CDA
∠ABC = ∠ADC (Opposite angles of a //gm are equal)
AC = AC (Common)
∠BAC = ∠DCA (Alternate angles; BA//CD, AC is the transversal)
Therefore Triangle ABC ≅ Triangle CDA by AAS criteria
Given: l bisects ∠A, therefore ∠QAB = ∠BAP, ∠Q = ∠P = 900
To prove: Triangle APB ≅ Triangle AQB and BP=BQ
Proof: In Triangles APB & AQC
∠AQB = ∠APB (both 900
; given)
AB = AB (Common)
∠QAB = ∠BAP (Given)
Therefore Triangle APB ≅ Triangle AQB by AAS criteria
BP = BQ (by CPCT)
Given: AC = AE, AB = AD and ∠BAD = ∠EAC
To prove: BC = DE
Proof:
∠BAD = ∠EAC
Adding ∠DAC on both sides
∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC= ∠EAD … (1)
In Triangles ABC & ADE
AC = AE (Given)
AB = AD (Given)
∠BAC= ∠EAD (Proved above)
Therefore Triangle ABC ≅ Triangle ADE by SAS criteria
BC = DE (by CPCT)
Given: AP = PB (p is the midpoint of AB), ∠EPA = ∠DPB
and ∠BAD = ∠ABE
Proof:
∠EPA = ∠DPB
Adding ∠EPD on both sides
∠EPA + ∠EPD = ∠DPB + ∠EPD
∠DPA = ∠EPB … (1)
∠BAD = ∠ABE, can also be written as ∠PAD = ∠PBE … (2)
In triangles DAP and EBP
∠DPA = ∠EPB (Proved above)
∠PAD = ∠PBE (Shown above)
AP = PB (Given)
Therefore, Triangle DAP ≅ Triangle EBP by ASA criteria
AD = BE (by CPCT)
In triangles AMC and BMD
DM=MC (Given)
AM = MB (M is the midpoint of AB; given)
∠DMB = ∠AMC (Vertically opposite angles)
Triangle AMC ≅ Triangle BMD by SAS criteria
AC = BD (By CPCT)
AC = BD (proved above)
Equal sides have opposite angles equal therefore,
∠DBC = ∠ACB = 900
Hence ∠DBC is a right angleIn triangles ACB and DBC
BC=BC (Given)
BD = AC (proved above)
∠DBC = ∠ACB (proved above)
Triangle ACB ≅ Triangle DBC by SAS criteria
AB = CD (By CPCT)
AB = CD
AB = CM + MD
But MD = CM
AB = 2CM
1/2 AB = CM or CM = 1/2 AB
NCERT solutions for class 9 Maths Chapter 7 Exercise 7.2
AB=AC
Therefore ∠B = ∠C (Equal sides have opposite angles equal)
∠B & ∠C are bisected by BO and CO respectively therefore
∠OBC = ∠OCB & ∠OBA = ∠OCA … (1)
In triangles OBA and OCA
∠OBA = ∠OCA (from 1)
AB=AC (Given)
OA= OA (Common)
Triangle OBA ≅ Triangle OCA by SSA criteria
OB=OC (by CPCT)
∠BAO = ∠CAO (by CPCT) , therefore, OA bisects ∠A
In Triangles ADB and ADC
AD = AD (Common)
AD = DC (AD bisects BC; given)
∠ADC = ∠ADB (Both 900; given)
Triangle ADB ≅ Triangle ADC by SAS criteria
AB = AC (by CPCT)
Hence proved that Triangle ABC is an isosceles triangle where AB = AC
In triangles ABE and ACF
AB = AC (Given)
∠AEB = ∠CFA (Both 900)
∠BAC = ∠CAB (Common angle)
Triangle ABE ≅ Triangle ACF by AAS criteria
BE = FC (by CPCT)
Hence proved that the altitudes are equal
In triangles ABE and ACF
BE = FC (Given)
∠AEB = ∠CFA (Both 900)
∠BAC = ∠CAB (Common angle)
Triangle ABE ≅ Triangle ACF by AAS criteria
AB = AC (by CPCT)
In triangle ABC
AB = AC (Given)
Therefore, ∠ABC = ∠ACB (Equal sides have opposite angles equal) …. (1)
In triangle BDC
BD = DC (Given)
Therefore, ∠DBC = ∠DCB (Equal sides have opposite angles equal) … (2)
Adding (1) and (2)
∠ABC + ∠DBC = ∠ACB + ∠DCB
∠ABD = ∠ACD
Hence proved
In triangle ABC
AB = AC (Given) … (1)
Therefore, ∠ABC = ∠ACB (Equal sides have opposite angles equal)
In triangle ADC
AD = AC (Given) … (2)
Therefore, ∠ADC = ∠ACD (Equal sides have opposite angles equal)
By (1) and (2)
AB = AC = AD
This means that ∠ABC = ∠ACB = ∠ADC = ∠ACD
Let each angles measure is x
In triangle ABC
∠ABC + ∠DCB +∠ADC = 1800 (Angle sum property of a triangle)
∠ABC + (∠DCA+ ∠BCA) +∠ADC = 1800
x+ x+ x+ x = 1800
4x= 1800
x = 450
Now, ∠DCB = ∠DCA+ ∠BCA
= 450 + 450 = 900
Hence proved
AB = AC (Given)
∠ACB = ∠ABC (Equal sides have opposite angles equal)
Let the measure of each angle be x
In triangle ABC
∠ABC + ∠BAC +∠ACB = 1800 (Angle sum property of a triangle)
x+ 900 + x = 1800
2x = 900
x = 450
∠ACB = ∠ABC = 450
Let ABC be the equilateral triangle
AB = BC = CA
Therefore ∠ABC = ∠BAC =∠ACB (Equal sides have opposite angles equal)
Let the measure of each angle be x
In triangle ABC
∠ABC + ∠BAC +∠ACB = 1800 (Angle sum property of a triangle)
x+ x + x = 1800
3x = 1800
x = 600
∠ABC = ∠BAC =∠ACB = 600
Hence proved
NCERT solutions for class 9 Maths Chapter 7 Exercise 7.3
In triangles ABD and ACD
AB = AC (ABC is an isosceles triangle)
BD = DC (DBC is an isosceles triangle)
AD = AD (Common)
Triangle ABD ≅ Triangle ACD by SSS criteria
∠BAP = ∠CAP (by CPCT) … (1)
In triangles ABP and ACP
AB = AC (ABC is an isosceles triangle)
∠BAP = ∠CAP (Proved above)
AP = AP (Common)
Triangle ABP ≅ Triangle ACP by SAS criteria
BP =PC (by CPCT) … (2)
In triangles BDP and CDP
BD = DC (ABC is an isosceles triangle)
BP = PC (Proved above)
DP = DP (Common)
Triangle BDP ≅ Triangle CDP by SSS criteria
∠BDP = ∠CDP (by CPCT) … (3)
By (1) & (3) it is clear that AP bisects both ∠A & ∠C
∠BPD = ∠CPD (by CPCT)
In triangle ABC
∠BPD = ∠CPD (Proved above)
∠BPD + ∠CPD = 1800 (Linear Pair)
x + x = 1800
2x= 1800
x= 900
∠BPD = ∠CPD = 900 … (4)
By (2) & (4), it is clear that AP is perpendicular bisector of BC
AD is the altitude, therefore ∠ADB = ∠ADC = 900
In triangles ADB and ADC
AB = AC (ABC is an isosceles triangle)
∠ADB = ∠ADC (Given)
AD = AD (Common)
Triangle ADB≅ Triangle ADC by SAS criteria
AD =DC (by CPCT)
Therefore AD bisects BC
∠DAB = ∠DAC (by CPCT)
Therefore AD bisects ∠A
In triangles ABM and PQN
AB = PQ (Given)
AM = PN (Given)
BM = QN (Given)
Triangle ABM≅ Triangle PQN by SSS criteria
∠ABM = ∠PQN (by CPCT)
It can also be written as ∠ABC = ∠PQR since C & M and N & R
are collinear
In triangles ABC and PQR
AB = PQ (Given)
∠ABC = ∠PQR (Proved above)
BM = QN (Given)
So 2BM = 2QN => BC = QR
Triangle ABC≅ Triangle PQR by SAS criteria
In triangles BEC and CFB
BC = BC (Common)
∠BFC = ∠CEB (Both 900 )
BE = CF (Given)
Triangle BEC≅ Triangle CFB by RHS criteria
∠FBC= ∠ECB (by CPCT)
In triangle ABC
∠B= ∠C (proved above)
Therefore AB = AC (Angles opposite to the equal sides of a
triangle are equal)
In triangles APB and APC
AP = AP (Common)
∠APC = ∠APB (Both 900 )
AB = AC (Given)
Triangle APB≅ Triangle APC by RHS criteria
∠B= ∠C (by CPCT)
NCERT solutions for class 9 Maths Chapter 7 Exercise 7.4
In triangle ABC right angled at B
∠B > ∠C
AC > AB ( Side opposite to larger angle is longer) … (1)
∠B > ∠A
AC > BC ( Side opposite to larger angle is longer) … (2)
By (1) & (2)
AC is longer than both AB and BC, this proves that
hypotenuse is the longest side of the right triangle
∠PBC <∠QCB … (1)
∠PBC and ∠ABP from linear pair
Similarly, ∠QCB and ∠ACB from linear pair
can also be written as
∠PBC <∠QCB
1800 - ∠ABP < 1800 - ∠ACB
- ∠ABP < - ∠ACB
∠ACB > ∠ABP
AB > AC ( Side opposite to larger angle is longer) … (2)
∠A> ∠B
OB > OA ( Side opposite to larger angle is longer) … (1)
∠D > ∠C
OC > OD ( Side opposite to larger angle is longer) … (2)
Adding (1) & (2)
OB + OC > OA + OD
BC > AD
Construct diagonal AC
∠BAC>∠BCA (Angle opposite to longer side is larger) … (1)
∠DAC>∠DCA (Angle opposite to longer side is larger) … (2)
Adding (1) & (2)
∠A>∠C (Angle opposite to longer side is larger)
Similarly construct diagonal BD
∠ABD > ∠ADB (Angle opposite to longer side is larger) … (3)
∠DBC>∠BDC (Angle opposite to longer side is larger) … (4)
Adding (3) & (4)
∠B>∠D (Angle opposite to longer side is larger)
PR> PQ
∠PQR> ∠PRQ (Angle opposite to longer side is larger) … (1)
∠QPS = ∠RPS (PS bisects ∠P) … (2)
In triangle PQS
∠PQS + ∠PSQ + ∠QPS = 1800 … (3)
In triangle PRS
∠PSR + ∠PRS + ∠SPR = 1800 … (4)
Using (1)
∠PQR> ∠PRQ, Or ∠PQS> ∠PRS
Using (2) and (3)
180 - ∠PSQ - ∠QPS > 180 - ∠PSR - ∠SPR
Using (3)
(-∠PSQ)> (-∠PSR)
∠PSR > ∠PSQ
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