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SEBA Class 9 Mathematics Chapter 7 Triangles
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Triangles
Chapter – 7
| Exercise 7.1 |
1. In quadrilateral ABCD AC = AD and AB bisects ∠A. Show that ∠ABC ≅ ∆ADB. What can you say about BC and RD?
Ans: In ∆ABC and ∆ABD
AC = AD (Given)
∠CAB = ∠DAB (Given)
AB = AB (Given)
Therefore,
By SAS congruence condition
∆ABC ≅ ∆ABD
So, BC = BD (By C.P.C.T)
2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠BCA Prove that
(i) ∆AB D ≅ ∆ BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC
Ans: (i) In ∆ ABD and ∆BAC
(ii) AD = BC (Given)
(iii) ∠DAB = ∠CBA (Given)
and AB = BA (Common)
By SAS Congruence Condition
∆ABD ≅ ∆BAC
(ii) BD = AC( By C.P.C.T)
(iii) ∠ABD = ∠ BAC
(Again by C.P.C.T)
3. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Ans: In ∆ AOD and ∆ BOC
AD = BC (Given)
∠OAD = ∠OBC (each 90⁰)
∠AOD = ∠BOC
(Vertically opposite angles)
Therefore, by ASA congruence condition.
So, OA = OB
(By C.P. С.Т.)
Hence, CD bisects line segment AB.
4. I and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC = ∆CDA
Ans: We have given that l || m and p || q
Therefore, In ∆ ABC and ∆ACDA
(Alternate interior angles as AB || CD)
∠ACB = ∠CAD
(Alternate interior angles as BC II DA)
AC = CA
So, By A-S-A congruence condition.
∆ ABC ≅ ∆CDA
5. Line 1 is the bisector of an angle ∠A and B is any point on 1. BP and BQ are perpendiculars from B to the arms of ∠A show that:
(i) ∆APB = ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A
Ans: (i) In ∆ABP and ∆ABQ
∠BAP = ∠BAQ (Given)
∠APB = ∠AQB (Each 90⁰)
AB = AB (Common)
By A – A – S congruence condition.
So, ∆ABP = ∆ABQ
(ii) BP = BQ (By C.P.C.T.)
6. In AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Ans: In ∆ BAC and ∆ DAE
AB = AD (Given)
AC = AB (Given)
∠BAD = ∠EAC …(1) (Given)
Adding ∠DAC both side in equation (1)
∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE
Therefore by S-A-S Congruency Condition
∆BAC ≅ ∆DAE
So, BC = DE (By C.P.C.T.)
7. AB is a line segment and P is its mid point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB show that
(i) ∆DA P ≅ EBP
Ans: In ∆DAP and ∆EBP
∠DAP = ∠EBP (Given)
∠APE = ∠DPB (Given)
∴ ∠APE + ∠ EPD = ∠DPB + ∠EPD
(Add ∠EPD both side)
∠APD =∠BPE
AP = BP
(Given P is the midpoint of AB)
By A-S-A Congruence Condition.
∆DAP = ∆EBP
(ii) AD = BE
Ans: AD = BE
(By C.P.C.T.)
8. In the right triangle ABC, right angled at C, M is the midpoint of hypotenuse AB, C is joined to M and produces a point D such that DM = CM. Point D is joined to point B. Show that
(i) ∆AMC ≅ ∆BMD
Ans: In ∆AMC and ∆BMD
AM = BM (Given)
CM = DM (Given)
∠AMC = ∠BMD
(Vertically opp. angles)
∴ By S-A-S Congruency Condition.
∆AMC ≅ ∆BMD
(ii) ∠DBC is right angle.
Ans: ∠CAM = ∠DBM (by C.P.C.T)
Also, ∠CAM+∠MBC = 90⁰ (Since ∠C 90⁰ )
∴ ∠DBM + ∠MBC = 90⁰ (∠CAM = ∠DBM)
or, ∠ DBC = 90⁰
(iii) ∆DBC ≅ ∆ACB
Ans: In ∆ DBC and ∆ ACB
BC = BC (Common)
DB = AC
(∴ ∆BMD = ∆AMC, by C.P.M.T)
and ∠DBC = ∠AVB (each 90 proved above)
Therefore, by S-A-S Congruency Condition.
∆ DBC ≅ ∠ ACB
(iv) CM = 1/2 AB
Ans: Since ∆ DBC ≅ ∠ ACB
DC = AB
CM = AM
| Exercise 7.2 |
1. In an isosceles triangle ABC, with AB = AC the bisectors of ∠ B and ∠ C intersect each other at O. Show that:
(i) OB = OC
(ii) AO bisects ∠ A
Ans:
Given: In an isoscelés triangle ABC, with AB = AC the bisectors of ∠B and ∠ C intersect each other at O.
Join A to O.
To Prove: (i) Ob = OC
(ii) AO bisects A
Proof: (i) AB = AC | Given
∴ ∠ A = ∠C |Angles opposite to equal sides of a triangle are equal
∴ BO and CO are the bisectors of ∠B and ∠C respectively
∴ OB = OC | Sides opposite to equal angles of a triangle are equal
(ii) In ∆OAB and ∆OAC
AB = AC | Given
OB = OC | Proved (i) above
AB = AC | Given
∴ ∠ B = ∠C | Angles opposite to equal sides of a triangle are equal
∴ ∠ ABO = ∠ ACO
∵ BO and CO are the bisectors of ∠B and ∠C respectively
∴ ∆OAB ≅ ∆OAC |By SAS Rule
∴ ∠OAB = ∠OAC |c.p.c.t.
∴ AO bisects ∠ A
2. In ∠ ABC, AD is the perpendicular bisector of BC. Show that ∆ABC is an isosceles triangle in which AB = AC.
Ans: Given: In ∆ ABC, AD is the perpendicular bisector of BC.
Το Prove: ∆ ABC is an isosceles triangle in which AB = AC
Proof: In ∆ ADB and ∆ ADC,
∠ ADB = ∠ ADC | Each = 90⁰
DB = DC | ∵ AD is the perpendicular bisector of BC
AD = AD | Common
∴ ∆ ADB ≅ ∆ADC | By SAS Rule
∴ ∆ABC is an isosceles triangle in which AB = AC
3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively. Show that these altitudes are equal.
Ans: Given: ABC is an isosceles triangle in which altitudes BE and CF are drawn on sides AC and AB respectively.
To Prove: BE = CF
Proof: ∵ ABC is an isosceles triangle
∴ AB = AC
∴ ∠ ABC = ∠ ACB … (1) | Angles opposite to equal sides of a triangle are equal
In ∆ BEC and ∆CFB
∠BEC = ∠CFB | Each = 90⁰
BC = CB | Common
∠ECB = ∠FBC | From (1)
∴ ∆ BEC ≅ CFB |By ASA Rule
∴ BE = CE | c.p.c.t.
4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that:
(i) ∆ABE ≅ ACF
(ii) AB =AC, i.e. ∆ ABC is an isosceles triangle.
Ans: Given: ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal
To Prove:
(i) ΔΑΒΕ ≅ ∆ACF
(ii) AB = AC, i.e. ∆ ABC is an isosceles triangle
Proof: (i) In ∆ABE and ∆ACF
BE = CF | Given
∠BAC = ∠BAE = ∠CAF | Common
∠AEB = ∠AFC | Each = 90⁰
∴ Δ ΑΒΕ ≅ ∆ACF | By AAS Rule
(ii) Δ ΑΒΕ ≅ ΔACF | Proved in (i) above
∴ AB = AC | c.p.c.t.
∴ ∆ ABC is an isosceles triangle.
5. ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.
Ans: Given: ABC and DBC are two isosceles triangles on the same base BC.
To Prove: ∠ABD = ∠ACD
Proof: ∵ ABC is an isosceles triangle on the base BC
∴ ∠ABC = ∠ACB …(1)
∵ DBC is an isosceles triangle on the base BC
∴ ∠DBC = ∠DCB …(2)
Adding the corresponding sides of (1) and (2), we get
∠ABC + ∠DBC = ∠ACB + ∠DCB
⇒ ∠ABD = ∠ACD
6. ∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.
Ans: Given: ∆ ABC is an isosceles triangle in which AB = AC
Side BA is produced to D such that AD = AB
To Prove: ∠ BCD is a a right angle.
Proof: ∵ ABC is an isosceles triangle
∴ ∠ ABC = ∠ ACB …(1)
∵ AB = AC and AD = AB
∴ AC = AD
∴ In ∆ACD
∠CDA = ∠ACD Angles opposite to equal sides of a triangle are equal
⇒ ∠CDB = ∠ACD …(2)
Adding the corresponding sides of (1) and (2), we get
∠ABC + ∠CDB = ∠ACB +∠ACD
⇒ ∠ABC + ∠CDB = ∠BCD …(3)
In ∆BCD,
∠BCD + ∠DBC + ∠CDB = 180⁰ | ∵ Sum of all the angles of a triangle is 180⁰
⇒ ∠BCD + ∠ABC + ∠CDB = 180⁰
⇒ ∠BCD + ∠BCD = 180⁰ | Using (3)
⇒ 2 ∠BCD = 180⁰
⇒ ∠BCD = 90⁰
⇒ ∠BCD is a right angle.
7. ABC is a right angled triangle in which ∠ A = 90⁰ and AB = AC. Find ∠ B and ∠ C.
Ans: ∵ In ∆ABC, AB = AC
∴ ∠ B = ∠C …(1) | Angles opposite to equal sides of a triangle are equal
In ∆ ABC,
∠A+∠B+ ∠C = 180⁰ Sum of all the angles of a triangle is 180⁰
⇒ 90⁰+∠B+∠C = 180⁰ | ∵ ∠ A = 90⁰ (given)
⇒ ∠ B+∠C = 90⁰ …(2)
From (1) and (2), we get, ∠ B = ∠C = 45⁰
8. Show that the angles of an equilateral triangle are 60⁰ each.
Ans: Given: An equilateral triangle ABC
To Prove: ∠ A = ∠B = ∠ C = 60⁰
Proof: ∵ ABC is an equilateral triangle
∴ AB = BC = CA …(1)
∵ AB = BC
∴ ∠ A =∠C …(2) | Angles opposite to equal sides of a triangle are equal
∵ BC = CA
∴ ∠A =∠B …(3) | Angles opposite to equal sides of a triangle are equal
From (2) and (3), we obtain
∠ A = ∠ B = ∠C …(4)
In ∆ABC, ∠ A+∠B+ ∠ C = 180⁰ …(5) | Sum of all the angles of a triangle is 180⁰
From (4) and (5), we get, ∠ A = ∠B = ∠ C = 60⁰
| Exercise 7.3 |
1. ∆ ABC and ∆DBE are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that:
(i) ∆ADB ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠ A as well as ∠ D
(iv) AP is the perpendicular bisector of BC
Ans: Given: ∆ ABC and ∆ DBE are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. AD is extended to intersect BC at P.
To Prove: (i) ∆AB D ≅ ACD
(ii) ∆ ABP ≅ ACP
(iii) AP bisects ∠ A as well as ∠ D
(iv) AP is the perpendicular bisector of BC
Proof:
(i) In ∆ABD and ∆ACD
AB = AC …(1) | ∆ ABC is an isosceles triangle
BD = CD …(2) | ∵ ∆DBC is an isosceles triangle
AD = AD …(3) | Common side SSS Rule
∴ ∆ ABD ≅ ∆ ACD
(ii) In ∆ABP and ∆ACP,
AB = AC …(4) | From (1)
∠ ABP = ∠ ACP …(5)
∵ AB = AC | From (1)
∴ ∠ ABP =∠ACP |Angles opposite to equal sides of a triangle are equal
∵ ∆ ABP ≅ ∆ ACD | Proved in (i) above
∴ ∠ BAP = ∠ CAP …(6) | c.p.c.t.
From (4), (5) and (6) we get
∆ ABP ≅ ∆ACP | ASA Rule
(iii) ∵ ∆ ABP ≅ ∆ACP | Proved in (ii) above c.p.c.t.
∴ ∠ BAP = ∠CAP
⇒ AP bisects ∠ A
In ∆BDP and ∆ CDP,
BD = CD …(7) | From (2)
DP = DP …(8) | Common
∵ ∆ ABP ≅ ∆ACP | Proved in (ii) above
∴ BP = CP …(9) | c.p.c.t.
From (7), (8) and (9) we get
∆ BDP ≅ ∆ CDP | SSS Rule
∴ ∠BDP = ∠ CDP | c.p.c.t.
⇒ DP bisects ∠ D
⇒ AP bisects ∠ D
(iv) ∵ ∆ BDP ≅ ∆ CDP | Proved in (iii) above
∴ BP = CP | c.p.c.t.
∠ BDP = ∠ CDP | c.p.c.t.
But ∠ BPD+∠CPD = 180⁰ | Linear Pair Axiom
∴ ∠ BPD = ∠ CPD = 90⁰ ….(11)
From (10) and (11) we get
AP is the perpendicular bisector of BC
2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:
(i) AD bisects BC
(ii) AD bisects ∠ A
Ans: Given: AD-is an altitude of an isosceles triangle ABC in which AB = AC
To Prove: (i) AD bisects BC
(ii) AD bisects ∠ A
Proof: (i) In right ∆ADB and right ∆ADC,
Hyp. AB = Hyp. AC | Given
Side AD = Side AD | Common
∴ ∆ ADB ≅ ∆ ADC | RHS Rule
∴ BD = CD | c.p.c.t.
⇒ AD bisects BC
(ii) ∆ ADB ≅ ∆ ADC | Proved in (i) above
∴ ∠ BAD = ∠ CAD | c.p.c.t
⇒ AD bisects ∠ A
3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of triangle PQR. Show that:
(i) ∆ ABM ≅ ∆ PQN
(i) ∆ ABC ≅ ∆PQR
Ans: Given: Two sides AB and BC and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆ PQR
To Prove: (i) ∆ ABM ≅ ∆ PQN
(ii) ∆ ABC ≅ ∆ PQR
Proof: (i) In ∆ ABM and ∆ PQN
AB = PQ …(1)
AM = PN …(2) | Given
BC = QR | Given
⇒ 2BM = 2QN
∵ M and N are the mid-points of BC and QR respectively
⇒ BM = QN …(3)
In view of (1), (2) and (3),
∆ ABM ≅ ∆ PQN | SSS Rule
(ii) ∵ ∆ ABM = ∆ PQN | Proved in (1) above
∴ ∠ABM = ∠PQN | c.p.c.t.
⇒ ∠ABC = ∠PQR …(4)
In ∆ ABC and ∆PQR,
AB = PQ | Given
BC = QR | Given
∠ABC = ∠PQR | From (4)
∴ Δ ABC ≅ ∆ PQR | SAS Rule
4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Ans: Given: BE and CF are two equal altitudes of a triangle ABC.
Το Prove: ∆ ABC is isosceles.
Proof: In right ∆ BEC and right ∆ CFB,
Side BY = Side CF | Given
Hyp. BC = Hyp. СВ | Common
∴ ∆ BEC ≅ ∆ CFB | RHS Rule
∴ ∠ BEC = ∠ CBF | c.p.c.t.
∴ AB = AC | Sides opposite to equal angles of a triangle are equal
∴ ∆ ABC is isosceles.
5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C
Ans: Given: ABC is an isosceles triangle with AB = AC
To Prove: ∠B = ∠C
Construction: Draw AP⊥BC
Proof: In the right triangle APB and right triangle APC,
Hyp. AB = Hyp. AC | Given
Side AP = Side AP | Common
∴ Δ ΑΡΒ ≅ ΔΑΡΕ | RHS Rule
∴ ∠ABP = ∠ APC | c.p.c.t.
⇒ ∠B = ∠C
| Exercise 7.4 |
1. Show that in a right angled triangle, the hypotenuse is long. east side.
Ans: Let ABC be a right angled triangle in which ∠ B = 90⁰
Then, ∠ A+ ∠ C = 90⁰ | ∵ Sum of all the angles of a triangle is 180⁰
∴ ∠B = ∠ A+∠C
∴ ∠B 〉∠ A and ∠ B 〉∠ C
∴ AC〉BC and AC〉AB | ∵ Side opposite to greater angle is longer
∴ AC is the longest side, i.e. hypotenuse is the longest side.
2. In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC ∠QCB. Show that AC 〉AB.
Ans: Given: Sides AB and AC of ∆ABC are extended to points P and Q respectively. Also ∠ PBC〈 ∠ QCB
To Prove: AC〉AB
Proof: ∠ PBC〈 ∠QCB | Given
⇒ – ∠ PBC -〉∠ QCB
⇒ 180⁰ – ∠ PBC 〉180⁰-∠QCB
⇒ ∠ABC 〉∠ ACB
∴ AC〉AB | ∵ Side opposite to the greater angle is longer
3. In figure, ∠ B〈 ∠A and ∠C〈∠D. Show that AD〈 BC
Ans: Given: In figure, ∠B〈 ∠ A and ∠ C〈 ∠ D
Prove: AD〈BC | Given
∴ ∠ B〈 ∠A
∴ OB〈 OA …(1) | Side opposite to greater angle is longer
∠C〈 ∠D|Given
∴ ∠D〉∠C
∴ OC〈OD …(2) | Side opposite to greater angle is longer
From (1) and (2), we get
OB+OC 〉OA+OD ⇒ BC 〉AD ⇒ AD〈 BC
4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that
∠A〉∠C and ∠B〉∠D
Ans: Given: AB and CD are respec-tively the smallest and longest sides of a quadrilateral ABCD
To Prove: ∠A〉∠C and ∠B〉∠D
Construction: Join AC
Proof: In ∆ ABC,
AB〈 BC | ∵ AB is the smallest side of quadrilateral ABCD
⇒ BC 〉AB
∴ ∠BAC〉∠ BCA …(1) | Angle opposite to longer side is greater
In ∆ACD,
CD〉AD | ∵ CD is the longest side of quadrilateral ABCD
∴ ∠CAD〉∠ ACD | Angle opposite to longer side is greater ..(2)
From (1) and (2), we obtain
∠BAC + ∠CAD〉∠ BCA + ∠ ACD
⇒ ∠A 〉∠C
Similarly, joining B to D, we can prove that ∠B 〉∠D
5. In figure, PR 〉PQ and PS bisects ∠QPR. Prove that ∠PSR〉∠PSQ
Ans: Given: In figure, PR〉PQ and PS bisects ∠QPR
To Prove: ∠PSR〉∠ PSQ
Proof: In ∆ PQR
PR〉PQ | Given
∴ ∠PQR 〉∠PRQ …(1) | Angle opposite to longer side is greater
∵ PS is the bisector of ∠QPR
∠QPS = RPS
In ∆ PQS, …(2)
∠PQR + ∠QPS + ∠PSQ = 180⁰…(3) | ∵ The sum of the three angles of a ∆ is 180⁰
In A PRS,
∠PRS + ∠SPR + ∠PSR = 180⁰ …(4) | ∵ The sum of the three angles of a ∆ is 180⁰
From (3) and (4),
∠PQR + ∠QPS + ∠ PSQ = ∠ PRS +∠SPR + ∠ PSR
⇒ ∠PQR + ∠PSQ = ∠ PRS + ∠ PSR
⇒ ∠ PRS + ∠PSR = ∠PQR + ∠PSQ
⇒ ∠ PRS + ∠PSR 〉∠PRQ + ∠PSQ | From (1)
⇒ ∠PRQ + ∠PSR〉∠PRS + ∠PSQ | ∵ ∠PRQ = ∠PRS
⇒ ∠PSR〉∠PSQ
6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Ans: Given: I is a line and P is a point not lying on l. PM⊥ l N is any point on I other than M.
To Prove: PM〈 PN
Proof: In ∆ PMN
∠M = 90⁰
∴ ∠N is an acute angle.
| Angle sum property of a triangle
∴ ∠M〉∠N
∴ PN〉PM | Side opposite to greater angle is greater
⇒ PM〈PN
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