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SEBA Class 9 Mathematics Chapter 7 Triangles
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Triangles
Chapter – 7
| Exercise 7.1 |
Q.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)
Q.2. ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ BCA Prove that
(i) ∆AB D ≅ ∆ BAC
Ans: In ∆ ABD and ∆BAC
(ii) BD = AC
Ans: AD = BC (Given)
(iii) ∠ ABD =∠BAC
Ans: ∠ DAB = ∠CBA
(Given)
(Common)
and AB = BA
By SAS Congruence Condition
∆ABD ≅ ∆BAC
(ii) BD=AC( By C.P.C.T)
(iii) ∠ABD = ∠ BAC
(Again by C.P.C.T)
Q.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.
Q.4. L 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
Q.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
Ans:
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 or B is equidistant from the arms of ∠ A
Ans: BP = BQ (By C.P.C.T.)
Q.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 Congruence Condition
∆ BAC ≅ ∆ DAE
So, BC = DE (By C.P.C.T.)
Q.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: ∆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.)
Q.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 Congruence 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 Congruence Condition.
∆ DBC ≅ ∠ ACB
(iv) CM = 1/2 AB
Ans: Since ∆ DBC ≅ ∠ ACB
DC = AB
CM = AM
| Exercise 7.2 |
Q.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
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
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) AO bisects ∠ A
Ans: In ∆OAB and ∆OAC
AB = AC
OB = OC
AB = AC
| Given Proved (i) above
| 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
∴ ∠ OAB =∠OAC
∴ AO bisects ∠ A
|By SAS Rule
|c.p.c.t.
Q.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
Q.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 tri- angle 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.
Q.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.
Q.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
Q.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. D
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 …(3) ⇒ ⇒ ⇒
⇒ ∠ BCD is a right angle.
Q.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 tri angle is 180⁰
⇒ 90⁰+∠B+∠C = 180⁰ | ∵ ∠ A = 90⁰(given)
⇒∠ B+∠C = 90⁰ …(2)
From (1) and (2), we get, ∠ B=∠C = 45⁰
Q.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 |
Q.1. ∆ ABC and ∆BDE 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
Ans: 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) ∆ ABP≅ACP
Ans:
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) AP bisects ∠ A as well as ∠ D
Ans: ∆ 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) AP is the perpendicular bisector of BC
Ans: ∆ 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
Q.2. AD is an altitude of an isosceles triangle ABC in which AB = AC.
Show that:
(i) AD bisects BC
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) AD bisects ∠ A
Ans: ∆ ADB ≅ ∆ ADC | Proved in (i) above
∴ ∠ BAD = ∠ CAD | c.p.c.t
⇒ AD bisects ∠ A
Q.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
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
Ans:
To Prove: ∆ ABM ≅ ∆ PQN
∆ 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) ∆ ABC ≅ ∆PQR
Ans: ∆ ABM = ∆ PQN | Proved in (1) above
∴ ∠ABM = ∠PQN | c.p.c.t. …(4)
⇒∠ABC = ∠PQR
In ∆ ABC and ∆PQR,
AB = PQ | Given
BC = QR | Given
∠ABC = ∠PQR | From (4)
∴ Δ ABC ≅∆ PQR | SAS Rule
Q.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.
Q.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 Ξ ΔΑΡΕ | RHS Rule
∴ ∠ABP = ∠ APC | c.p.c.t.
⇒ ∠B = ∠C
| Exercise 7.4 |
Q.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.
Q.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
Q.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
Q.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
Q.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
Q.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
