KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1

In this chapter, we provide KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 pdf, free KSEEB solutions for v book pdf download. Now you will get step by step solution to each question.

Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.1

Question 1.
In quadrilateral ACBD, AC = AD, and AB bisect ∠A. Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?
Solution:
Data: In a quadrilateral ABCD,
AC = AD and AB bisect ∠A.

To Prove: ∆ABC ≅ ∆ABD
Proof: In ∆ABC and ∆ABD
AC = AD (Data)
∠CAB = ∠DAB .
∠A bisected
AB is common.
Here Side, angle, side rule is there.
∴ ∆ABC ≅ ∆ABD.

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that
(1) ∆ABD ≅ ∆BAC
(2) BD = AC
(3) ∠ABD = ∠BAC A
Solution:
Data: ABCD is a quadrilateral.
AD = BC and ∠DAB = ∠CBA.

To Prove:
(1) ∆ABD ≅ ∆BAC
(2) BD = AC
(3) ∠ABD = ∠BAC
Proof: (i) In ∆ABD and ∆BAC
AD = BC (Data)
∠DAB = ∠CBA (Data)
AB is common SAS postulate.
∴ ∆ABD ≅ ∆BAC.

(ii) ∆ABD ≅ ∆BAC.
∵ Corresponding sides are equal.
∴ BD = AC.

(iii) ∆ABD ≅ ∆BAC. (Proved)
Equal sides of Adjacent angles are equal.
As we have AD = BC,
The adjacent angle for AD is ABD
The adjacent angle for BC is BAC
∴∠ABD = ∠BAC.

Question 3.
AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Solution:
Data: AD and BC are equal perpendiculars to a line segment AB.

To Prove: CD Bisects AB.
Proof: In ∆CBO and ∆DAO,
BC = AD
∠CBO = ∠DAO = 90°
∠BOC = ∠AOD (Vertically opposite angles)
SAS postulate,
∴ ∆CBO = ∆DAO ∴ OA = OB
∴ CD bisects AB at ‘O’.

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC ≅ ∆CDA.

Solution:
Data: l and m are two parallel lines intersected by another pair of parallel lines p and q.
To Prove: ∆ABC ≅ ∆CDA
Proof: In AABC and ACDA,
∠ACB = ∠DAC [∵ Alternate angles.]
∠BAC = ∠ACD
AC is common.
A.S.A. postulate.
∆ABC ≅ ∆CDA

Question 5.
Line 1 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. Show that :

(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Solution:
Data: 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.
To Prove:
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Proof: In ∆APB and ∆AQB
∠APB = ∠AQB = 90°
∠PAB = ∠QAB (∵ l bisects ∠A).
AB is common.
∴ ∆APB ≅ ∆AQB (ASA postulate)
∴ BP = BQ.
B is equidistant from the arms of ∠A.

Question 6.
In AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:
Data: AC = AE, AB = AD and ∠BAD = ∠EAC
To Prove : BC = DE
Proof: In ∆ABC and ∆EAD,
AB = AD (Data)
AC = AE (Data)
∆BAC = ∆EAD
(∵∠BAD + ∠DAC = AEAC + ∠DAC and ∠DAC = ∠DAC.)
Side – Angle – Side (SAS) Postulate
∴ ∆ABC = ∆EAD
∴ BC = DE.

Question 7.
AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB. Show that

(i) ∆DAP ≅ ∆EBP
(ii) AD = BE.
Solution:
Data: 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.
To Prove :
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE
Proof : (i) In ∆DAP and ∆EBP,
AP = BP (∵ P is the mid-point of AB)
∠BAD = ∠ABE (Data)
∠APD = ∠BPE
∵ (∠EPA = ∠DPB Adding ∠EPD to both sides)
∠EPA + ∠EPD = ∠DPB + ∠EPD
∴ ∠APD = -BPE.
Now, Angle, Side, Angle postulate.
∴ ∆DAP ≅ ∆EBP.

(ii) As it is ∆DAP ≅ ∆EBP,
Three sides and three angles are equal to each other.
∴ AD = BE.

Question 8.
In the 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. Show that :

(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle.
(iii) ∆DBC ≅ ∆ACB
(iv) CM = (frac{1}{2}) AB.
Solution:
Data: In a 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.
To Prove:
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle.
(iii) ∆DBC ≅ ∆ACB
(iv) CM = (frac{1}{2}) AB.
Proof: (i) In ∆AMC and ∆BMD,
BM = AM (∵ M is the mid-point of AB)
DM = CM (Data)
∠BMD = ∠AMC (Vertically opposite angles)
Now, Side, Angle, Side postulate.
∴ ∆AMC ≅ ∆BMD

(ii) ∆AMC ≅ ∆BMD (Proved)
Adding AMBC to both sides,
∆BMD + ∆MBC = ∆AMC + ∆MBC
∆DBC ≅ ∆ACB.
∴ AB Hypotenuse = DC Hypotenuse
∠ACB = ∠DBC = 90°
∴ ∠DBC is a right angle.

(iii) In ∆DBC and ∆ACB,
DB = AC
∵ ∆DMB ≅ DMC (Proved)
∠DBC = ∠ACB (Proved)
BC is Common.
Side, angle, side postulate.
∴ ∠DBC ≅ ∆ACB.

(iv) ∆DBC ≅ ∆ACB (proved)
Hypotenuse AB = Hypotenuse DC
(frac{1}{2}) AB = (frac{1}{2}) DC
(frac{1}{2}) AB = CM
∴ CM = (frac{1}{2})AB.

All Chapter KSEEB Solutions For Class 9 Maths

—————————————————————————–

All Subject KSEEB Solutions For Class 9

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share kseebsolutionsfor.com to your friends.

Best of Luck!!

Leave a Comment

Your email address will not be published.