In this chapter, we provide KSEEB SSLC Class 9 Maths Chapter 5 Triangles Ex 5.4 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 9 Maths Chapter 5 Triangles Ex 5.4 pdf, free KSEEB SSLC Class 9 Maths Chapter 5 Triangles Ex 5.4 pdf download. Now you will get step by step solution to each question.

**Karnataka State Syllabus Class 9 Maths Chapter 5 Triangles Ex 5.4**

## Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.4

Question 1.

Show that in a right angled triangle, the hypotenuse is the longest side.

Solution:

In ∆ABC, ∠ABC = 90°

∠A + ∠C = 90°

∠ABC > ∠BAC and ∠ABC < ∠BCA

∴ D is the largest side of ∠ABC.

∴ AC is opposite side of larger angle.

∴ AC hypoternuse is largest side of ∆ABC.

Question 2.

In sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Solution:

Data: AB and AC are the sides of ∆ABC and AB and AC are produced to P and Q respectively.

∠PBC < ∠QCB.

To Prove: AC > AB

Proof: ∠PBC < ∠QCB

Now, ∠PBC + ∠ABC = 180°

∠ABC = 180 – ∠PBC ………. (i)

Similarly, ∠QCB + ∠ACB = 180°

∠ACB = 180 – ∠QCB …………. (ii)

But, ∠PBC < ∠QCB (Data)

∴ ∠ABC > ∠ACB Comparing (i) and (ii), AC > AB

(∵ Angle opposite to larger side is larger)

Question 3.

In fig ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:

Data : ∠B < ∠A and ∠C < ∠D.

To Prove: AD < BC

Proof: ∠B < ∠A

∴ OA < OB …………. (i)

Similarly, ∠C < ∠D

∴ OD < OC …………. (ii)

Adding (i) and (ii), we have

OA + OD < OB + OC

∴ AD < BC.

Question 4.

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.

Solution:

Data : AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD.

To Prove :

(i) ∠A > ∠C

(ii) ∠B > ∠D

Construction: Join AC and BD

Proof: In ∆ABC, AB < BC (data)

∴ ∠3 < ∠4 ………. (i)

In ∆ADC, AD < CD (data)

∴ ∠2 <∠1 …………. (ii)

Adding (i) and (ii),

∠3 + ∠2 < ∠4 + ∠1

∠C < ∠A

∠A > ∠C

By adding B,

In ∆ABD, AB < AD

∠5 < ∠8

In ∆BCD, BC < CD

∠6 < ∠7

∴ ∠5 + ∠6 < ∠8 + ∠7

∠D < ∠B

OR ∠B > ∠D.

Question 5.

In Fig. PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.

Solution:

Data: PR > PQ and PS bisect ∠QPR.

To Prove: ∠PSR > ∠PSQ

Proof: In ∆PQR, PR > PQ

∴ ∠PQR > ∠PRQ ………. (i)

PS bisects ∠QPR.

∴ ∠QPS = ∠RPS ………… (ii)

By adding (i) and (ii),

∠PQR + ∠QPS > ∠PRQ + ∠RPS ………. (iii)

But, in ∆PQS, ∠PSR is an exterior angle.

∴ ∠PSR = ∠PQR + ∠QPS ………… (iv)

Similarly, in ∆PRS, ∠PSQ is the exterior angle.

∴ ∠PSQ = ∠PRS + ∠RPS ……… (v)

In (iii), subtracting (iv) and (v),

∠PSR > ∠PSQ.

Question 6.

Show that of all line segments drawn from a given point, not on it, the perpendicular line segment is the shortest.

Solution:

P is a point on straight line l.

PR is the segment for l drawn such that PQ ⊥ l.

Now, in ∆PQR,

∠PQR = 90° (Construction)

∴ ∠QPR + ∠QRP = 90°

∴ ∠QRP < ∠PQR

∴ PQ < PR

Any line segment is drawn from P to l they form a small angle.

∴ PQ ⊥ l is smaller.

**All Chapter KSEEB Solutions For Class 9 Maths**

—————————————————————————–**All Subject KSEEB Solutions For Class 9**

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