## MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.4

Question 1.

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

Solution:

Given

ABC is a right angle A.

To prove:

AC > AB and AC > BC

Proof:

In ∆ABC

∠B = 90°

∴ ∠A + ∠C = 90° (by ASP)

and so ∠B > ∠A and ∠B > ∠C

AC > BC and AC > AB

(In a A, sides opposite to large angle are always longer).

Question 2.

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

Solution:

Given

∠PRC < ∠QCB To prove AC > AB

Proof:

In ∆ ABC

Exterior angle is equal to sum of two opposite angles

∠PBC = ∠1 + ∠3 and

∠QCB = ∠1 + ∠2

∠QCB > ∠PBC (given)

⇒ ∠1 + ∠2 > ∠1 + ∠3

⇒ ∠2 > ∠3

∴ AC > AB

(∴ In a A, side opposite to larger angle is always longer).

Question 3.

In Fig. below, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:

Given

∠B < ∠A i.e., ∠A > ∠B

∠C < ∠D i.e., ∠D> ∠C

To prove: AD > BC

i.e., BC > AD

Proof:

In ∆OCD

∠D > ∠C

OC > OD

(∴ In a ∆, sides opposite to larger angle are always longer) …(1)

In ∆OBA

∠A > AB

OB > OA

(∴ In a A, sides oppositedo larger angle are always longer) …(2)

Adding (1) and (2), we get

OC + OB > OD + OA

BC > AD

Question 4.

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

Solution:

Given

AB is the smallest siBe and CD is the longest side

To prove:

- ∠A > ∠C and
- ∠B > ∠D

Construction:

Join AC

1. Proof:

In, ∆ABC

BC > AB (∴ AB is the smallest side) ∠1 > ∠3

(∴ In a ∆ angle opposite to longer side is always larger) …..(1)

In ∆ACD

CD > AD (∴ CD is the largest side) ∠2 > ∠4

(∴ In a ∆ angles opposite to longer side are always larger) …(2)

Adding (1) and (2), we get

∠1 + ∠2 > ∠3 + ∠4

∠A > ∠C

2. To prove: AB > AD

Construction:

Join BD

Proof:

In ∆ABD

AD > AB (AB is the smallest side)

∠5 > ∠7 …(3)

In ∆BCD

CD > BC (CD is the longest side)

∠6 > ∠8 …(4)

Adding (3) and (4), we get

∠5 + ∠6 > ∠7 + ∠8

∠B > ∠D

Question 5.

In Fig. below, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Sol.

Given:

PR > PQ

∠1 = ∠2

To prove:

∠PSR > ∠PSQ

Proof:

In ∆PSQ

∠PSR = ∠1 + ∠Q (EAP)

In ∆PSR

∠PSQ = ∠2 + ∠R (EAP)

= ∠1 + ∠R (∠1 = ∠2)

In ∆PQR

PR > PQ (given)

∠Q > ∠R

(∴ In a ∆, angle opposite to longer side is always larger)

Adding ∠1 on both sides

∠Q + ∠l > ∠R + ∠1

∴ ∠PSR > ∠PSQ

(∴ ∠PSR = ∠1 + ∠Q and ∠PSQ = ∠1 + ∠R)

Question 6.

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

Solution:

Given

Let us consider the ∆PMN such that ∠M = 90°

Since, ∠M + ∠N + ∠P = 180° [Sum of angles of a triangle]

∠M = 90° [PM ⊥ l]

∠N < ∠M

PM < PN …..(1)

Similarly PM < PN_{1} …..(2)

PM < PN_{2} …..(3)

From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l.

Thus, the perpendicular segment is the shortest line segment drawn on a line from a point not on it.