Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.4 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

Question 1.

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

Answer:

In ∆ ABC, ∠ B is a right angle.

∴ ∠ B = 90° and AC is the hypotenuse.

In ∆ ABC,

∠A + ∠B + ∠C = 180°

∴ ∠ A + 90° + ∠ C = 180°

∴ ∠ A + ∠ C = 90°

Now, ∠ A and ∠ C are both positive (in degrees) and their sum is 90°.

∴ ∠ A < 90° and ∠ C < 90°

∴ ∠ A < Z B and ∠ C < ∠ B

∴ BC < AC and AB < AC (Theorem 7.7)

Hence, hypotenuse AC is greater than each of the other two sides BC and AB. Thus, the hypotenuse is the longest side in a right angled triangle.

Question 2.

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

Answer:

∠ PBC < ∠ QCB (Given)

∴ – ∠ PBC > – ∠ QCB (Multiplying an inequality by (-1), it reverses)

∴ 180° – ∠ PBC > 180° – ∠ QCB (Adding 180° on both the sides) …………. (1)

Now, ∠ ABC and ∠ PBC as well as ∠ ACB and ∠ QCB from a linear pair.

∴ ∠ ABC = 180° – ∠ PBC and

∠ ACB = 180° – ∠ QCB

Substituting these values in (1), we get

∠ ABC > ∠ ACB

Now, in ∆ ABC, ∠ ABC > ∠ ACB A

∴ AC > AB (Theorem 7.7)

Question 3.

In the given figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.

Answer:

In ∆ OAB, ∠ B < ∠ A

∴ OA < OB (Theorem 7.7) …………….. (1)

In ∆ OCD, ∠ C < ∠ D

∴ OD < OC (Theorem 7.7) ……………… (2)

Adding (1) and (2),

OA + OD < OB + OC

∴ AD < BC (As O is the point of intersection of AD and BC, it lies on both the line segments.)

Question 4.

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

Answer:

Construction: In quadrilateral ABCD, draw diagonal AC.

AB is the smallest side of ABCD and CD is the longest side of ABCD.

∴ AB < BC and AD < CD.

In ∆ ABC, AB < BC

∴ ∠ ACB < ∠ BAC …… (1)

In ∆ CDA, AD < CD

∴ ∠ DCA < ∠ DAC ……… (2)

Adding (1) and (2),

∠ ACB + ∠ DCA < ∠ BAC + ∠ DAC

∴ ∠ BCD < ∠ BAD

∴ ∠ BAD > ∠ BCD

Thus, in quadrilateral ABCD, ∠ A > ∠ C. Similarly, after constructing diagonal BD and using the inequalities in A ABD and A CBD, it can be proved that ∠ B > ∠ D.

Question 5.

In the given figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.

Answer:

PS is the bisector of ∠QPR.

∴ ∠QPS = ∠RPS = \(\frac{1}{2}\) ∠QPR ……………. (1)

∠ PSR is an exterior angle of A PQS and ∠ PSQ is an exterior angle of A PRS.

∴ PSR = ∠ Q + ∠QPS and

∠PSQ = ∠R + ∠RPS …………….. (2)

Now, in A PQR, PR > PQ

∴ ∠ Q > ∠ R

∴ ∠Q + \(\frac{1}{2}\) ∠ QPR > ∠R + \(\frac{1}{2}\) ∠ QPR

∴ ∠Q + ∠QPS > ∠R + ∠RPS [from (1)]

∴ ∠PSR > ∠PSQ

Question 6.

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

Answer:

AB is a line and P is a point not on AB.

PM is the perpendicular line segment drawn from P to line AB.

N is any point on AB, other than M.

In ∆ PMN, ∠ M = 90°

∴ ∠ N < 90°

Thus, in ∆ PMN, ZN < ZM.

∴ PM < PN

This is true for any location of point N.

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