Haryana State Board HBSE 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 7 Triangles Exercise 7.4

Question 1.

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

Solution :

Given: A right-angled Î”ABC in which âˆ B = 90Â°

To prove: Hypotenuse is the longest side i.e.

(i) AC > BC

(ii) AC > AB.

Proof: (i) In Î”ABC, we have

âˆ B = 90Â°

But we know that sum of interior angles of a triangle is 180Â°

âˆ´ âˆ A + âˆ B + âˆ C = 180Â°

â‡’ âˆ A + 90Â° + âˆ C = 180Â°

â‡’ âˆ A + âˆ C = 180Â° – 90Â° = 90Â°

â‡’ âˆ A and âˆ C are acute angles

â‡’ âˆ B > âˆ A

â‡’ AC > BC,

(Side opposite to greater angle is larger)

(ii) Similarly. âˆ B > âˆ C

â‡’ AC > AB

Hence, hypotenuse is the longest side.

Proved

Question 2.

In the figure 7.73, sides AB and AC of AABC are extended to points P and Q respectively. Also, âˆ PBC < âˆ QCB. Show that AC > AB

Solution:

In Î”ABC, we have

âˆ PBC = âˆ BAC + âˆ ACB …(i)

and âˆ QCB = âˆ BAC + âˆ ABC …(ii)

Now, âˆ PBC < âˆ QCB (Given)

â‡’ âˆ BAC + âˆ ACB < âˆ BAC + âˆ ABC [Using (i) and (ii)]

â‡’ âˆ ACB < âˆ ABC â‡’ âˆ ABC > âˆ ACB

â‡’ AC > AB,

(side opposite to greater angle is larger)

Hence proved

Question 3.

In figure âˆ B < âˆ A and âˆ C < âˆ D. Show that AD < BC.

Solution:

In Î”AOB

âˆ B < âˆ A, (Given) â‡’ âˆ A > âˆ B

â‡’ BO > AO …..(i)

(Side opposite to greater angle is larger)

In Î”COD, âˆ C > âˆ D, (Given)

â‡’ âˆ D > âˆ C

â‡’ OC > OD ….(ii)

(Side opposite to greater angle is larger)

Adding (i) and (ii), we get

BO + OC > AO + OD

â‡’ BC > AD

â‡’ AD < BC.

Hence proved

Question 4.

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure 7.75). Show that âˆ A > âˆ C and âˆ B > âˆ D (NCERT Exemplar Problems]

Solution :

Join AC and BD.

In Î”ABC, BC > AB

(âˆµ AB is the smallest side)

â‡’ âˆ BAC > âˆ ACB ………(i)

(Angle opposite to greater side is larger)

In Î”ADC, CD > AD,

[âˆµ CD is the largest side]

â‡’ âˆ CAD > âˆ ACD ………(ii)

[Angle opposite to greater side is larger]

Adding (i) and (ii), we get

âˆ BAC + âˆ CAD > âˆ ACB + âˆ ACD

â‡’ âˆ A > âˆ C

In Î”ABD, AD > AB,

(âˆµ AB is the smallest side)

â‡’ ABD > ZADB …(iii) [Angle opposite to longer side is larger]

In Î”BCD, CD > BC,

(âˆµ CD is the greatest side)

â‡’ âˆ CBD > âˆ CDB …(iv) (Angle opposite to longer side is larger)

Adding (iii) and (iv), we get

âˆ ABD + âˆ CBD > âˆ ADB + âˆ CDB

â‡’ âˆ B > âˆ D

Hence, âˆ A > âˆ C and âˆ B > âˆ D.

Hence Proved

Question 5.

In figure 7.77, PR > PQ and PS bisects âˆ QPR. Prove that âˆ PSR > âˆ PSQ.

Solution:

In Î”PQR, PR > PQ. (Given)

â‡’ âˆ PQR > âˆ PRQ ……(i)

[âˆµ Angle opposite to longer side is larger]

â‡’ âˆ PQR + âˆ QPS > âˆ PRQ + âˆ QPS,

[Adding âˆ QPS on both sides)

â‡’ âˆ PQR + âˆ QPS > âˆ PRQ + âˆ RPS,

âˆµ PS is the bisector of âˆ QPR

âˆ´ âˆ QPS = âˆ RPS

â‡’ âˆ PQS + âˆ QPS > âˆ PRS + âˆ RPS,

[âˆµ âˆ PQR = âˆ PQS

and âˆ PRQ = âˆ PRS] …… (ii)

In Î”PQS, we have

âˆ PSR = âˆ PQS + âˆ QPS, …(iii)

(By theorem 6.8)

In Î”PRS, we have

âˆ PSQ = âˆ PRS + âˆ RPS, …(iv)

(By theorem 6.8)

From (ii), (iii) and (iv), we get

âˆ PSR > âˆ PSQ. Hence proved

Question 6.

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

Solution:

Given: l is a line and P is a point not lying on it. PM âŠ¥ l and N is any point on l other than M.

To prove: PM < PN.

Proof: In Î”PNM,

âˆ PMN = 90Â°, (âˆµ PM âŠ¥ l)

âˆ´ âˆ PNM + âˆ MPN = 90Â°

â‡’ âˆ PNM < 90Â°

â‡’ âˆ PNM < âˆ PMN â‡’ âˆ PMN >âˆ PNM

â‡’ PN > PM (Side opposite to greater angle is longer)

â‡’ PM