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 ΔABC in which ∠B = 90°

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

(i) AC > BC

(ii) AC > AB.

Proof: (i) In ΔABC, we have

∠B = 90°

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

∴ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

⇒ ∠A + ∠C = 180° – 90° = 90°

⇒ ∠A and ∠C are acute angles

⇒ ∠B > ∠A

⇒ AC > BC,

(Side opposite to greater angle is larger)

(ii) Similarly. ∠B > ∠C

⇒ 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, ∠PBC < ∠QCB. Show that AC > AB

Solution:

In ΔABC, we have

∠PBC = ∠BAC + ∠ACB …(i)

and ∠QCB = ∠BAC + ∠ABC …(ii)

Now, ∠PBC < ∠QCB (Given)

⇒ ∠BAC + ∠ACB < ∠BAC + ∠ABC [Using (i) and (ii)]

⇒ ∠ACB < ∠ABC ⇒ ∠ABC > ∠ACB

⇒ AC > AB,

(side opposite to greater angle is larger)

Hence proved

Question 3.

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

Solution:

In ΔAOB

∠B < ∠A, (Given) ⇒ ∠A > ∠B

⇒ BO > AO …..(i)

(Side opposite to greater angle is larger)

In ΔCOD, ∠C > ∠D, (Given)

⇒ ∠D > ∠C

⇒ OC > OD ….(ii)

(Side opposite to greater angle is larger)

Adding (i) and (ii), we get

BO + OC > AO + OD

⇒ BC > AD

⇒ 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 ∠A > ∠C and ∠B > ∠D (NCERT Exemplar Problems]

Solution :

Join AC and BD.

In ΔABC, BC > AB

(∵ AB is the smallest side)

⇒ ∠BAC > ∠ACB ………(i)

(Angle opposite to greater side is larger)

In ΔADC, CD > AD,

[∵ CD is the largest side]

⇒ ∠CAD > ∠ACD ………(ii)

[Angle opposite to greater side is larger]

Adding (i) and (ii), we get

∠BAC + ∠CAD > ∠ACB + ∠ACD

⇒ ∠A > ∠C

In ΔABD, AD > AB,

(∵ AB is the smallest side)

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

In ΔBCD, CD > BC,

(∵ CD is the greatest side)

⇒ ∠CBD > ∠CDB …(iv) (Angle opposite to longer side is larger)

Adding (iii) and (iv), we get

∠ABD + ∠CBD > ∠ADB + ∠CDB

⇒ ∠B > ∠D

Hence, ∠A > ∠C and ∠B > ∠D.

Hence Proved

Question 5.

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

Solution:

In ΔPQR, PR > PQ. (Given)

⇒ ∠PQR > ∠PRQ ……(i)

[∵ Angle opposite to longer side is larger]

⇒ ∠PQR + ∠QPS > ∠PRQ + ∠QPS,

[Adding ∠QPS on both sides)

⇒ ∠PQR + ∠QPS > ∠PRQ + ∠RPS,

∵ PS is the bisector of ∠QPR

∴ ∠QPS = ∠RPS

⇒ ∠PQS + ∠QPS > ∠PRS + ∠RPS,

[∵ ∠PQR = ∠PQS

and ∠PRQ = ∠PRS] …… (ii)

In ΔPQS, we have

∠PSR = ∠PQS + ∠QPS, …(iii)

(By theorem 6.8)

In ΔPRS, we have

∠PSQ = ∠PRS + ∠RPS, …(iv)

(By theorem 6.8)

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

∠PSR > ∠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 ⊥ l and N is any point on l other than M.

To prove: PM < PN.

Proof: In ΔPNM,

∠PMN = 90°, (∵ PM ⊥ l)

∴ ∠PNM + ∠MPN = 90°

⇒ ∠PNM < 90°

⇒ ∠PNM < ∠PMN ⇒ ∠PMN >∠PNM

⇒ PN > PM (Side opposite to greater angle is longer)

⇒ PM