Jharkhand Board JAC Class 9 Maths Solutions Chapter 7 Triangles Ex 7.4 Textbook Exercise Questions and Answers.

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

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Question 1.

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

Answer:

ABC is a triangle right angled at B.

Now,

∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠C = 90° (∵ ∠B = 90°)

So ∠A < 90° and ∠C < 90°

⇒ ∠B > ∠A and ∠B > ∠C

⇒ AC > BC and AC > AB [As side opposite to larger angle is longer]

Question 2.

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

Answer:

Given: ∠PBC < ∠QCB

Proof: ∠ABC + ∠PBC = 180°

⇒ ∠PBC = 180° – ∠ABC …(i)

Also, ∠ACB + ∠QCB = 180°

⇒ ∠QCB = 180° -∠ACB

⇒ ∠PBC < ∠QCB (Given)

⇒ 180° – ∠ABC < 180° – ∠ACB

(From (i) and (ii))

⇒ – ∠ABC < – ∠ACB

⇒ ∠ABC > ∠ACB

⇒ AC > AB (As sides opposite to larger angle is longer.)

Question 3.

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

Answer:

Given: ∠B < ∠A and ∠C < ∠D

Proof: In ΔAOB,

∠B < ∠A ⇒ AO < BO ……….(i)

(Side opposite to smaller angle is smaller)

In ΔCOD

∠C < ∠D ⇒ DO < CO…(ii)

(Side opposite to smaller angle is smaller)

Adding (i) and (ii)

AO + OD < BO + OC

⇒ AD < BC

Question 4.

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

Answer:

In ΔABD,

AB < AD (As AB is the shortest side)

∴ ∠ADB < ∠ABD ……..(i) (Angle opposite to longer side is larger.)

Now, in ΔBCD,

BC < DC (As CD is the longest side)

∴ ∠BDC < ∠CBD …..(ii) (Angle opposite to longer side is larger.)

Adding (i) and (ii), we get

∠ADB + ∠BDC < ∠ABD + ∠CBD

⇒ ∠ADC < ∠ABC

⇒ ∠B > ∠D

Similarly, in Δ ABC,

AB < BC (As AB is the smallest side)

∠ACB < ∠B AC …(iii) (Angle opposite to longer side is larger)

Now, in ΔADC,

AD < DC (As CD is the longest side)

∠DCA < ∠DAC …(iv) (Angle opposite to longer side is larger)

Adding (iii) and (iv), we get

∠ACB + ∠DCA < ∠BAC + ∠DAC

⇒ ∠BCD < ∠BAD

⇒ ∠A > ∠C

Question 5.

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

Answer:

Given: PR > PQ and PS bisects ∠QPR

To prove: ∠PSR > ∠PSQ

Proof: ∠PQR > ∠PRQ …….(i)

(PR > PQ as angle opposite to longer side is larger)

∠QPS = ∠RPS …(ii) (PSbisects ∠QPR)

∠PSR = ∠PQR + ∠QPS ………(iii)

(exterior angle of a triangle equals the sum of opposite interior angles)

∠PSQ = ∠PRQ + ∠RPS …(iv)

(exterior angle of a triangle equals the sum of opposite interior angles)

Adding (i) and (ii)

∠PQR + ∠QPS > ∠PRQ + ∠RPS

⇒ ∠PSR > ∠PSQ [from (iii) and (iv)]

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Question 6.

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

Answer:

Given: Let l is a line segment and B is a point lying on it. We draw a line segment AB perpendicular to 1. Let C be any other point on 1.

To prove: AB < AC

Proof: In ∆ABC,

∠B = 90°

Now, ∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠C = 90°

⇒ ∠C < 90° (∠B = 90°)

⇒ ∠C < ∠B

∴ ∠C must be acute angle or ∠C < ∠B

⇒ AB < AC (Side opposite to the larger angle is longer.)