**Download NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.4 – Triangles**

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

**Solution :**

Let us consider a right-angled triangle ABC, right-angled at B.

By using angle sum property, In ∆ABC

∠A + ∠B + ∠C = 180°

∠A + 90° + ∠C = 180°

∠A + ∠C = 90°

=> The other two angles have to be acute (i.e., less than 90º).

∠B is the largest angle in ∆ABC.

∠B > ∠A and ∠B > ∠C

AC > BC and AC > AB

[In any triangle, the side opposite to the larger (greater) angle is longer.]

=> AC is the largest side in ∆ABC.

Also, AC is the hypotenuse of ∆ABC.

Hence, Hypotenuse is the longest side in a right-angled triangle.

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

**Solution :**

In the given figure,

∠PBC + ∠ABC = 180° (Linear pair)

=> ∠PBC = 180° − ∠ABC —- (1)

Also,

∠QCB + ∠ACB = 180°

=> ∠QCB = 180° − ∠ACB —- (2)

As ∠PBC < ∠QCB, 180° − ∠ABC > 180° − ∠ACB

∠ABC > ∠ACB

=> AC > AB (Side opposite to the larger(greater) angle is longer.)

Hence Proved

**3. In Figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.**

**Solution :**

In ∆ABO,

∠B < ∠A (Given) => AO < BO (Side opposite to smaller angle is smaller) — (1)

In ∆CDO,

∠C < ∠D (Given) => OD < OC (Side opposite to smaller angle is smaller) — (2)

On adding Equations (1) and (2), we get

AO + DO < BO + CO => AD < BC

Hence Proved

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

**Solution :**

Construction :

Let us join AC.

In ∆ABC,

AB < BC (AB is the smallest side of quadrilateral ABCD)

=> ∠2 < ∠1 (Angle opposite to the smaller side is smaller) — (1)

In ∆ADC,

AD < CD (CD is the largest side of quadrilateral ABCD)

=> ∠4 < ∠3 (Angle opposite to the smaller side is smaller) — (2)

On adding Equations (1) and (2), we get

∠2 + ∠4 < ∠1 + ∠3

=> ∠C < ∠A

=> ∠A > ∠C

Hence Proved.

Construction :

Let us join BD

In ∆ABD,

AB < AD (AB is the smallest side of quadrilateral ABCD)

=> ∠8 < ∠5 (Angle opposite to the smaller side is smaller) — (3)

In ∆BDC,

BC < CD (CD is the largest side of quadrilateral ABCD)

=> ∠7 < ∠6 (Angle opposite to the smaller side is smaller) — (4)

On adding Equations (3) and (4),

we obtain,

∠8 + ∠7 < ∠5 + ∠6 => ∠D < ∠B => ∠B > ∠D

Hence Proved.

**5. In Figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.**

**Solution :**

In ∆PQR,

PR > PQ (Given)

∠PQR > ∠PRQ (Angle opposite to larger side is larger) — (1)

Since, PS is the bisector of ∠QPR.

=> ∠QPS = ∠RPS —- (2)

And, ∠PSR is the exterior angle of ∆PQS.

=> ∠PSR = ∠PQR + ∠QPS —- (3)

Also, ∠PSQ is the exterior angle of ∆PRS.

=> ∠PSQ = ∠PRQ + ∠RPS —- (4)

On adding Equations (1) and (2), we get

∠PQR + ∠QPS > ∠PRQ + ∠RPS

∠PSR > ∠PSQ [Using Equations (3) and (4)]

Hence Proved.

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

**Solution : **

Let us take a line l and from point A (i.e., not on line l),

Draw two line segments AC and AB.

Let, AC be perpendicular to line l and AB is drawn at some other angle.

In ∆ACB,

∠C = 90˚

∠A + ∠C + ∠B = 180˚ (Angle sum property of a triangle)

=> ∠A + ∠B = 90˚

Clearly,

∠B is an acute angle.

=> ∠B < ∠C => AC < AB (Side opposite to the smaller angle is smaller)

Similarly, By drawing more line segments from A to l, it can be proved that AC is smaller in comparison to them.

=> It can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.**Hence Proved.**

**Download NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.4 – Triangles**

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