# NCERT solution class 9 chapter 7 Triangles exercise 7.4 mathematics

## EXERCISE 7.4

#### Question 1:

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

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

In ΔABC,

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

∠A + 90º + ∠C = 180°

∠A + ∠C = 90°

Hence, the other two angles have to be acute (i.e., less than 90º).

∴ ∠B is the largest angle in ΔABC.

⇒ ∠B > ∠A and ∠B > ∠C

⇒ AC > BC and AC > AB

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

Therefore, AC is the largest side in ΔABC.

However, AC is the hypotenuse of ΔABC. Therefore, hypotenuse is the longest side in a right-angled triangle.

#### Question 2:

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

In the given figure,

∠ABC + ∠PBC = 180° (Linear pair)

⇒ ∠ABC = 180° − ∠PBC … (1)

Also,

∠ACB + ∠QCB = 180°

∠ACB = 180° − ∠QCB … (2)

As ∠PBC < ∠QCB,

⇒ 180º − ∠PBC > 180º − ∠QCB

⇒ ∠ABC > ∠ACB [From equations (1) and (2)]

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

#### Question 3:

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

In ΔAOB,

∠B < ∠A

⇒ AO < BO (Side opposite to smaller angle is smaller) … (1)

In ΔCOD,

∠C < ∠D

⇒ OD < OC (Side opposite to smaller angle is smaller) … (2)

On adding equations (1) and (2), we obtain

AO + OD < BO + OC

#### 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.

Let us join AC.

In ΔABC,

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

∴ ∠2 < ∠1 (Angle opposite to the smaller side is smaller) … (1)

∴ ∠4 < ∠3 (Angle opposite to the smaller side is smaller) … (2)

On adding equations (1) and (2), we obtain

∠2 + ∠4 < ∠1 + ∠3

⇒ ∠C < ∠A

⇒ ∠A > ∠C

Let us join BD.

In ΔABD,

∴ ∠8 < ∠5 (Angle opposite to the smaller side is smaller) … (3)

In ΔBDC,

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

#### Question 5:

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

As PR > PQ,

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

PS is the bisector of ∠QPR.

∴∠QPS = ∠RPS … (2)

∠PSR is the exterior angle of ΔPQS.

∴ ∠PSR = ∠PQR + ∠QPS … (3)

∠PSQ is the exterior angle of ΔPRS.

∴ ∠PSQ = ∠PRQ + ∠RPS … (4)

Adding equations (1) and (2), we obtain

∠PQR + ∠QPS > ∠PRQ + ∠RPS

⇒ ∠PSR > ∠PSQ [Using the values of equations (3) and (4)]

#### Question 6:

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

Let us take a line l and from point P (i.e., not on line l), draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.

In ΔPNM,

∠N = 90º

∠P + ∠N + ∠M = 180º (Angle sum property of a triangle)

∠P + ∠M = 90º

Clearly, ∠M is an acute angle.

∴ ∠M < ∠N

⇒ PN < PM (Side opposite to the smaller angle is smaller)

Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to them.

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