NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

# NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

## NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 7 Triangles Ex 7.4.

Ex 7.4 Class 9 Maths Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.

Solution:
Let us consider a ∆ABC in which B = 90°.
A + B + C = 180°
A + 90° + C = 180°
A + C = 90°
⇒ ∠A + C = B
B > A and B > C

Side opposite to B is longer than the side opposite to A
Therefore, AC > BC.
Similarly, AC > AB.
Hence, AC is the longest side of the ∆ABC. But AC is the hypotenuse of the triangle. Thus, the hypotenuse is the longest side.

Ex 7.4 Class 9 Maths Question 2.
In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB.

Solution:
ABC + PBC = 180°           [Linear pair]
and
ACB + QCB = 180°     [Linear pair]
But
PBC < QCB                  [Given]

180° – PBC > 180° – QCB
ABC > ACB
AC > AB                         [The side opposite to ABC > the side opposite to ACB]

Ex 7.4 Class 9 Maths Question 3.
In figure, B < A and C < D. Show that AD < BC.

Solution:

Since A > B                [Given]
OB > OA                    …(1)                     [Side opposite to greater angle is longer]
Similarly, OC > OD      …(2)
Adding (1) and (2), we get
OB + OC > OA + OD

Ex 7.4 Class 9 Maths Question 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:
Let us join AC.
Now, in ∆ABC, AB < BC [ AB is the smallest side of the quadrilateral ABCD]

BC > AB
BAC > BCA             …(1)           [Angle opposite to longer side of a ∆ is greater]

Again, in ∆ACD, CD > AD                  [CD is the longest side of the quadrilateral ABCD]
CAD > ACD             …(2)          [Angle opposite to longer side of a ∆ is greater]
Adding eqs. (1) and (2), we get
BAC + CAD > BCA + ACD
A > C
Similarly, by joining BD, we can prove that
B > D.

Ex 7.4 Class 9 Maths Question 5.
In figure, PR > PQ and PS bisect QPR. Prove that PSR > PSQ.

Solution:
In ∆PQR, PS bisects QPR              [Given]
QPS = RPS            …(1)
and PR > PQ                                     [Given]
PQS > PRS           …(2)           [Angle opposite to longer side of a ∆ is greater]
Adding eqs. (1) and (2), we get

PQS + QPS > PRS + RPS       …(3)

Exterior PSR = [PQS + QPS]
and exterior
PSQ = [PRS + RPS]
[An exterior angle is equal to the sum of interior opposite angles]
Now, from (3), we have
PSR > PSQ.

Ex 7.4 Class 9 Maths Question 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 consider the ∆PMN such that M = 90°

Since, M + N + P = 180°
[Sum of angles of a triangle is 180°]
M = 90° [PM MN]
So,
N + P = M
N < M
PM < PN                 …(1)
Similarly, PM < PN1    …(2)
and PM < PN2             …(3)
From (1), (2) and (3), we have PM is the smallest line segment drawn from a given point P on the line l.

Thus, the perpendicular line segment is the shortest line segment drawn on a line from a point not on it.

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