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
⇒ BC > AD

Or, AD < BC

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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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

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