Ex 7.4 NCERT Solutions- Triangles Class 9 Notes | EduRev

Mathematics (Maths) Class 9

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Class 9 : Ex 7.4 NCERT Solutions- Triangles Class 9 Notes | EduRev

The document Ex 7.4 NCERT Solutions- Triangles Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
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Question 1. Show that in a right angled triangle, the hypotenuse is the longest side
 Solution:
Let us consider ΔABC such that ∠B = 90º
∴ ∠A + ∠B + ∠C = 180º
∴ [∠A + ∠C] + ∠B = 180º
⇒ ∠A + ∠C = 90º
⇒ ∠A + ∠C= ∠B
∴ ∠B > ∠A and ∠B > ∠C

Ex 7.4 NCERT Solutions- Triangles Class 9 Notes | EduRev

⇒ Side opposite to ∠B is longer than the side opposite to ∠A.
i.e. AC > BC                      ...(1)
Similarly, AC > AB             ...(2)
From (1) and (2), we get AC is the longest side.
But AC is the hypotenuse of the triangle.
Thus, hypotenuse is the longest side.
 

Question 2. In the adjoining figure, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
 Solution:
∠ABC + ∠PBC = 180º                [Linear pair]

Ex 7.4 NCERT Solutions- Triangles Class 9 Notes | EduRev

and ∠ACB + ∠QCB = 180º                 [Linear pair]
∴ ∠ABC + ∠PBC = ∠ACB + ∠QCB
But ∠PBC < ∠QCB             [Given]
∴ ∠ABC > ∠ACB
⇒ [The side opposite to ∠ABC] > [The side opposite to ∠ACB]
⇒ AC > AB


Question 3. In the figure, ∠B < ∠A and ∠C < ∠D. Show that AD > BC.
 Solution:
∵ ∠B< ∠A                [Given]
⇒ ∠A> ∠B
∴ OB > OA [Side opposite to greater angle is longer]                    ...(1)

Ex 7.4 NCERT Solutions- Triangles Class 9 Notes | EduRev

Similarly, OC > OD              ...(2)
From (1) and (2), we have
[OB + OC] > [OA + OD]
⇒ BC > AD or AD < BC


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.

Ex 7.4 NCERT Solutions- Triangles Class 9 Notes | EduRev

Now, in ΔABC, AB < BC          [∵ AB is the smallest side of quadrilateral ABCD]

⇒ BC > AB
∴ [Angle opposite to BC] < [Angle opposite to AB]
⇒ ∠BAC > ∠BCA                 ...(1)
Again, in ΔACD,
CD > AD            [∵ CD is the longest side of the quadrilateral ABCD]
∴ [Angle opposite to CD] > [Angle opposite to AD]
⇒ ∠CAD > ∠ACD                           ...(2)

Ex 7.4 NCERT Solutions- Triangles Class 9 Notes | EduRev

Adding (1) and (2), we get
[∠BAC + CAD] > [∠BCA + ∠ACD]
⇒ ∠A> ∠C
Similarly, by joining BD, we have ∠B> ∠D
 

Question 5. In the figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.
 Solution: 
In ΔPQR, PS bisects ∠QPR                  [Given]
∴ ∠QPS = ∠RPS
∵ PR > PQ                     [Given]

Ex 7.4 NCERT Solutions- Triangles Class 9 Notes | EduRev

∴ [Angle opposite to PR] >           [Angle opposite to PQ]
⇒ ∠PQS > ∠PRS
⇒ [∠PQS + ∠QPS] > [∠PRS + ∠RPS]                ...(1)
[∵ ∠QPS = ∠RPS]
∵ Exterior ∠PSR = [∠PQS + ∠QPS]     
[∵ An exterior angle is equal to the sum of interior opposite angles] and  
Exterior ∠PSQ = [∠PRS + ∠RPS]
Now, from (1), we have ∠PSR > ∠PSQ

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 ΔPMN such that ∠M = 90º
Since, ∠M + ∠N + ∠P = 180º

Ex 7.4 NCERT Solutions- Triangles Class 9 Notes | EduRev

[Sum of angles of a triangle] ∵
∠M = 90º [∵ PM ⊥ ℓ]
⇒ ∠N< ∠M ⇒ PM < PN              ...(1)
Similarly, PM < PN1                  ...(2)
PM < PN2                                  ...(3)
From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l.
Thus, the perpendicular segment is the shortest line segment drawn on a line from a point not on it.

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