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7. Triangles Mathematics class 9 in English Medium ncert book solutions Exercise 7.4
7. Triangles Exercise 7.4 – Complete NCERT Book Solutions for Class 9 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 7. Triangles Exercise 7.4 to help you master concepts and score higher.
7. Triangles Mathematics class 9 in English Medium ncert book solutions Exercise 7.4
NCERT Solutions for Class 9 Mathematics play an important role in helping students understand the concepts of the chapter 7. Triangles clearly. This chapter includes the topic Exercise 7.4, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 9 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 7. Triangles is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 7.4. By studying these updated NCERT Solutions for Class 9 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
7. Triangles
Exercise 7.4
Q1: Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
Given:Let us consider a right-angled triangle ABC, right-angled at B.
To prove: AC is the longest side.
Proof: In ΔABC,
∠ A + ∠ B + ∠ C = 180° (Angle sum property of a
∠ 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.
Q2 : 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.
Answer : Given: AB and AC of ΔABC are extended to points PandQ respectively. Also, ∠PBC < ∠QCB.
To prove: AC > AB
Proof: 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.)
Q3 : In the given figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.
Solution:
Given: ∠ B < ∠ A and ∠ C < ∠ D.
To prove: AD < BC
Proof: 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
AD < BC
Q4 :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.
Solution :
Given: AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD .
To prove: ∠ A > ∠ C and ∠ B > ∠ D.
Construction: Let us join AC.
Proof: In ΔABC,
AB < BC (AB is the smallest side of quadrilateral ABCD)
∴ ∠ 2 < ∠ 1 (Angle opposite to the smaller side is smaller) ... (1)
In ΔADC,
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 obtain
∠ 2 + ∠ 4 < ∠ 1 + ∠ 3
⇒ ∠ C < ∠ A
⇒ ∠ A > ∠ C
Let us join BD.
In ΔABD,
AB < AD (AB is the smallest side of quadrilateral ABCD)
∴ ∠ 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
Q5 : In the given figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR >∠ SQP.
Solution :
Given: PR > PQ and PS bisects ∠ QPR.
To prove: ∠ PSR >∠ SQP.
Proof: 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)]
Q6 : Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution:
Given: PNM is a right angled triangle at N.
To prove: PN < PM.
Proof: 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.
See other sub-topics of this chapter:
1. Exercise 7.1 2. Exercise 7.2 3. Exercise 7.3 4. Exercise 7.4