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7. Triangles Class 9 Mathematics Solutions English Medium-Exercise 7.4
7. Triangles Class 9 Mathematics Solutions English Medium-Exercise 7.4 Get chapter-wise detailed explanations, step-by-step answers, important questions and exam-ready study material in Hindi and English medium.
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7. Triangles Class 9 Mathematics Solutions English Medium-Exercise 7.4
NCERT Solutions for Class 9 are specially prepared according to the latest CBSE syllabus (2026-27) to help students understand every concept clearly. These solutions provide step-by-step explanations, accurate answers, and exam-oriented guidance for all chapters. Class 9 students can improve their problem-solving skills, strengthen conceptual understanding, and prepare confidently for school as well as board examinations. All questions are solved in a simple and easy-to-understand language for both Hindi and English medium learners.
7. Triangles Class 9 Mathematics Solutions English Medium-Exercise 7.4
NCERT Solutions Class 9 Mathematics English Medium
7. Triangles
Topic: 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.
All Topics From 7. Triangles
See other sub-topics of this chapter:
1. Exercise 7.1 2. Exercise 7.2 3. Exercise 7.3 4. Exercise 7.4NCERT Solutions Class 9 Hindi and English Medium – Complete Study Material
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