7. Triangles Mathematics Exercise - 7.4 class 9 Maths in English - CBSE Study
NCERT Solutions for Class 9 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 7. Triangles with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 7.4 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 9 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 9 English Medium Mathematics All Chapters:
7. Triangles
4. 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.
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