In the previous exercises of Maths chapter 7, students of Class 9 have studied the congruence of sides and angles of a triangle. This must have helped them in understanding the similarities between the two triangles. But what if two triangles are different from each other. Then how do we compare them? For this, we need to know the concept of Inequalities in triangles. This concept is beautifully explained with the help of activity in the NCERT Class 9 Maths Textbook. Students must read it to get in-depth understanding of this topic. Also, the theorems based on the Inequalities in triangles has been illustrated before the exercise problems.
It often happens that even though students have gone through all the theorems and activities of the NCERT books, they are still unable to solve the questions given in the exercises or get stuck in them. So, to help them we have provided the step by step NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4 in PDF format. Students can download the complete solution pdf of Exercise 7.4 from the links given below.
Access Other Exercise Solutions of Class 9 Maths Chapter 7 – Triangles
Students can have a look at the solution of all the remaining exercises of this chapter by clicking on the link below.
Exercise 7.1 Solution 8 Questions (6 Short Answer Questions, 2 Long Answer Question)
Exercise 7.2 Solution 8 Questions (6 Short Answer Questions, 2 Long Answer Question)
Exercise 7.3 Solution 5 Questions (3 Short Answer Questions, 2 Long Answer Question)
Exercise 7.5 (Optional) Solution 4 Questions
Students can also have a look at the NCERT Solutions of Class 9 for Science subject and other NCERT study material and notes.
Access Answers to NCERT Class 9 Maths Chapter 7 – Triangles Exercise 7.4
1. Show that in a right-angled triangle, the hypotenuse is the longest side.
It is known that ABC is a triangle right angled at B.
We know that,
∠A +∠B+∠C = 180°
Now, if ∠B+∠C = 90° then ∠A has to be 90°.
Since A is the largest angle of the triangle, the side opposite to it must be the largest.
So, AB is the hypotenuse which will be the largest side of the above right-angled triangle i.e. ΔABC.
2. In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
It is given that ∠PBC < ∠QCB
We know that ∠ABC + ∠PBC = 180°
So, ∠ABC = 180°-∠PBC
∠ACB +∠QCB = 180°
Therefore ∠ACB = 180° -∠QCB
Now, since ∠PBC < ∠QCB,
∴ ∠ABC > ∠ACB
Hence, AC > AB as sides opposite to the larger angle is always larger.
3. In Fig. 7.49, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
In the question, it is mentioned that angles B and angle C is smaller than angles A and D respectively i.e. ∠B < ∠A and ∠C < ∠D.
Since the side opposite to the smaller angle is always smaller
AO < BO — (i)
And OD < OC —(ii)
By adding equation (i) and equation (ii) we get
AO+OD < BO + OC
So, AD < BC
4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).
Show that ∠A > ∠C and ∠B > ∠D.
In ΔABD, we see that
AB < AD < BD
So, ∠ADB < ∠ABD — (i) (Since angle opposite to longer side is always larger)
Now, in ΔBCD,
BC < DC < BD
Hence, it can be concluded that
∠BDC < ∠CBD — (ii)
Now, by adding equation (i) and equation (ii) we get,
∠ADB + ∠BDC < ∠ABD + ∠CBD
∠ADC < ∠ABC
∠B > ∠D
Similarly, In triangle ABC,
∠ACB < ∠BAC — (iii) (Since the angle opposite to the longer side is always larger)
Now, In ΔADC,
∠DCA < ∠DAC — (iv)
By adding equation (iii) and equation (iv) we get,
∠ACB + ∠DCA < ∠BAC+∠DAC
⇒ ∠BCD < ∠BAD
∴ ∠A > ∠C
5. In Fig 7.51, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.
It is given that PR > PQ and PS bisects ∠QPR
Now we will have to prove that angle PSR is smaller than PSQ i.e. ∠PSR > ∠PSQ
∠QPS = ∠RPS — (ii) (As PS bisects ∠QPR)
∠PQR > ∠PRQ — (i) (Since PR > PQ as angle opposite to the larger side is always larger)
∠PSR = ∠PQR + ∠QPS — (iii) (Since the exterior angle of a triangle equals to the sum of opposite interior angles)
∠PSQ = ∠PRQ + ∠RPS — (iv) (As the exterior angle of a triangle equals to the sum of opposite interior angles)
By adding (i) and (ii)
∠PQR +∠QPS > ∠PRQ +∠RPS
Thus, from (i), (ii), (iii) and (iv), we get
∠PSR > ∠PSQ
6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
First, let “l” be a line segment and “B” be a point lying on it. A line AB perpendicular to l is now drawn. Also, let C be any other point on l. The diagram will be as follows:
AB < AC
In ΔABC, ∠B = 90°
Now, we know that
∠A+∠B+∠C = 180°
∴ ∠A +∠C = 90°
Hence, ∠C must be an acute angle which implies ∠C < ∠B
So, AB < AC (As the side opposite to the larger angle is always larger)
This exercise covers the theorem related to the “Inequalities in a Triangle” and questions related to it. There are 3 theorems that are explained in this exercise. These are mentioned below:
- Theorem 7.6: If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater).
- Theorem 7.7: In any triangle, the side opposite to the larger (greater) angle is longer.
- Theorem 7.8: The sum of any two sides of a triangle is greater than the third side.
Exercise 7.4 contains a total of 6 questions which are based on the above theorems only. So, before starting to solve the exercise questions firstly understand these theorems in depth. Also, try to write the answers in steps because the questions will be based on proving. The solution provided in PDF will also help students in writing the answers in a better way so that they could score high marks in first term exams.
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