NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4

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 understand 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 Solutions. Students must read it to get an in-depth understanding of this topic. Also, the theorems based on the Inequalities in triangles have 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 with 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.

NCERT Solutions for Class 9 Maths Chapter 7 – Triangles Exercise 7.4

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Access Other Exercise Solutions of Class 9 Maths Chapter 7 – Triangles

Students can have a look at the solution to 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.

Solution:

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.

Solution:

It is given that ∠PBC < ∠QCB

We know that ∠ABC + ∠PBC = 180°

So, ∠ABC = 180°-∠PBC

Also,

∠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.

Solution:

In the question, it is mentioned that angles B and  C are smaller than angles A and D, respectively, i.e., ∠B < ∠A and ∠C < ∠D.

Now,

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.

Solution:

In ΔABD, we see that

AB < AD < BD

So, ∠ADB < ∠ABD — (i) (Since the angle opposite to the 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.

Solution:

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

Proof:

∠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.

Solution:

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:

To prove:

AB < AC

Proof:

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 theorems related to the “Inequalities in a Triangle” and questions related to it. 3 theorems are explained in this exercise. They are mentioned below:

1. Theorem 7.6: If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater).
2. Theorem 7.7: In any triangle, the side opposite to the larger (greater) angle is longer.
3. 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, first, understand these theorems in depth. Also, try to write the answers in steps because the questions will be based on proving. The NCERT solutions provided in PDF will also help students in writing the answers in a better way so that they can score high marks in the board exams.

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