NCERT Solutions for Class 7 Maths Chapter 6: The Triangle and its Properties - Exercise 6.4

Exercise 6.4

Refer to page 1-6 for exercise 6.4 in the PDF.

1. Is it possible to have a triangle with the following sides?

i) Is it possible to have a triangle with the following sides 2 cm, 3 cm, 5 cm?

Ans: A triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. Take the two sides as 2 cm and 3 cm. Since $2 + 3 = 5$ and the third side is also 5 cm, it is not possible to have a triangle with the sides 2 cm, 3 cm and 5 cm.

ii) Is it possible to have a triangle with the following sides 3 cm, 6 cm, 7 cm?

Ans: A triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. First, take the two sides as 3 cm and 6 cm. Since $3 + 6 = 9$ and $9 > 7$, the property of the triangle is satisfied. Now take the two sides as 6 cm and 7 cm. Since $6 + 7 = 13$ and $13 > 3$, the property of the triangle is satisfied.  Now, take the two sides as 7 cm and 3 cm. Since $7 + 3 = 10$ and $10 > 6$, the property of the triangle is satisfied. Hence, it is possible to have a triangle with the sides 3 cm, 6 cm and 7 cm.

iii) Is it possible to have a triangle with the following sides 6 cm, 3 cm, 2 cm?

Ans: A triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. Take the two sides as 2 cm and 3 cm. Since $2 + 3 = 5$ and 5 is not greater than 6, it is not possible to have a triangle with the sides 6 cm, 3 cm and 2 cm.

2. Take any point $O$ in the interior of a triangle $PQR$. Is the following inequality satisfied?

i) Is $OP + OQ > PQ$?

Ans: In the given triangle, join $OR$, $OQ$ and $OP$.

From the diagram, it can be seen that $OPQ$ is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. So, $OP + OQ > PQ$. Yes, $OP + OQ > PQ$.

ii) Is $OQ + OR > QR$?

Ans: In the given triangle, join $OR$, $OQ$ and $OP$.

From the diagram, it can be seen that $OQR$ is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. So, $OQ + OR > QR$. Yes, $OQ + OR > QR$.

iii) Is $OR + OP > RP$?

Ans: In the given triangle, join $OR$, $OQ$ and $OP$.

From the diagram, it can be seen that $OPR$ is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. So, $OR + OP > RP$. Yes, $OR + OP > RP$.

3. AM is the median of a triangle $ABC$. Is $AB + BC + CA > 2AM$?

Ans: The sum of two sides of a triangle is always greater than the third side of the triangle.

In $\Delta ABM$, $AB + BM > AM$ and in $\Delta AMC$, $AC + MC > AM$.

Add both the inequalities and simplify.

$AB + BM + AC + MC > AM + AM$

$ \Rightarrow AB + AC + \left( {BM + MC} \right) > 2AM$

Substitute $BC$ for $BM + MC$.

$ \Rightarrow AB + AC + BC > 2AM$

Hence, $AB + BC + CA > 2AM$ is true.

3. $ABCD$ is a quadrilateral. Is $AB + BC + CD + DA > AC + BD$?

Ans: The sum of two sides of a triangle is always greater than the third side of the triangle.

In $\vartriangle ABC$, $AB + BC > AC$.

In $\vartriangle ADC$, $AD + DC > AC$.

In $\vartriangle DCB$, $DC + CB > DB$.

In $\vartriangle ADB$, $AD + AB > DB$.

Add all the four inequalities and simplify.

$AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB + DB $

$\Rightarrow \left( {AB + AB} \right) + \left( {BC + BC} \right) + \left( {AD + AD} \right) + \left( {DC + DC} \right) > 2AC + 2DB $

$\Rightarrow 2AB + 2BC + 2AD + 2DC > 2\left( {AC + DB} \right) $

$\Rightarrow 2\left( {AB + BC + AD + DC} \right) > 2\left( {AC + DB} \right) $

Divide both sides by 2 and simplify.

$\Rightarrow \dfrac{2}{2}\left( {AB + BC + AD + DC} \right) > \dfrac{2}{2}\left( {AC + DB} \right) $

$\Rightarrow AB + BC + AD + DC > AC + DB$

Hence, $AB + BC + CD + DA > AC + BD$ is true.

4. $ABCD$ is a quadrilateral. Is $AB + BC + CD + DA < 2\left( {AC + BD} \right)$?

Ans: Draw a quadrilateral $ABCD$. Join $AC$ and $BD$.

The sum of two sides of a triangle is always greater than the third side of the triangle.

In $\vartriangle AOB$, $AB < OA + OB$.

In $\vartriangle BOC$, $BC < OB + OC$.

In $\vartriangle COD$, $DC < OC + OD$.

In $\vartriangle AOD$, $DA < OD + OA$.

Add all the four inequalities and simplify.

\[ AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA \]

\[ \Rightarrow AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD \]

\[ \Rightarrow AB + BC + CD + DA < 2\left[ {\left( {AO + OC} \right) + \left( {DO + OB} \right)} \right] \]

Substitute $AC$ for $AO + OC$ and $BD$ for $DO + OB$.

\[ \Rightarrow AB + BC + CD + DA < 2\left[ {AC + BD} \right]\]

Hence, $AB + BC + CD + DA < 2\left( {AC + BD} \right)$ is true.

5. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

Ans: It is given that two sides of a triangle are 12 cm and 15 cm. The sum of two sides of a triangle is always greater than the third side of the triangle.

Therefore, the third side of the triangle should be less than $12 + 15 = 27$ cm.

Also, the third side cannot be less than the difference of two sides of a triangle. Therefore, The third side should be more than $15 - 12 = 3$ cm. 

Therefore, the length of the third side should be between 3 cm and 27 cm.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Exercise 6.4

Opting for the NCERT solutions for Ex 6.4 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 6.4 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 7 Maths Chapter 6 Exercise 6.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 7 Maths Chapter 6 Exercise 6.4, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 7 Maths Chapter 6 Exercise 6.4 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.