NCERT Solutions for Class 7 Maths Chapter 6 The Triangle And Its Properties Ex 6.4

Exercise 6.4

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.

4. $ABCD$ 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.

5. $ABCD$ 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.

6. 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.

Conclusion

In Class 7 Maths Chapter 6.4, we have reinforced our understanding of key geometric principles. By exploring the Exterior Angle Property and the Angle Sum Property, we learned how to determine unknown angles in triangles. These properties, which are covered in Class 7 Maths Chapter 6 Exercise 6.4 Solutions are fundamental in geometry, providing the foundation for more advanced concepts. Mastering these principles allows you to solve various geometric problems with confidence.

Class 7 Maths Chapter 6: Exercises Breakdown

CBSE Class 7 Maths Chapter 6 Other Study Materials

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