Exercise 6.4 (Revised) - Chapter 6 - Triangles & Its Properties - Ncert Solutions class 7 - Maths

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Chapter 6 - Triangles & Its Properties | NCERT Solutions Class 7 Maths

Ex 6.4 Question 1.

Is it possible to have a triangle with the following sides?
1. $2 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}$
2. $3 \mathrm{~cm}, 6 \mathrm{~cm}, 7 \mathrm{~cm}$
3. $6 \mathrm{~cm}, 3 \mathrm{~cm}, 2 \mathrm{~cm}$

Answer:

Since, a triangle is possible whose sum of the lengths of any two sides should be greater than the length of third side.
(i) $2 \mathrm{~cm}, 3 \mathrm{~cm}, 5 \mathrm{~cm}$
$
\begin{aligned}
& 2+3=5 \text { No } \\
& 2+5>3 \text { Yes } \\
& 3+5>2 \text { Yes }
\end{aligned}
$

This triangle is not possible.
(ii) $3 \mathrm{~cm}, 6 \mathrm{~cm}, 7 \mathrm{~cm}$
$
\begin{aligned}
& 3+6>7 \text { Yes } \\
& 6+7>3 \mathrm{Yes} \\
& 3+7>6 \mathrm{Yes}
\end{aligned}
$

This triangle is possible.

(iii) $6 \mathrm{~cm}, 3 \mathrm{~cm}, 2 \mathrm{~cm}$
$
\begin{aligned}
& 6+3>2 \text { Yes } \\
& 6+2>3 \text { Yes }
\end{aligned}
$

$
2+3<6 \mathrm{No}
$

This triangle is not possible.
Ex 6.4 Question 2.

Take any point $\mathrm{O}$ in the interior of a triangle $\mathrm{PQR}$. Is:

1. $\mathrm{OP}+\mathrm{OQ}>\mathrm{PQ}$ ?
2. $\mathrm{OQ}+\mathrm{OR}>\mathrm{QR}$ ?
3. $\mathrm{OR}+\mathrm{OP}>\mathrm{RP}$ ?

Since sum of two sides is greater than third side.
Answer:

Join OR, OQ and OP.

(i) Is OP $+\mathrm{OQ}>\mathrm{PQ}$ ?

Yes, POQ form a triangle.
(ii) Is $\mathrm{OQ}+\mathrm{OR}>\mathrm{QR}$ ?

Yes, RQO form a triangle.
(iii) Is $\mathrm{OR}+\mathrm{OP}>\mathrm{RP}$ ?

Yes, ROP form a triangle.

Ex 6.4 Question 3.

AM is a median of a triangle $\mathrm{ABC}$. Is $\mathrm{AB}+\mathrm{BC}+\mathrm{CA}>2 \mathrm{AM}$ ?
(Consider the sides of triangles ABM and AMC.)

Answer:

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore, In $\triangle \mathrm{ABM}, \mathrm{AB}+\mathrm{BM}>\mathrm{AM}$ $\qquad$
In $\triangle \mathrm{AMC}, \mathrm{AC}+\mathrm{MC}>\mathrm{AM}$ $\qquad$
Adding eq. (i) and (ii),
$
\begin{aligned}
& \mathrm{AB}+\mathrm{BM}+\mathrm{AC}+\mathrm{MC}>\mathrm{AM}+\mathrm{AM} \\
& \Rightarrow \mathrm{AB}+\mathrm{AC}+(\mathrm{BM}+\mathrm{MC})>2 \mathrm{AM} \\
& \Rightarrow \mathrm{AB}+\mathrm{AC}+\mathrm{BC}>2 \mathrm{AM}
\end{aligned}
$

Hence, it is true.

Ex 6.4 Question 4.

$\mathrm{ABCD}$ is a quadrilateral. Is $\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}>\mathrm{AC}+\mathrm{BD}$ ?

Answer:

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore, In $\triangle \mathrm{ABC}, \mathrm{AB}+\mathrm{BC}>\mathrm{AC}$.
In $\triangle \mathrm{ADC}, \mathrm{AD}+\mathrm{DC}>\mathrm{AC}$. $\qquad$ (ii)

In $\Delta \mathrm{DCB}, \mathrm{DC}+\mathrm{CB}>\mathrm{DB}$ $\qquad$
In $\Delta \mathrm{ADB}, \mathrm{AD}+\mathrm{AB}>\mathrm{DB}$ $\qquad$ (iv)

Adding eq. (i), (ii), (iii) and (iv),
$
\begin{aligned}
& \mathrm{AB}+\mathrm{BC}+\mathrm{AD}+\mathrm{DC}+\mathrm{DC}+\mathrm{CB}+\mathrm{AD}+\mathrm{AB}>\mathrm{AC}+\mathrm{AC}+\mathrm{DB}+\mathrm{DB} \\
& \Rightarrow(\mathrm{AB}+\mathrm{AB})+(\mathrm{BC}+\mathrm{BC})+(\mathrm{AD}+\mathrm{AD})+(\mathrm{DC}+\mathrm{DC})>2 \mathrm{AC}+2 \mathrm{DB} \\
& \Rightarrow 2 \mathrm{AB}+2 \mathrm{BC}+2 \mathrm{AD}+2 \mathrm{DC}>2(\mathrm{AC}+\mathrm{DB}) \\
& \Rightarrow 2(\mathrm{AB}+\mathrm{BC}+\mathrm{AD}+\mathrm{DC})>2(\mathrm{AC}+\mathrm{DB}) \\
& \Rightarrow \mathrm{AB}+\mathrm{BC}+\mathrm{AD}+\mathrm{DC}>\mathrm{AC}+\mathrm{DB} \\
& \Rightarrow \mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}>\mathrm{AC}+\mathrm{DB}
\end{aligned}
$

Hence, it is true.
Ex 6.4 Question 5.

$\mathrm{ABCD}$ is quadrilateral. Is $\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<2(\mathrm{AC}+\mathrm{BD})$ ?
Answer:

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore, In $\Delta \mathrm{AOB}, \mathrm{AB}<\mathrm{OA}+\mathrm{OB}$ $\qquad$
In $\Delta \mathrm{BOC}, \mathrm{BC}<\mathrm{OB}+\mathrm{OC}$ $\qquad$
In $\Delta \mathrm{COD}, \mathrm{CD}<\mathrm{OC}+\mathrm{OD}$ $\qquad$
In $\triangle \mathrm{AOD}, \mathrm{DA}<\mathrm{OD}+\mathrm{OA}$ $\qquad$
Adding eq. (i), (ii), (iii) and (iv),
$\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<\mathrm{OA}+\mathrm{OB}+\mathrm{OB}+\mathrm{OC}+\mathrm{OC}+\mathrm{OD}+\mathrm{OD}+\mathrm{OA}$
$\Rightarrow \mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<2 \mathrm{OA}+2 \mathrm{OB}+2 \mathrm{OC}+2 \mathrm{OD}$
$\Rightarrow \mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<2[(\mathrm{AO}+\mathrm{OC})+(\mathrm{DO}+\mathrm{OB})]$

$
\Rightarrow \mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<2(\mathrm{AC}+\mathrm{BD})
$

Hence, it is proved.
Ex 6.4 Question 6.

The lengths of two sides of a triangle are $12 \mathrm{~cm}$ and $15 \mathrm{~cm}$. Between what two measures should the length of the third side fall?

Answer:

Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

It is given that two sides of triangle are $12 \mathrm{~cm}$ and $15 \mathrm{~cm}$.
Therefore, the third side should be less than $12+15=27 \mathrm{~cm}$.

And also the third side cannot be less than the difference of the two sides.
Therefore, the third side has to be more than $15-12=3 \mathrm{~cm}$.
Therefore, the third side could be the length more than $3 \mathrm{~cm}$ and less than $27 \mathrm{~cm}$.