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

Q.1

$\begin{array}{l}\mathrm{Is}\mathrm{it}\mathrm{possible}\mathrm{to}\mathrm{have}\mathrm{a}\mathrm{triangle}\mathrm{with}\mathrm{the}\mathrm{following}\mathrm{sides}?\\ \left(\mathrm{i}\right)2\mathrm{cm},3\mathrm{cm},5\mathrm{cm}\\ \left(\mathrm{ii}\right)3\mathrm{cm},6\mathrm{cm},7\mathrm{cm}\\ \left(\mathrm{iii}\right)6\mathrm{cm},3\mathrm{cm},2\mathrm{cm}\end{array}$

Ans.

\begin{array}{l}\text{For a triangle, the sum of lengths of either two sides is always}\\ \text{greater than the third side}\text{.}\\ \text{(i) Given sides of the triangle are 2 cm, 3 cm and 5 cm}\text{.}\\ \text{2 cm + 3 cm = 5 cm}\\ \text{So, the sum of two side is not greater than the third side}\text{.}\\ \text{Therefore, this triangle is not possible}\text{.}\\ \text{(ii) Given sides of the triangle are 3 cm, 6 cm and 7 cm}\text{.}\\ \text{3 cm + 6 cm = 9 cm}\\ \text{So, the sum of two side is greater than the third side}\text{.}\\ \text{Therefore, this triangle is possible}\text{.}\\ \text{(iii) Given sides of the triangle are 6 cm, 3 cm and 2 cm}\text{.}\\ \text{2 cm + 3 cm = 5 cm}\\ \text{So, the sum of two side is not greater than the third side}\text{.}\\ \text{Therefore, this triangle is not possible}\text{.}\end{array}

Q.2

Take any point O in the interior of a triangle PQR. Is
(i) OP + OQ > PQ?
(II) OQ + OR > QR?
( iii ) OR + OP > RP?

Ans.

\begin{array}{l}\text{If O is a point in the interior of a given triangle, then three}\\ \text{triangles}\Delta \text{OPQ,}\Delta \text{OQR,}\text{and}\Delta \text{ORP can be constructed}\text{.}\\ \text{In a triangle, the sum of the lengths of either two sides}\\ \text{is always greater than the third side}\text{.}\\ \text{(i) Yes, as}\Delta \text{OPQ is a triangle with side OP, OQ and PQ}\text{.}\\ \text{So, OP + OQ > PQ}\\ \text{(ii) Yes, as}\Delta \text{OQR is a triangle with side OP, OQ and PQ}\text{.}\\ \text{So, OQ + OR > QR}\\ \text{(iii)}\text{Yes, as}\Delta \text{ORP is a triangle with side OP, OQ and PQ}\text{.}\\ \text{So, OP + OR > PR}\end{array}

Q.3

$\begin{array}{l}\mathrm{AM}\mathrm{is}\mathrm{a}\mathrm{median}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{ABC}.\\ \mathrm{Is}\mathrm{AB}+\mathrm{BC}+\mathrm{CA}>\; 2\mathrm{AM}?\end{array}$

(Consider the sides of triangles ΔABM and ΔAMC.)

Ans.

\begin{array}{l}\text{Since, in a triangle, the sum of either two sides is always}\\ \text{greater than the third side}\text{.}\\ \text{So, in}\Delta \text{ABM}\\ \text{AB}+\text{BM}>\text{AM}\dots \left(\text{i}\right)\\ \text{Similary in}\Delta \text{ACM, we get}\\ \text{AC}+\text{CM}>\text{AM}\dots \left(\text{ii}\right)\\ \text{Adding (i) and (ii) to get}\\ \text{AB}+\text{BM}+\text{AC}+\text{CM}>\text{2AM}\\ \text{AB}+\text{CA}+\text{(BM}+\text{CM)}>\text{2AM}\\ \text{AB}+\text{AC}+\text{BC}>\text{2AM}\\ \boxed{\text{AB}+\text{BC}+\text{CA}>\text{2AM}}\\ \text{Therefore, the given expression is true}\text{.}\end{array}

Q.4

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

Ans.

\begin{array}{l}\text{Since, in a triangle, the sum of either two sides is always}\\ \text{greater than the third side}\text{.}\\ \text{So, in}\Delta \text{ABC}\\ \text{AB}+\text{BC}>\text{CA}\dots \left(\text{i}\right)\\ \text{Similary in}\Delta \text{BCD, we get}\\ \text{BC}+\text{CD}>\text{DB}\dots \left(\text{ii}\right)\\ \text{In}\Delta \text{CDA, we get}\\ \text{CD}+\text{DA}>\text{AC}\dots \left(\text{iii}\right)\\ \text{In}\Delta \text{DAB, we get}\\ \text{DA}+\text{AB}>\text{DB}\dots \left(\text{iv}\right)\\ \text{Adding (i), (ii), (iii) and (iv) to get}\\ \text{AB}+\text{BC}+\text{BC}+\text{CD}+\text{CD}+\text{DA}+\text{DA}+\text{AB}>\text{CA}+\text{DB}+\text{AC}+\text{DB}\\ \text{2}\left(\text{AB}+\text{BC}+\text{CD}+\text{DA}\right)+\text{2(AC}+\text{DB)}\\ \boxed{\text{AB}+\text{BC}+\text{CD}+\text{DA}>\text{AC+BD}}\\ \text{Therefore, the given expression is true}\text{.}\end{array}

Q.5

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

Ans.

Consider a quadrilateral ABCD.

Since, in a triangle, the sum of either two sides is always greater than the third side. So, in ΔOAB OA+OB>AB …( i )
Similary in ΔOBC, we get OC+OB>BC …( ii )
In ΔOCD, we get OD+OC>CD …( iii )
In ΔODA, we get OA+OD>DA …( iv )
Adding (i), (ii), (iii) and (iv) to get OA+OB+OC+OB+OD+OC+OA+OD>AB+BC+CD+DA
2( OA+OB+OC+OD )>2( AC+BD ) ( OA+OB+OC+OD )>( AC+BD )
Therefore, the given expression is true.

Q.6

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

Ans.

\begin{array}{l}\text{In a triangle, the sum of either two sides is always greater}\\ \text{than the third side and also, the difference of the lengths of}\\ \text{either two sides is always lesser than the third side}\text{.}\\ \text{Here, the third side will be lesser than the sum of these two}\\ \text{(i}\text{.e}\text{.,}12+15=27\text{) and also, it will be greater than the difference}\\ \text{of these two (i}\text{.e}\text{.,}15-12=3\text{)}\text{.}\\ \text{Therefore, those two measures are 27 cm and 3 cm}\text{.}\end{array}