NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.4) Exercise 6.4
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Examinations are an important part of a student’s academic journey. It is a way to measure their progress and see how they are doing academically.With the help of examinations, students are able to learn about their strengths and weaknesses. They are a way of evaluating students’ academic knowledge and abilities. They may be oral or written and may be administered by the student’s teacher, a standardised testing organisation, or a college or university. Examinations are essential because they allow teachers to measure students’ progress and understand their capabilities. Through examinations, teachers can identify which concepts students have learned and which ones they need more help with. In addition, examinations allow teachers to evaluate the effectiveness of their teaching methods. Through examinations, students can improve their study methods and get better scores. The NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.4 can be used by students, for their preparation for the Mathematics exam.
The NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.4, advised by Extramarks, include answers to all questions in the textbook. NCERT Solutions increases students’ interest in the subjects that they study.The NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.4 are provided for students of Class 7. The Extramarks website and mobile application have the best NCERT Solutions that students can access from the internet.
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NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.4) Exercise 6.4
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NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Exercise 6.4
Chapter 6 of the Mathematics subject in Class 7 covers the concepts of Triangles and their Properties and introduces topics like the Medians and Altitudes of a Triangle, the Exterior Angle of a Triangle, the Angle Sum Property of a Triangle, Two Special Triangles which are Equilateral and Isosceles, Sum of the lengths of Two sides of a Triangle, and Rightangled Triangles and Pythagoras Property. All the questions and exercises can be decoded by students with the guidance of the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.4.
Furthermore, Exercise 6.4 covers the following topics:
1) Students have to specify whether the given numbers can be considered to make a triangle.
2) Other questions are related to proving which value is greater or smaller than the other.
Using Extramarks’ smartphone application or website, students can have an interactive learning experience. NCERT Solutions similar to the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.4 help students be wellprepared for their examination. Students facing difficulties while decoding Mathematics topics or questions may find the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.4 to be helpful to them. They can utilise Extramarks’ services and features by visiting the Extramarks website or mobile application. Some of the features that students may access are:
 Learning application – for students
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 NCERT Solutions like NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.4
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 Sample papers
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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
(i) OP + OQ > PQ
(II) OQ + OR > QR?
(
Ans.
\begin{array}{l}\text{If O is a point in the interior of a given triangle, then three}\\ \text{triangles}\Delta \text{OPQ,}\text{\hspace{0.17em}}\Delta \text{OQR,}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\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{\hspace{0.17em}}\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
$\begin{array}{l}\mathrm{AM}\mathrm{is}\mathrm{a}\mathrm{median}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{ABC}.\\ \mathrm{Is}\mathrm{AB}+\mathrm{BC}+\mathrm{CA}>\; 2\mathrm{AM}?\end{array}$
(
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}\text{}\dots \left(\text{i}\right)\\ \text{Similary in}\Delta \text{ACM, we get}\\ \text{AC}+\text{CM}>\text{AM}\text{}\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}\\ \overline{)\text{AB}+\text{BC}+\text{CA}>\text{2AM}}\\ \text{Therefore, the given expression is true}\text{.}\end{array}
Q.4
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}\text{}\dots \left(\text{i}\right)\\ \text{Similary in}\Delta \text{BCD, we get}\\ \text{BC}+\text{CD}>\text{DB}\text{}\dots \left(\text{ii}\right)\\ \text{In}\Delta \text{CDA, we get}\\ \text{CD}+\text{DA}>\text{AC}\text{}\dots \left(\text{iii}\right)\\ \text{In}\Delta \text{DAB, we get}\\ \text{DA}+\text{AB}>\text{DB}\text{}\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)}\\ \overline{)\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.
Similary in
In
Adding (i), (ii), (iii) and (iv) to get
2
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{.,}1512=3\text{)}\text{.}\\ \text{Therefore, those two measures are 27 cm and 3 cm}\text{.}\end{array}