NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.5) Exercise 6.5
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Mathematics requires constant practice. Students are expected not only to practise questions and problems provided as a part of their prescribed NCERT textbooks for Mathematics. They are also strongly recommended that they seek out other sources like past years’ question papers, meticulously crafted sample question papers and other kinds of welldesigned practise assessments. Practising with the aid of a wide variety of academic resources would have an evident positive effect on the academic performance of students. It would also give them the advantage of familiarising themselves with the pattern of questions which could appear in the examinations. Once they have mastered this level of reading comprehension, they will be able to determine the best course of action to take when faced with specific types of problems.The system of examinations which is prevalent in the national education system established, regulated, and monitored by the NCERT is a thorough and complex one. A distinct schema of mark distribution characterises the question papers, and students are expected to frame their answers while keeping this scheme in mind.
Students can gain several advantages by using the PDF of Extramarks’ NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5. All the questions have been addressed to help students easily understand all the concepts. Students will undoubtedly benefit from practising with the help of these NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 to prepare for exams and score well. Students can understand complex topics more clearly and simply with the help of NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5. Extramarks’ NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 will benefit the students by clarifying their doubts. Extramarks provides simple explanations while compiling the solutions to help as many students as possible.
The solutions for Class 7 Maths Chapter 6 Exercise 6.5 have been created for students who feel intimidated by Mathematics and require a thorough study resource to prepare for their exams. The solutions for Maths Class 7 Chapter 6 Exercise 6.5 are compiled by Extramarks’ experts while considering each student’s requirements. Students can download the PDF of NCERT Class 7 Maths Chapter 6 Exercise 6.5, to gain a firm understanding of the material.
Extramarks is one of the best learning platforms where students can access various learning tools like the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5. Students can also download the PDFs of these study materials to study offline. Moreover, students can also use the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 to understand the chapter deeply. Students can also study offline with the help of the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5. The NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 are created using simple and easy language. Students can also recheck their answers with the help of NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5.
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.5) Exercise 6.5
The NCERT Solutions For Class 7 Maths Chapter 6 Exercise 6.5 contains two important formulas that students must remember. One is that a Triangle’s Interior Angles add up to 180 degrees, and the other is the Pythagorean theorem, which states that the square of the Hypotenuse is equal to the sum of the Squares of the Base and Perpendicular of a RightAngled Triangle.
There is a thorough explanation of RightAngled Triangles and the Pythagorean theorem in NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5. A very helpful theorem in Mathematics is the Pythagoras theorem. This exercise consists of 8 questions, 5 of which are fairly simple, 2 of medium difficulty, and 1 for which students will need some time to gather the information. Students can take help from the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 to understand the chapter deeply.
When studying shapes in earlier classes, students must have learned about Triangles. Students can learn about Triangles in this lesson, including their types, angles, and characteristics. Triangles can be divided into two groups according to their sides and angles. Equilateral, Scalene, and Isosceles are three of them. These are based on Triangles with Sides and Triangles with Acute, Obtuse, and Right Angles. To learn about a Triangle’s Altitudes, students must ascertain the Triangle’s height. It is possible to define the Altitude as a Line Segment with two ends—one at a Vertex and the other at the line on the other side. For better understanding, students might need to work through some problems based on these.
The Pythagoras theorem application and proof should be conceptually clear to the students. Moreover, one must remember that the Triangle is rightangled if the Pythagoras theorem is accurate.
Students can easily remember and recall questions and concepts when they are presented to them in visually appealing ways or when they are depicted in pictures. Therefore, students should spend a lot of time practising with the diagrams in the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 as these will help them understand the question, particularly in this chapter.
To ensure accuracy in their calculations and logic, students should be wellversed in the Pythagoras Theorem and practice the solved examples independently. Students will excel in their understanding of Chapter 6 if they consistently review the facts and theorems in the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5.
Access NCERT Solutions for Class 7 Chapter 6 Triangles and its Properties
When three lines come together, the Triangle is the first and most basic geometrical figure that results. The NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 focuses on the characteristics of Triangles. In order for students to have a solid understanding of the Sides and Angles of a Triangle in their minds, as well as the classification that these side lengths and angle measurements give to the Triangles, it is necessary to start by describing the fundamental components of Triangles.
For example, Triangles are categorized as Scalene, Isosceles, and Equilateral based on their various side measurements. Triangles are divided into AcuteAngled, RightAngled, and ObtuseAngled categories based on the angles. The NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 discusses examples from the practice of the ideas of Median, Altitude, Angle Sum Property, and the Pythagoras theorem.
Students should focus on the Properties of Triangles and how they differ from one another, as well as the crucial fact that the sum of any given Triangle’s angles is 180 degrees, which is covered in NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5. This will be very beneficial for resolving the exercise questions that can be downloaded in PDF format.
The explanations in the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5, about Triangle and its Properties, will help deepen the student’s understanding of triangles.
Due to a clearly defined geometrical feature, Triangles can be used as templates for other shapes, such as Hexagons and Quadrilaterals. Therefore, understanding Triangles will provide a solid foundation for further Geometry study. The NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5, should be used by the students to strengthen their foundational knowledge. Triangles and their various components, categorising Triangles according to their side lengths, and angle measurements are the topics that the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5. Additionally, the terms Median, Altitude, Angle Sum Theorem, and Pythagoras theorem are defined with pertinent examples.
NCERT Solutions For Class 7 Maths Chapter 6 The Triangle and its Properties Exercise 6.5
When it comes to exam preparation, choosing the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 is the best choice for CBSE students. Numerous exercises are included in this chapter. Extramarks provides the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 in PDF format. Students can study these solutions directly from Extramarks’ website or mobile application or download it as needed. The problems and questions from the exercise have been meticulously solved by subjectmatter experts while adhering to all CBSE guidelines.
The NCERT Solutions for Class 7 Maths Ch 6 Ex 6.5 and all chapter exercises compiled in one location, prepared by expert teachers following NCERT (CBSE) Book guidelines, are available for free PDF download. The solutions for Class 7th Maths Chapter 6 Exercise 6.5 will assist students in reviewing the entire syllabus and earning higher marks. Students can register to receive emails with all the exercise solutions. To make studying easy and engaging, the NCERT solution is offered. If students can access the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 their preparation will become simple.
Any student in Class 7 who is thorough with all the ideas from the Mathematics textbook and knowledgeable about all the exercises can easily score the highest marks on the final exam. Students can learn the pattern of questions that may be asked in the exam from this chapter and the chapter’s weightage with the help of these NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 so that they can adequately prepare for the final exam.
Numerous exercises in this chapter include numerous questions in addition to the ones mentioned in the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5. Understanding all of the textbook concepts and completing the exercises provided next to the students is crucial for achieving the highest grade possible in the exams.
From Extramarks’ website, students can download the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 for better exam preparation. Students can also download the same with the Extramarks application installed on their phones. The best feature of the NCERT Solutions Class 7 Maths Chapter 6 Exercise 6.5 is that they can be used offline and online.
Q.1
$\begin{array}{l}\text{PQR is a triangle right angled at P. If PQ =10 cm}\\ \text{and PR=24cm,find\hspace{0.33em}QR.}\end{array}$
Ans.
100
676
QR
$\sqrt{676}$
Q.2
$\begin{array}{l}\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{triangle}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{C}.\mathrm{If}\mathrm{AB}=\; 25\mathrm{cm}\mathrm{and}\\ \mathrm{AC}=\; 7\mathrm{cm},\mathrm{find}\mathrm{BC}.\end{array}$
Ans.
\begin{array}{l}\text{By Pythagoras theorem in}\Delta \text{ABC, we get}\\ {\left(\text{AC}\right)}^{2}+{\left(BC\right)}^{2}={\left(AB\right)}^{2}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(BC\right)}^{2}={\left(AB\right)}^{2}{\left(AC\right)}^{2}\\ \text{}\text{}\text{}={24}^{2}{7}^{2}\\ \text{}\text{}\text{}=62549=576\\ \text{}\text{}\overline{)\text{BC}=\text{24}\text{\hspace{0.17em}}\text{cm}}\end{array}
Q.3
$\begin{array}{l}\mathrm{A}15\mathrm{m}\mathrm{long}\mathrm{ladder}\mathrm{reached}\mathrm{a}\mathrm{window}12\mathrm{m}\mathrm{high}\mathrm{from}\mathrm{the}\\ \mathrm{ground}\mathrm{on}\mathrm{placing}\mathrm{it}\mathrm{against}\mathrm{a}\mathrm{wall}\mathrm{at}\mathrm{a}\mathrm{distance}\u2018\mathrm{a}\u2018.\mathrm{Find}\mathrm{the}\\ \mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{foot}\mathrm{of}\mathrm{the}\mathrm{ladder}\mathrm{from}\mathrm{the}\mathrm{wall}.\end{array}$
Ans.
\begin{array}{l}\text{By Pythagoras theorem}\\ {\left(15\right)}^{2}={\left(12\right)}^{2}+{a}^{2}\\ 225144={a}^{2}\\ 81={a}^{2}\\ \overline{)a=9\text{\hspace{0.17em}}\text{cm}}\\ \text{Therefore, the distance of the foot of the ladder from the wall}\\ \text{is 9 cm}\text{.}\end{array}
Q.4
$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{can}\mathrm{be}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{right}\mathrm{triangle}?\\ \left(\mathrm{i}\right)\mathrm{\hspace{0.33em}}2.5\mathrm{cm},6.5\mathrm{cm},6\mathrm{cm}.\\ \left(\mathrm{ii}\right)\mathrm{\hspace{0.33em}}2\mathrm{cm},2\mathrm{cm},5\mathrm{cm}.\\ \left(\mathrm{iii}\right)\mathrm{\hspace{0.33em}}1.5\mathrm{cm},2\mathrm{cm},2.5\mathrm{cm}\end{array}$
Ans.
\begin{array}{l}\left(\text{i}\right)\text{2}.\text{5 cm},\text{6}.\text{5 cm},\text{6 cm}.\\ {2.5}^{2}=6.25,{6.5}^{2}=42.25{\text{and 6}}^{2}=36\\ \text{Here, 2}{\text{.5}}^{2}+{6}^{2}={6.5}^{2}\\ \text{So, the square of the length of one side is the sum of the}\\ \text{squares of the lengths of remaining two sides}\text{.}\\ \text{Hence, these are the sides of a right}\text{angled triangle}\text{.}\\ \left(\text{ii}\right)\text{2 cm},\text{2 cm},\text{5 cm}.\\ {2}^{2}=4,{2}^{2}=4{\text{and 5}}^{2}=25\\ {\text{Here, 2}}^{2}+{2}^{2}\ne {5}^{2}\\ \text{So, the square of the length of one side is not the sum of the}\\ \text{squares of the lengths of remaining two sides}\text{.}\\ \text{Hence, these are not the sides of a right}\text{angled triangle}\text{.}\\ \left(\text{iii}\right)\text{1}.\text{5 cm},\text{2cm},\text{2}.\text{5 cm}\\ {1.5}^{2}=2.25,{2}^{2}=4\text{and 2}{\text{.5}}^{2}=6.25\\ \text{Here, 1}{\text{.5}}^{2}+{2}^{2}={2.5}^{2}\\ \text{So, the square of the length of one side is the sum of the}\\ \text{squares of the lengths of remaining two sides}\text{.}\\ \text{Hence, these are the sides of a right}\text{angled triangle}\text{.}\end{array}
Q.5
$\begin{array}{l}\text{A tree is broken at a height of 5 m from the ground and its}\\ \text{top touches the ground at a distance of 12 m from the}\\ \text{Base of the tree.Find the original height of the tree.}\end{array}$
Ans.
\begin{array}{l}\text{In above figure, BC represents the unbroken part of the tree}\text{.}\\ \text{Point C represents the point where the tree broke and CA}\\ \text{represents the broken part of the tree}\text{.}\\ \text{Triangle ABC thus formed is a rightangled triangle}\text{.}\\ \text{So, applying Pythagoras theorewm to get}\\ {\text{AC}}^{2}={\text{AB}}^{2}+{\text{BC}}^{2}\\ \text{}={12}^{2}+{5}^{2}\\ \text{}=144+25\\ \text{}=169=13\text{\hspace{0.17em}}\text{cm}\\ \text{Thus,the original height of the tree is}=\text{AC}+\text{CB}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{13 cm}+\text{5 cm}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\overline{)\text{18 cm}}\end{array}
Q.6
$\begin{array}{l}\mathrm{Angles}\mathrm{Q}\mathrm{and}\mathrm{R}\mathrm{of}\mathrm{a}\mathrm{\Delta PQR}\mathrm{are}25\xba\mathrm{and}65\xba.\\ \mathrm{Write}\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{is}\mathrm{true}:\\ \left(\mathrm{i}\right){\mathrm{PQ}}^{2}+{\mathrm{QR}}^{2}={\mathrm{RP}}^{2}\\ \left(\mathrm{ii}\right){\mathrm{PQ}}^{2}+{\mathrm{RP}}^{2}={\mathrm{QR}}^{2}\\ \left(\mathrm{iii}\right){\mathrm{RP}}^{2}+{\mathrm{QR}}^{2}={\mathrm{PQ}}^{2}\end{array}$
Ans.
\begin{array}{l}\text{Since, the sum of all interior angle is 180}\xb0\text{.}\\ \text{So, 25}\xb0\text{+ 65}\xb0+\angle \text{QPR}=180\xb0\\ \angle \text{QPR}=180\xb090\xb0\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=90\xb0\\ \text{Therefore,}\Delta \text{PQR is a rigfht angled triangle at P}\text{.}\\ \text{Thus,}\overline{){\text{PQ}}^{\text{2}}+{\text{RP}}^{\text{2}}={\text{QR}}^{\text{2}}}\\ \text{So, (ii) is true}\text{.}\end{array}
Q.7
$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{perimeter}\mathrm{of}\mathrm{the}\mathrm{rectangle}\mathrm{whose}\mathrm{length}\mathrm{is}40\mathrm{cm}\\ \mathrm{and}\mathrm{a}\mathrm{diagonal}\mathrm{is}41\mathrm{cm}.\end{array}$
Ans.
\begin{array}{l}\text{In a rectangle},\text{all interior angles are of 9}0\xba\text{measure}.\\ \text{Therefore},\text{Pythagoras theorem can be applied here}.\\ \text{}\text{}{\left(\text{41}\right)}^{\text{2}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(\text{4}0\right)}^{\text{2}}+{x}^{\text{2}}\\ \text{}\text{}\text{1681}=\text{16}00\text{}+{x}^{\text{2}}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{\text{2}}=\text{1681}\text{16}00\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{81}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\text{9 cm}\\ \text{So, Perimeter}=\text{2}\left(\text{Length}+\text{Breadth}\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}\left(x+\text{4}0\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{2}\left(\text{9}+\text{4}0\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{}\overline{)\text{98 cm}}\end{array}
Q.8
$\begin{array}{l}\mathrm{The}\mathrm{diagonals}\mathrm{of}\mathrm{a}\mathrm{rhombus}\mathrm{measure}16\mathrm{cm}\mathrm{and}30\mathrm{cm}.\\ \mathrm{Find}\mathrm{its}\mathrm{perimeter}.\end{array}$
Ans.
\begin{array}{l}\text{Let ABCD be a rhombus}\left(\text{all sides are of equal length}\right)\text{and its}\\ \text{diagonals},\text{AC and BD},\text{are intersecting each other at point O}.\\ \text{Since, diagonals of a rhombus bisect each other at 9}0\xb0.\\ \text{So,}\\ \text{By applying Pythagoras theorem in}\Delta \text{AOB},\\ {\text{OA}}^{\text{2}}+{\text{OB}}^{\text{2}}={\text{AB}}^{\text{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{8}}^{\text{2}}+{\text{15}}^{\text{2}}={\text{AB}}^{\text{2}}\\ \text{\hspace{0.17em}}\text{64}+\text{225}={\text{AB}}^{\text{2}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{289}={\text{AB}}^{\text{2}}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AB}=\text{17 cm}\\ \text{Therefore},\text{the length of the side of rhombus is 17 cm}.\\ \text{Perimeter of rhombus}=\text{4}\times \text{Side of the rhombus}\\ \text{}\text{}\text{}\text{}\text{}\text{}=\text{4}\times \text{17}\\ \text{}\text{}\text{}\text{}\text{}\text{}=\text{}\overline{)\text{68 cm}}\end{array}