NCERT Solutions for Class 9 Maths Chapter 7 Triangles (Ex 7.5) Exercise 7.5
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The Central Board of Secondary Education, or CBSE, is one of the most popular and highly esteemed educational boards that operate in India. It has various schools affiliated with it that are present all over India. CBSE deploys a very objective approach to its syllabus and curriculum. This means that students who want to score well in their exams must take their subjects very seriously and learn how to sift through the redundant information in each chapter and focus on the core ideas. CBSE is widely known for its vast syllabus, and students find it very daunting. Although teachers at Extramarks have shared their experience and provided students with efficient tricks and hacks that seem to have helped students greatly.
The National Council of Educational Research and Training, NCERT, is a centralized organization set up by the Indian Government in 1961. NCERT and CBSE work in coordination and everything that CBSE does is dictated by NCERT. NCERT helps the government make wellresearched decisions regarding issues related to the country’s educational development. NCERT sets out its own rules and regulations, and therefore CBSE follows all of them. NCERT also has its own publishing house and therefore publishes its own books, which adhere completely to the rules and regulations. NCERT also has its own syllabus, and the books follow the syllabus. Therefore, teachers advise students to strictly follow the NCERT textbooks because the more they refer to the books, the more they become familiar with the paper pattern. Teachers have reassured students that there is nothing for them to worry about if they incorporate little changes in their daily routines.
Mathematics is one of the most important subjects that a student in Class 9 has because mathematics echoes the most in other subjects, therefore before students choose their stream after their Class 10 boards. CBSE in their Mathematics curriculum focuses a lot on developing the capability of students to solve mathematical calculations by themselves, and therefore the sums are solved in very interesting ways. Although not all students are familiar with these techniques and therefore the teachers at Extramarks have strived tirelessly and finally releases the NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5.
NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 provide solutions for the fifth exercise of the 7th chapter of the Class 9 Mathematics syllabus. The NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5, provide solutions to every unsolved question in the NCERT textbook. These solutions are provided by highly qualified teachers with years of experience teaching CBSE kids. The Class 9 curriculum is pretty challenging, so students are advised to take it very seriously. NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 details every step that is involved in the problemsolving process. This way, students generally understand almost all the solutions. The NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5, are made with great precision, making sure that all the solutions provided in them are accurate. Teachers have shared that the solutions are highly reliable.
Class 9 Maths Chapter 7 Exercise 7.5 is a challenging exercise and therefore students ought to be careful.
The NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 strictly adheres to all the rules and regulations that were put forward by the NCERT. Since CBSE has a very distinct question paper pattern, when students use the NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 they get used to it. When students refer to NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 they gradually learn about the right way they must answer all the questions in their CBSE exam.
H2 – NCERT Solutions for Class 9 Maths Chapter 7 Triangles (Ex 7.5) Exercise 7.5
NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5, provide solutions for Chapter 7, Triangles. This is a very important chapter in the syllabus. The ideas taught in this chapter will be discussed more in the later classes, so it is crucial for students to take this chapter seriously. When students use the NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5, it provides them with a very formidable grasp of the subject.
Click the link below to access the NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5.
H2 – Access NCERT Solutions for Class 9 Mathematics Chapter 7 – Triangles
The NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 is released by Extramarks. The NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 are available on their website. They also have a mobile application which is made available. The NCERT solutions can be accessed through the mobile application as well.
Class 9 Maths Ex 7.5 discuss properties of triangles.
H2 – NCERT Solutions for Class 9 Maths Chapter 7 Triangles
The NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5, are compiled with a lot of care. These solutions are one of the best tools that can be possessed by a student. Teachers have shared the various different uses of the NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5. Students have noted that they use them while they are solving a chapter for the first time. Teachers at Extramarks have shared that instead of waiting for someone else to resolve their doubts, they can directly refer to the solutions, which saves them a lot of time. Students have used the NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5, while they are revising old chapters and problems. When students have made significant progress in their syllabus, they are generally advised to revise the ones they have solved in the past. When students refer to the NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 while revising often the doubts that remain point to the areas of doubt that a student has and give them a direction on how to proceed. NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 has also helped students who are very prone to making silly errors. When students solve their own doubts by themselves without any external help, it boosts their selfesteem. This helps them stay calm during exams while significantly reducing their chance of making silly errors. Exercise 7.5 Class 9 Maths should thus be dealt with care.
H3 – NCERT Solutions for Class 9
Extramarks have released the NCERT solutions for every class. It has all the solutions to every subject in the Class 9 curriculum. The solutions follow the exact pattern and take from reliable resources.
To access NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 and other resources available for Class 9 click on the link below.
H3 – CBSE Study Materials for Class 9
The NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5, are translated into Hindi for students whose vernacular is Hindi. CBSE has the provision and allowance for students to appear in their exams in two different languages: Hindi and English. Therefore, the NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 becomes one of the most efficient resources for them. There is a general dearth of reliable resources in Hindi, so these welltranslated solutions provide great help to these students.
H3 – CBSE Study Materials
Extramarks releases various different kinds of resources for students to access. They can access these solutions anytime and anywhere they need help. Since these PDFs are available online, they can be accessed in every possible way. There are mobile applications for Extramarks that make education even more tangible and accessible. Since these solutions are provided by trained and educated professionals, teachers find great relief knowing that students use the NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5. To assist students of all classes, Extramarks provides NCERT Solutions.
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Q.1 ABC is a triangle. Locate a point in the interior of Δ ABC which is equidistant from all the vertices of ΔABC.
Ans
Circumcentre of a triangle is always equidistant from all the vertices of that triangle.
Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together.
In ∆ABC, we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of ∆ABC.
Q.2 In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Ans
A point in the interior of a triangle which is equidistant from all the sides is in centre. In centre is a point which is obtained by the intersection of angle bisectors.
Here, in ∆ABC, we can find the in centre of this triangle by drawing the angle bisectors of the interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. Therefore, I is the point equidistant from all the sides of ∆ABC.
Q.3 In a huge park, people are concentrated at three points (see figure below):
A: where there are different slides and swings for children,
B: near which a manmade lake is situated,
C: which is near to a large parking and exit. Where should an icecream parlour be set up so that maximum number of persons can approach it?
Ans
Icecream parlour should be set up at equidistant point from three points A, B and C so that maximum number of people will approach there.
When we join three points A, B and C, we get a triangle. A point which is equidistant from three vertex is called Circumcentre.
In figure, Circumcentre O is obtained by intersection of perpendicular bisectors of sides of triangle ABC. At this point of icecream parlour, maximum number of people will approach.
Q.4 Complete the hexagonal and star shaped Rangolies [see the figure below (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
Ans
\begin{array}{l}\text{\hspace{0.17em}}\left(i\right)Area\left(\Delta AOB\right)=\frac{\sqrt{3}}{4}{\left(side\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(5\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{25\sqrt{3}}{4}{\text{cm}}^{\text{2}}\end{array}
\begin{array}{l}\text{Area of hexagonal \u2013 shaped}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{rangoli}=6\times \text{\hspace{0.17em}}\text{\hspace{0.17em}}Area\left(\Delta AOB\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6\times \text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{25\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{75\sqrt{3}}{2}{\text{cm}}^{\text{2}}\\ Area\text{}of\text{equilateral triangle having its side as 1 cm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(side\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(1\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ Number\text{of equilateral triangles of 1 cm side that can be filled}\\ \text{in hexagonalshaped rangoli}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\left(\frac{75\sqrt{3}}{2}\right)}{\left(\frac{\sqrt{3}}{4}\right)}=150\end{array}
\begin{array}{l}\end{array} $\begin{array}{l}\text{Star}\text{shaped rangoli has 12 equilateral triangles of side}\text{5 cm in it}.\\ \text{Area of equilateral triangle}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(side\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(5\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{25\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ \text{Area of starshaped rangoli}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=12\times \frac{25\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=75\sqrt{5}{\text{cm}}^{\text{2}}\\ \text{Area}of\text{equilateral triangle having its side as 1 cm}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(side\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\left(1\right)}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\sqrt{3}}{4}{\text{cm}}^{\text{2}}\\ \text{Number of equilateral triangles of 1 cm side that can be filled}\\ \text{in hexagonalshaped rangoli}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\left(75\sqrt{3}{\text{cm}}^{\text{2}}\right)}{\left(\frac{\sqrt{3}}{4}\right)}=300\\ \text{Therefore},\text{starshaped rangoli has more equilateral}\\ \text{triangles in it}.\end{array}$
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1. How can a student use the NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5?
The NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 can be used dynamically, it can be used while solving a chapter for the first time. It can be referred to while students are revising old chapters that they had solved long back. Students can also use them just before their exams for quick revision.
2. Are the NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 correct?
The solutions in NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5 are absolutely correct. Before the release, the solutions are properly reviewed and checked to make sure that there are no inaccuracies in them.
Click on the link below to access the Ex 7.5 Class 9 Maths NCERT Solutions.