NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry (EX 10.2) Exercise 10.2
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Chapter 10 of Class 7 Mathematics in the NCERT textbook is Practical Geometry. Exercise 10.2 of the chapter includes the Construction of Triangles. It includes the Construction of a Triangle when the lengths of its three sides are known. The Class 7 Maths Chapter 10 Exercise 10.2 provided by the Extramarks website is accurate and welldetailed so that students can have access to authentic and credible study material. Extramarks recommends they download Class 7 Maths Chapter 10 Exercise 10.2 Solutions to understand the chapter’s concepts and resolve their queries. The concepts of the chapter can be challenging for some students. They can easily understand the complicated problems related to Chapter 10; Practical Geometry using Class 7th Math Exercise 10.2 which is available on the Extramarks website.
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NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry (EX 10.2) Exercise 10.2
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Access NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry
The NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.2 provided by Extramarks is easily accessible online and offline. As a result, students can review the solutions whenever they want. They may find the steps in NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.2 difficult at times.They need to be very careful with the steps involved in this exercise. Some portals might not provide students with accurate and credible NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.2. It is important to access these solutions from a credible source in order to make sure that the learning process is authentic. Students can rely on the Extramarks website if they want to access the NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.2. Extramarks also provides students with the NCERT solutions for all the exercises of the chapter. Students who find Class 7 Chapter 10 challenging can find the study material available on the Extramarks website helpful. Extramarks provides students with learning tools such as the NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.2 for structuring their fundamentals so that they do not struggle with the subject any further. They are provided with a flexible learning environment with Extramarks, which provides them with a number of benefits. Students can access the portal from anywhere and at any time. They can access authentic study material without seeking help elsewhere. The Extramarks website provides students with comprehensive courses which are affordable. As a result, students perform better in their studies and have better career prospects.
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.2
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Class 7 Maths 10.2 consists of four questions based on the concepts of line segment and triangle construction.
Students in Class 7 should have a solid understanding of the NCERT curriculum. These books are the first important step for students to build their fundamentals of the Mathematics Class 7 curriculum. It is essential for students to have the solutions to NCERT questions available all the time so that they do not waste time searching for them. Extramarks provides them with the NCERT Solutions Class 7 Maths Chapter 10 Exercise 10.2 so that they can have access to comprehensive study material without having to look anywhere else. Extramarks provides students with complete learning material.
Q.1
$\mathrm{Construct}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}\mathrm{of}\mathrm{side}5.5\mathrm{cm}.$
Ans.
\begin{array}{l}\text{Since, we need to construct an equilateral triangle so all sides}\\ \text{should be equal}\text{.}\\ \text{So, AB}=\text{BC}=\text{CA}=\text{5}\text{.5 cm}\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment BC of length 5}\text{.5 cm}\text{.}\end{array}
\text{(ii) Taking B as centre, draw an arc of 5}\text{.5 cm radius}\text{.}
\begin{array}{l}\text{(iii) Taking C as centre, draw an arc of 5}\text{.5 cm radius to}\\ \text{meet previous arc at point A}\text{.}\end{array} \text{}
\text{(iv) Join A}\text{}\text{to B and C}\text{.}
\text{Therefore, ABC is the required triangle}\text{.}
Q.2
$\begin{array}{l}\mathrm{Draw}\mathrm{\Delta PQR}\mathrm{with}\mathrm{PQ}=4\mathrm{cm},\mathrm{QR}=3.5\mathrm{cm}\mathrm{and}\mathrm{PR}=4\mathrm{cm}.\\ \mathrm{What}\mathrm{type}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{this}?\end{array}$
Ans.
\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment QR of length 3}\text{.5 cm}\text{.}\end{array}
\text{(ii) Taking Q as centre, draw an arc of 4 cm radius}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Taking R as centre, draw an arc of 4 cm radius to}\\ \text{meet previous arc at point P}\text{.}\end{array} \text{}
\text{(iv) Join P to Q and R}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{Therefore, PQR is the required triangle}\text{.}\\ \text{Since two sides of required triangle are equal,}\\ \text{so it is an isosceles triangle}\text{.}\end{array}
Q.3
$\begin{array}{l}\mathrm{Draw}\mathrm{\Delta PQR}\mathrm{with}\mathrm{PQ}=4\mathrm{cm},\mathrm{QR}=3.5\mathrm{cm}\mathrm{and}\mathrm{PR}=4\mathrm{cm}.\\ \mathrm{What}\mathrm{type}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{this}?\end{array}$
Ans.
\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment QR of length 3}\text{.5 cm}\text{.}\end{array}
\text{(ii) Taking Q as centre, draw an arc of 4 cm radius}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Taking R as centre, draw an arc of 4 cm radius to}\\ \text{meet previous arc at point P}\text{.}\end{array} \text{}
\text{(iv) Join P to Q and R}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{Therefore, PQR is the required triangle}\text{.}\\ \text{Since two sides of required triangle are equal,}\\ \text{so it is an isosceles triangle}\text{.}\end{array}
Q.4
$\begin{array}{l}\mathrm{Construct}\mathrm{\Delta ABC}\mathrm{such}\mathrm{that}\mathrm{AB}=2.5\mathrm{cm},\mathrm{BC}=6\mathrm{cm}\mathrm{and}\\ \mathrm{AC}=6.5\mathrm{cm}.\mathrm{Measure}\angle \mathrm{B}.\end{array}$
Ans.
\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment BC of length 6 cm}\text{.}\end{array}
\text{(ii) Taking C as centre, draw an arc of 6}\text{.5 cm radius}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{(iii) Taking B as centre, draw an arc of 2}\text{.5 cm radius to}\\ \text{meet previous arc at point A}\text{.}\end{array}
\text{(iv) Join A}\text{}\text{to B and C}\text{.} \begin{array}{l}\text{}\end{array}
\begin{array}{l}\text{Therefore, ABC is the required triangle}\text{.}\\ \angle \text{B can be measured with the help of a protractor and}\\ \text{it comes out to be 90}\xb0\text{.}\end{array}