# NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry (EX 10.2) Exercise 10.2

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 well-detailed 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

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### 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.

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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.

$\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment BC of length 6 cm}\text{.}\end{array}$

$\text{(ii) Taking C as centre, draw an arc of 6}\text{.5 cm radius}\text{.}$ $\begin{array}{l}\text{}\end{array}$

$\begin{array}{l}\text{(iii) Taking B as centre, draw an arc of 2}\text{.5 cm radius to}\\ \text{meet previous arc at point A}\text{.}\end{array}$

$\text{(iv) Join A}\text{​}\text{to B and C}\text{.}$ $\begin{array}{l}\text{}\end{array}$

$\begin{array}{l}\text{Therefore, ABC is the required triangle}\text{.}\\ \angle \text{B can be measured with the help of a protractor and}\\ \text{it comes out to be 90}°\text{.}\end{array}$