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

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{(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}\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{(iv) Join P to Q and R}\text{.} \begin{array}{l}\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}\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{(iv) Join P to Q and R}\text{.} \begin{array}{l}\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}\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}\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}