NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry (EX 10.5) Exercise 10.5

Q.1

$\mathrm{Construct}\mathrm{the}\mathrm{right}\mathrm{angled}\mathrm{\Delta PQR},\mathrm{where}\mathrm{m}\angle \mathrm{Q}=90\mathrm{°},\mathrm{QR}=8\mathrm{cm}\mathrm{and}\mathrm{PR}=10\mathrm{cm}.$

Ans.

\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment QR of length 8 cm}\text{.}\end{array}

\text{(ii) At point Q, draw a ray QX making 90}°\text{with QR}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking R as centre, draw an arc of 10 cm radius to}\\ \text{intersect ray QX at point P}\text{.}\end{array}

\text{(iv) Join P to R}\text{.}

\text{Thus,}\Delta \text{PQR is the required right-angled triangle}\text{.}

Q.2

Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.

Ans.

\begin{array}{l}\text{A right-angled triangle ABC with hypotenuse 6 cm and one}\\ \text{of the legs as 4 cm has to be constructed}\text{.}\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment BC of length 4 cm}\text{.}\end{array}

\text{(ii) At point B, draw a ray BX making 90}°\text{with BC}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking C as centre, draw an arc of 6 cm radius to}\\ \text{intersect ray BX at point A}\text{.}\end{array}

\text{(iv) Join A to C}\text{.}

\text{Thus,}\Delta \text{ABC is the required right-angled triangle}\text{.}

Q.3

$$\begin{gathered} \mathrm{Construct}\mathrm{an}\mathrm{isosceles}\mathrm{right}–\mathrm{angled}\mathrm{triangle}\mathrm{ABC}, \\ \mathrm{where}\mathrm{m}\angle \mathrm{ACB}=90\mathrm{°}\mathrm{and}\mathrm{AC}=6\mathrm{cm}. \end{gathered}$$

Ans.

\begin{array}{l}\text{Since, in a isosceles triangle, the length of any two side}\\ \text{are equal}\text{. So, AB}=\text{BC}=\text{6cm}\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment AC of length 6 cm}\text{.}\end{array}

\text{(ii) At point C, draw a ray CX making 90}°\text{with AC}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking C as centre, draw an arc of 6 cm radius to}\\ \text{intersect ray CX at point B}\text{.}\end{array}

\text{(iv) Join A to B}\text{.}

\text{Thus,}\Delta \text{ABC is the required triangle}\text{.}