NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry (EX 10.4) Exercise 10.4

Q.1

$\mathrm{Construct}\mathrm{\Delta ABC},\mathrm{given}\mathrm{m}\angle \mathrm{A}=60\mathrm{°},\mathrm{m}\angle \mathrm{B}=30\mathrm{°}\mathrm{and}\mathrm{AB}=5.8\mathrm{cm}.$

Ans.

\text{(i) Draw a line segment AB of length 5}\text{.8 cm}\text{.}

\text{(ii) At point A, draw a ray AX making 60}°\text{angle with AB}\text{.}

\text{(iii) At point B, draw a ray AX making 30}°\text{angle with AB}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iv) Point C has to lie on both the rays, AX and BY}\text{. Therefore,}\\ \text{C is the point of intersection of these two rays}\text{.}\end{array}

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

Q.2

$\begin{array}{l}\mathrm{Construct}\mathrm{\Delta PQR}\mathrm{if}\mathrm{PQ}=5\mathrm{cm},\mathrm{m}\angle \mathrm{PQR}=105\mathrm{°}\mathrm{and}\mathrm{m}\angle \mathrm{QRP}=40\mathrm{°}.\\ (\mathrm{Hint}:\mathrm{Recallangle}–\mathrm{sum}\mathrm{property}\mathrm{of}\mathrm{a}\mathrm{triangle}).\end{array}$

Ans.

\begin{array}{l}\text{To construct triangle PQR, we need to find}\angle \text{RPQ}\text{.}\\ \text{Using angle sum property of triangles to get}\\ \angle \text{PQR}+\angle \text{PRQ}+\angle \text{RPQ}=180°\\ 105°+40°+\angle \text{RPQ}=180°\\ \angle \text{RPQ}=180°-145°\\ =35°\\ \text{The steps of construction are as follows:}\\ \text{(i) Draw a line segment PQ of length 5 cm}\text{.}\end{array}

\text{(ii) At P, draw a ray PX making an angle of 35}°\text{with PQ}\text{.}

\text{(iii) At Q, draw a ray QY making an angle of 105}°\text{with PQ}\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iv) Point R has to lie on both the ray, PX and QY}\text{. Therefore,}\\ \text{R is the point of intersectin of these two rays}\text{.}\end{array}

\text{Thus, PQR is the required triangle}\text{.}

Q.3

$\begin{array}{l}\mathrm{Examine}\mathrm{whether}\mathrm{you}\mathrm{can}\mathrm{construct}\mathrm{\Delta DEF}\mathrm{such}\mathrm{that}\\ \mathrm{EF}=7.2\mathrm{cm},\mathrm{m}\angle \mathrm{E}=110\mathrm{°}\mathrm{and}\mathrm{m}\angle \mathrm{F}=80\mathrm{°}.\mathrm{Justify}\mathrm{your}\mathrm{answer}.\end{array}$

Ans.

\begin{array}{l}\text{We are given,}\\ \angle \text{E}=\text{110}°\text{and}\angle \text{F}=\text{80}°\\ \text{Using angle sum property to get}\\ \angle \text{D+}\angle \text{E+}\angle \text{F=180}°\\ \angle D+110°+80°=180°\\ \angle D+190°=180°\\ \text{Since}\angle \text{D}\text{can not be zero and it does not satisfy angle sum}\\ \text{property}\text{.}\\ \text{Thus, we can not construct}\Delta \text{DEF}\text{.}\end{array}

\begin{array}{l}\text{Also, it can be observed that point D should lie on both rays,}\\ \text{EX and FY, for the constructing the required triangle}\text{.}\\ \text{However, both rays are not intersecting each other}\text{.}\\ \text{Therefore, the required triangle can not be formed}\text{.}\end{array}