NCERT Solutions for Class 7 Maths Chapter 10 – Algebraic Expressions Ex 10.1

Q.1 Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.

Ans.

\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line AB}\text{. Take a point P on it}\text{. Take a point C outside}\\ \text{this line}\text{. Join C to P}\text{.}\end{array}

$$\begin{gathered} \text{(ii) Taking P as centre and with a convenient radius, draw arc intersecting line AB} \\ \text{at point D and PC at point E}\text{.} \end{gathered}$$

\text{(iii) Taking C as centre and with the same radius as before, draw an arc FG intersecting PC at H}\text{.}

\begin{array}{l}\text{(iv) Adjust the compasses up to the length of DE}\text{. Without}\\ \text{changing the opening of the compasses and taking H as the}\\ \text{cenre, draw an arc to intersect the previously draw arc}\\ \text{FG at po}\mathrm{int}\text{I}\text{.}\end{array}

\text{(v) Join the points C and I to draw a line}l.

\text{This is the required line which is parallel to AB}\text{.}

Q.2

$\begin{array}{l}\mathrm{Draw}\mathrm{a}\mathrm{line}\mathrm{l}.\mathrm{Draw}\mathrm{a}\mathrm{perpendicular}\mathrm{to}\mathrm{l}\mathrm{at}\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{l}.\\ \mathrm{On}\mathrm{this}\mathrm{perpendicular}\mathrm{choose}\mathrm{a}\mathrm{point}\mathrm{X},4\mathrm{cm}\mathrm{away}\mathrm{from}\mathrm{l}.\\ \mathrm{Through}\mathrm{X},\mathrm{draw}\mathrm{a}\mathrm{line}\mathrm{m}\mathrm{parallel}\mathrm{to}\mathrm{l}.\end{array}$

Ans.

\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line}l\text{and take a point P on the line}l\text{. Then draw a}\\ \text{perpendicular at point P}\text{.}\end{array}

\begin{array}{l}\text{(ii) Adjusting the compasses up to the length of 4 cm, draw an}\\ \text{arc to intersect this perpendicular at point X}\text{. Choose any point}\\ \text{Y on the line}l\text{. Join X to Y}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking Y as centre and with a convenient radius, draw}\\ \text{an arc intersecting}l\text{at A and XY at B}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iv) Taking X as centre and with same radius as before, draw}\\ \text{an arc CD cutting XY at E}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(v) Adjust the compasses upto the length of AB}\text{. Wothout}\\ \text{changing the opening of the compasses and taking E as the}\\ \text{centre, draw an arc intersecting the previous arc CD t F}\text{.}\end{array}

\text{(vi) Join the point X and F to draw a line}m\text{.} \text{Line}m\text{is the required line parallel to}l\text{.}

Q.3

$\begin{array}{l}\mathrm{Let}\mathrm{l}\mathrm{be}\mathrm{a}\mathrm{line}\mathrm{and}\mathrm{P}\mathrm{be}\mathrm{a}\mathrm{point}\mathrm{not}\mathrm{on}\mathrm{l}.\mathrm{Through}\mathrm{P},\mathrm{draw}\\ \mathrm{a}\mathrm{line}\mathrm{m}\mathrm{parallel}\mathrm{to}\mathrm{l}.\mathrm{Now}\mathrm{join}\mathrm{P}\mathrm{to}\mathrm{any}\mathrm{point}\mathrm{Q}\mathrm{on}\mathrm{l}.\mathrm{Choose}\\ \mathrm{any}\mathrm{other}\mathrm{point}\mathrm{R}\mathrm{on}\mathrm{m}.\mathrm{Through}\mathrm{R},\mathrm{draw}\mathrm{a}\mathrm{line}\mathrm{parallel}\mathrm{to}\mathrm{PQ}.\\ \mathrm{Let}\mathrm{this}\mathrm{meet}\mathrm{l}\mathrm{at}\mathrm{S}.\mathrm{What}\mathrm{shape}\mathrm{do}\mathrm{the}\mathrm{two}\mathrm{sets}\mathrm{of}\mathrm{parallel}\\ \mathrm{lines}\mathrm{enclose}?\end{array}$

Ans.

\begin{array}{l}\text{The steps of construction are as follows:}\\ \text{(i) Draw a line l and take a point A}\text{}\text{on it}\text{. Take a point P not}\\ \text{on}l\text{and join A to P}\text{.}\end{array}

\begin{array}{l}\text{(ii) Taking A as centre and with a convenient radius, draw an}\\ \text{arc cutting}l\text{at B and AP at C}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iii) Taking P as centre and with the same radius as before,}\\ \text{draw an arc DE to intersect AP at F}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{(iv) Adust the compasses up to the length of BC}\text{. Without}\\ \text{changing the opening of compasses and taking F as the}\\ \text{centre, draw an arc to intersect the previously draw arc}\\ \text{DE at point G}\text{.}\end{array}

\text{(v) Join P to G to draw a line m}\text{. Line}m\text{will be parallel to line}l\text{.} \begin{array}{l}\end{array}

\begin{array}{l}\text{(vi) Join P to any point Q on line}\text{. Choose another point R on}\\ \text{line}m\text{. Similarly, a line can be drawn through point R and}\\ \text{parallel to PQ}\text{.}\end{array} \begin{array}{l}\end{array}

\begin{array}{l}\text{Let it meet line}l\text{at point S}\text{. In quadrilateral PQRS, opposite}\\ \text{lines are parallel to each other}\text{. PQ}\parallel \text{RS and PR}\parallel \text{QS}\text{.}\\ \text{Thus PQRS is a parallelogram}\text{.}\end{array}