Question 12 Marks
Illustrate, if possible, one of the following with a rough diagram: An open curve made up entirely of line segments.
Answer
View full question & answer→Rough diagram of an open curve made up entirely of line segments. 
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Question 22 Marks
Lines $p, q$ are coplanar. So are the lines $p, r.$ Can we conclude that the lines $p, q, r$ are coplanar$?$
Answer
View full question & answer→No, $p, q$ and $r$ are not necessarily coplanar. Example: If we take $p$ as intersecting line of two consecutive walls of a room, $q$ as a line on the first wall and $r$ on the second wall whose (both walls) intersection is line $p.$ Thus we can see that $p, q$ and $r$ are not coplanar.
Question 32 Marks
From write concurrent lines and their points of concurrence.
Answer
View full question & answer→From the given figure, we have: Concurrent lines can be defined as three or more lines which share the same meeting point. Clearly lines, $n, q,$ and $l$ are concurrent with A as the point of concurrence. Lines, $m, q$ and $p$ are concurrent with $B$ as the point of concurrence.
Question 42 Marks
Using a ruler, check whether the following points given in collinear or not:
$i.\ D, A$ and $C$
$ii.\ A, B$ and $C$
$iii.\ A, B$ and $E$
$iv.\ B, C$ and $E$
$i.\ D, A$ and $C$
$ii.\ A, B$ and $C$
$iii.\ A, B$ and $E$
$iv.\ B, C$ and $E$
Answer
View full question & answer→$i.\ D, A$ and $C$ are collinear points.
$ii.\ A, B$ and $C$ are non$-$collinear points.
$iii.\ A, B$ and $E$ are collinear points.
$iv.\ B, C$ and $E$ are non$-$collinear points.
$ii.\ A, B$ and $C$ are non$-$collinear points.
$iii.\ A, B$ and $E$ are collinear points.
$iv.\ B, C$ and $E$ are non$-$collinear points.
Question 52 Marks
Draw a polygon and shade its interior. Also draw its diagonals.
Answer
View full question & answer→$ABCD$ is a polygon and $AC$ and $BD$ are its two diagonals.
Question 62 Marks
Is it ever possible for exactly one line to pass through three points?
Answer
View full question & answer→ Yes, it is possible if three points lie on a straight line,
Question 72 Marks
In the following cases, state whether you can draw line segments on the given surfaces: The curved surface of a cone.
Answer
View full question & answer→Yes, we can draw line segments on the curved surface of the cone. Every line segment joining the vertex of a cone and any point on the circumference of the cone will be a line segment.
Question 82 Marks
Illustrate, if possible, one of the following with a rough diagram: A closed curve that is not a polygon.
Answer
View full question & answer→A circle is a simple closed curve but not a polygon. A polygon has line segments, but a circle has only curve. 
Question 92 Marks
Mark the following points on a sheet of a paper. Tell how many line segments can be obtained in case: Two points $A, B.$
Answer
View full question & answer→If there are n points in a plane and no three of them are collinear,
the number of line segments obtained by joining these points is equal to $\frac{\text{n(n-1)}}{2}.$
On applying the above formula, we get:
For two points $A$ and $B:$ Number of line segments $=\frac{2(2-1)}{2}=1.$
the number of line segments obtained by joining these points is equal to $\frac{\text{n(n-1)}}{2}.$
On applying the above formula, we get:
For two points $A$ and $B:$ Number of line segments $=\frac{2(2-1)}{2}=1.$
Question 102 Marks
Give three examples of line segments from your environment.
Answer
View full question & answer→Examples of line segments in our home are:
$i.\ $Grout lines in the tile floors.
$ii.\ $Groves where wooden flooring connects.
$iii.\ $Metal outline of a sliding glass door.
$i.\ $Grout lines in the tile floors.
$ii.\ $Groves where wooden flooring connects.
$iii.\ $Metal outline of a sliding glass door.
Question 112 Marks
Mark the following points on a sheet of a paper. Tell how many line segments can be obtained in case: Three non-collinear points $A, B, C.$
Answer
View full question & answer→If there are $n$ points in a plane and no three of them are collinear, the number of line segments obtained by joining these points is equal to $\frac{\text{n(n-1)}}{2}.$ On applying the above formula, we get: For three non-collinear points $A, B$ and $C:$ Number of line segments $=\frac{3(3-1)}{2}=\frac{3\times2}{2}=3.$
Question 122 Marks
Coplanar points are the points that are in the same plane. Thus, Can $150$ points be coplanar?
Answer
View full question & answer→Yes, A group of points that lie in the same plane are called co planar points. Thus, it is possible that $150$ points can be co-planar.
Question 132 Marks
In the following cases, state whether you can draw line segments on the given surfaces: The curved surface of a cylinder.
Answer
View full question & answer→Yes, we can draw line segments on the curved surface of a cylinder. Every line segment parallel to the axis of a cylinder on the curved surface will be a line segment.
Question 142 Marks
How many lines may pass through one given point, two given points, any three collinear points?
Answer
View full question & answer→ Lines passing through one point - unlimited, 
Lines passing through two points - one, 
Lines passing through two points - one, 
Question 152 Marks
Give three examples of: Intersecting lines.
Answer
View full question & answer→Intersecting lines: 
Question 162 Marks
Mark three non-collinear points $A, B$ and $C$ in your note book. Draw lines through these points taking two at a time. Name these lines. How many such different lines can be drawn?
Answer
View full question & answer→These are three non-collinear points $A, B, C,$
Three lines can be drawn through these points. These three lines are $AB, BC$ and $AC.$
Three lines can be drawn through these points. These three lines are $AB, BC$ and $AC.$
Question 172 Marks
Mark the following points on a sheet of a paper. Tell how many line segments can be obtained in case:
Four points such that no three of them boelong to the same line.
Four points such that no three of them boelong to the same line.
Answer
View full question & answer→ If there are n points in a plane and no three of them are collinear, the number of line segments obtained by joining these points is equal to $\frac{\text{n(n-1)}}{2}.$
On applying the above formula, we get:
For four points such that no three of them belong to the same line: Number of line segments $=\frac{4(4-1)}{2}=\frac{4\times3}{2}=3.$
On applying the above formula, we get:
For four points such that no three of them belong to the same line: Number of line segments $=\frac{4(4-1)}{2}=\frac{4\times3}{2}=3.$
Question 182 Marks
Mark any two points $P$ and $Q$ in your note book and draw a line passing through the points. How many liens can you draw passing through both the points?
Answer
View full question & answer→We have two points $P$ and $Q$ and we draw a line passing through these two points,
Only one line can be drawn passing through these two points.
Only one line can be drawn passing through these two points.
Question 192 Marks
Give three examples of:
Parallel lines from your environment.
Parallel lines from your environment.
Answer
View full question & answer→Parallel lines from your environment: 
Question 202 Marks
Coplanar points are the points that are in the same plane. Thus, Can $3$ points be non-coplanar$?$
Answer
View full question & answer→No, Three points will be coplanar because we can have a plane that can contain $3$ points on it. Thus, it is not possible that $3$ points will be non-coplanar.
Question 212 Marks
In the following cases, state whether you can draw line segments on the given surfaces: The base of a cone.
Answer
View full question & answer→Yes, we can draw line segments on the base of the cone. Yes, we can draw line segments on the curved surface area of the cone. Every line segment joining the vertex of a cone and any point on the circumference of the cone will form a line segment.
Question 222 Marks
Count the number of line segments. 
Answer
View full question & answer→Count the number of line segments. 
Question 232 Marks
Mark the following points on a sheet of a paper. Tell how many line segments can be obtained in case: Any five points so that no three of them are collinear.
Answer
View full question & answer→If there are $n$ points in a plane and no three of them are collinear, the number of line segments obtained by joining these points is equal to $\frac{\text{n(n-1)}}{2}.$ On applying the above formula, we get: For any five points so that no three of them are collinear: Number of line segments $=\frac{5(5-1)}{2}=\frac{5\times4}{2}=10.$
Question 242 Marks
Explain why it is not possible for a line to have a mid-point.
Answer
View full question & answer→The length of the line is infinite. Thus, it is not possible to find its mid-point. On the other hand, we can find, we can find the mid-point of a line segment.