RD Sharma 2013 Solutions for Class 6 Math Chapter 18 Basic Geometrical Tools are provided here with simple step-by-step explanations. These solutions for Basic Geometrical Tools are extremely popular among class 6 students for Math Basic Geometrical Tools Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2013 Book of class 6 Math Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s RD Sharma 2013 Solutions. All RD Sharma 2013 Solutions for class 6 Math are prepared by experts and are 100% accurate.

#### Page No 18.3:

#### Question 1:

Construct the following angles using set-squares:

(i) 45°

(ii) 90°

(iii) 60°

(iv) 105°

(v) 75°

(vi) 150°

#### Answer:

(i) Place 45° set-square.

Draw two rays AB and AC along the edges from the vertex of 45° angle of the set-square.

The angle so formed is a 45° angle.

∠BAC = 45°

(ii) Place 45° set-square as shown in the figure.

Draw two rays BC and BA along the edge from the vertex of 90° angle.

The angle so formed is 90° angle.

∠ABC = 90°

(iii) Place 30° set-square as shown in the figure.

Draw the rays BA and BC along the edges from the vertex of 60°.

The angle so formed is 60°.

∠ABC = 60°

(iv) Place 30° set-square and make an angle of 60° by drawing the rays BA and BC as shown in the figure.

Now, place the vertex of 45° of the set-square on the ray BA as shown in the figure and draw the ray BD.

The angle so formed is 105°

∴ ∠DBC = 105°

(v) Place 45° set-square and make an angle of 45° by drawing the rays BD and BC as shown in the figure.

Now, place the vertex of 30° of the set-square on the ray BD as shown in the figure and draw the ray BA.

The angle so formed is 75°.

∴ ∠ABC = 75°

(Line BD is hidden)

(vi) Place the vertex of 45° of the set-square and make angle of 90° by drawing the rays BD and BC as shown in the figure.

Now, place the vertex of 30° of the set square on the ray BD as shown in the figure and draw the ray BA.

The angle so formed is 150°.

∴ ∠ABC = 150°

#### Page No 18.3:

#### Question 2:

Given a line *BC* and a point *A* on it, construct a ray *AD* using set squares so that ∠*DAC** *is

(i) 30°

(ii)60°

#### Answer:

(i) Draw a line BC and take a point A on it. Place 30° set-square on the line BC such that its vertex of 30° angle lies on point A and one edge coincides with the ray AC as shown in the figure.

Draw the ray AD.

Thus, $\angle $DAC is the required angle of 30°.

(ii) Draw a line BC and take a point A on it. Place 30° set-square on the line BC such that its vertex of 30° angle lies on point A and one edge coincides with the ray AB as shown in the figure.

Draw the ray AD.

$\angle $DAB = 30°

We know that the angles an one side of the straight line will always add to 180°.

∴ $\angle $DAB + $\angle $DAC = 180°

30° + $\angle $DAC = 180°

$\angle $DAC = 180° − 30°

∴ $\angle $DAC = 150°

#### Page No 18.5:

#### Question 1:

Mark two points, *A* and *B* on a piece of paper and join them. Measure this length. For each of the following draw a line segment *CD* that is:

(i) equal to the segment* AB*

(ii) twice *AB*

(iii) three times *AB*

(iv) half *AB*

(v) collinear with *AB* and is equal to it.

#### Answer:

Mark two points, A and B on a piece of paper and join them as follows:

To measure the length of AB, place the ruler with its edge along AB, such that the zero mark of the cm side of the ruler coincides with point A, as shown in the figure.

Now, read the mark on the ruler, which corresponds to the point B.

The reading on the ruler at point B is the length of the line segment AB.

Here, AB = 5.6 cm

(i) To draw the line segment CD equal to AB, take a divider and open it, such that the end-point of one of its arms is at A and the end-point of the second arm is at B, as shown in the figure.

Then, lift the divider and without disturbing its opening, place the end-points of both hands on the paper, where we have to draw CD.

(ii) To draw the line segment twice AB, draw a line *l* and take a point C on it. Now, take a divider and open it such that the end points of both its arms are at A and B.

Then, lift the divider and without disturbing its opening, place one end-point at C and the other end-point on the line *l,* as shown in the figure. Lift the divider and place one end-point at E and the other end-point on the line *l, *opposite C. Name this point D.

(iii) To draw the line segment three times A, we draw a line *l* and take a point C on it. Now take a divider and open it, such that the end-points of both its arms are at A and B.

Then, we lift the divider and place one end-point at C and the other end-point on the line *l,* as shown in the figure.

Let this point be E.

Again, lift the divider and place one end-pint at E and the other end-point on the line *l,* opposite to C.

Let this point be F.

Again, lift the divider and place one end-point at F and the other end-point on the line *l,* opposite to C. Name this point D.

(iv) To draw the line segment that is half AB, we draw a line *l* and take a point C on it.

Now, using a ruler, we measure the line segment AB and here, AB = 5.6 cm

$\mathrm{Half}\mathrm{of}\mathrm{AB}=\frac{5.6}{2}=2.8\mathrm{cm}$

Now, we take a divider and open it so much that its end of one hand is at 0 and end of the another hand is at 2.8 cm

Then, we lift the divider and place one end at C and the other end on the line *l *at point D.

(v) To draw a line segment CD collinear with AB and equal to AB, we take a ruler along AB and draw the line extended to AB, as shown in the figure.

We take a divider and open it such that the end-points of both its arms are at A and B.

Then, we lift the divider and place the end-points of both its hands on the extended line of AB and mark them as C and D.

#### Page No 18.5:

#### Question 2:

The end-point *P* of a line-segment *PQ* is against 4 cm mark and the end-point *Q* is against the mark indicating 14.8 cm on a ruler. What is the length of the segment *PQ*?

#### Answer:

Extend the line segment QP towards point zero of the ruler and take a point O on the extended line QP corresponding to point zero on the ruler.

From the figure, we can say:

OP = 4 cm and OQ = 14.8 cm

Now,

PQ = OQ − OP

= (14.8 − 4) cm

= 10.8 cm

#### Page No 18.5:

#### Question 3:

Draw a line segment *CD*. Produce it to *CE* such that *CE*=3*CD*.

#### Answer:

We draw a line *l* and take two points C and D on it.

Take a divider and open it such that its end of both arms are at C and D. Then, we lift the divider and place its one end at D and other end on the line *l* opposite to C as shown in the figure.

Let this point be A.

Lift the divider again and place its one end at A and other end on the line *l* opposite to C.

Name this point as E.

Here

CD = DE = AE

∴ CE = CD + DE + AE

= CD + CD + CD (As, CD ± DE = AE)

or, CE = 3CD

#### Page No 18.5:

#### Question 4:

If AB = 7.5 cm and *CD* = 2.5 cm, construct a segment whose length is equal to

(i) *AB* − *CD*

(ii) 2*AB*

(iii) 3*CD*

(iv) *AB* + *CD*

(v) 2*AB* + 3*CD*

#### Answer:

Given:

AB = 7.5 cm and CD = 2.5 cm

Draw AB and CD

(i) Draw a line *l* and take a point E on it.

Now, take a divider and open it such that the ends of both the arms are at A and B. Then, we lift the divider and place its one end at E and other end (F) on the line *l* as shown in the figure. Now, reset the divider in such a way that the end of its one hand is at C and the end of other hand is at D.

Then, we lift the divider and place its one end at E and other end (G) on the line *l* as shown in the figure. FG is required line segment, whose length is equal to (AB − CD).

(ii) Draw a line *l* and take a point E on it. Now take a divider and open it such that the ends of both its arms are at A and B.

Then, we lift the divider and place its one end at E and other end (say F) on the line *l* as shown in the figure.

Again, lift the divider and place its one end at F and other end on the line *l,* opposite to E.

Let this point be G.

EG is required line segment, whose length is equal to 2AB.

(iii) We draw a line *l* and take a point E on it. Now, take a divider and open it such that the ends of both its arms are at C and D.

Then, we lift the divider and place its one end at E and other end (F) on the *l,* as shown in the figure.

Again, lift the divider end (G) on the line l opposite to C.

Again, lift the divider and place its one end at G and another end (H) on the line *l,* opposite to E.

EH is required line segment, whose length is equal to 3CD.

(iv) We draw a line *l* and take a point E on it. Now, take a divider and open it such that the ends of both its arms are at A and B. Then, we lift the divider end (F) on the line *l,* as shown in the figure.

Now, reset the divider in such a way that the end of its one hand is at C and the end of other hand is at D.

Then, we lift the divider and place its one end at F and another end (G) on the line *l* opposite to E, as shown in the figure.

EG is required line segment, whose length is equal to (AB + CD).

(v) Draw a line *l* and take point E on it. Now, take a divider and open it such that the ends of both its arms are at A and B. Then, we lift the divider and place its one end at E and other end (say F) on the line *l,* as shown in the figure.

Again, lift the divider and place its one end at F and another end (G) on the line *l,* opposite to E.

Now, reset the divider in such a way that the ends of its one hand is at C and the end of other hand is at D.

Then, we lift the divider and place its one end at G and another end (say H) on the line *l,* opposite to E as shown in the figure.

Again, lift the divider and place its one end at H and other end (say I) on the line *l,* opposite to E as shown in the figure.

Again, lift the divider and place its one end at I and another end (say J) on the line *l,* opposite to E as shown in the figure.

EG is required line segment, whose length is equal to (2AB + 3CD).

#### Page No 18.5:

#### Question 5:

Fill in the blanks:

(i) A part of a line with two end-points is called is a........

(ii) segment *AB *is .......segment *BA*.

(iii) The length of a line segment is the ........distance between two end points.

(iv) Two segments are concurrent only if they have.......

(v) Two segments of the same length are said to be.......

#### Answer:

(i) line segment

Explanation: A line segment is a part of a line that is bounded by two distinct end points.

Example: Sides of a triangle or any polygon

A and B are definite fixed points.

∴ AB is a line segment.

(ii) equal to

Explanation: While naming a line segment, the order may not be same but the length will be equal.

If AB = 4 cm, then, BA = 4 cm

(iii) shortest

Explanation: We measure the length between two points as the length of the line segment between the two points. The length between two points is the straight line, which is the shortest distance between the two end points.

(iv) equal length

(v) congruent

Explanation: Line segments are congruent if they have the same length.

If AB = 4 cm and CD = 4 cm

Then, we say segment AB is congruent to segment CD.

#### Page No 18.5:

#### Question 6:

Match the following statements:

Column A |
Column B |
||

i | Line segment has | a | at a point |

ii | Two segments may intersect | b | if they have equal lengths |

iii | Two segments are congruent | c | two end-point |

iv | Line segment is | d | portion of a line |

#### Answer:

Column A |
Column B |
||

i | Line segment has | c | two end-point |

ii | Two segments may intersect | a | at a point |

iii | Two segments are congruent | b | if they have equal lengths |

iv | Line segment is | d | portion of a line |

Explanation:

(i) A line segment is a part of a line that is bounded by two distinct end points.

(ii) Two line segments will either not intersect at all or intersect at one point. It can never intersect at more than one point.

(iii) Line segments are congruent if they have the same lengths.

If AB = 6 cm and CD = 6 cm

Then, AB and CD are congruent.

(iv) A line segment is a part of a line that is bounded by two distinct end points.

Here, AB is a line and CD is a line segment

So, segment CD is a portion of a line AB.

#### Page No 18.5:

#### Question 7:

Tell which of the following statements are true (T) and which are false (F).

(i) The intersection of two segments may be a segment.

(ii) Two segments may intersect at a point which is not any end-point of either segments containing it.

(iii) Every ray is a segment.

(iv) Every segment is a ray.

#### Answer:

(i) False

Explanation: Two line segments can intersect maximum at one point and one point can not make a line segment.

(ii) True

Explanation: If two line segments intersect, then point of intersection will not be any of the end points.

(iii) True

Explanation: A portion of line that starts at a point and has no end point is called a ray, whereas aline segment has both its end points fixed. So every ray is a segment.

(iv) False

Both ends points of a line segment are fixed but ray has only one end fixed. Thus, a segment can never be a ray.

#### Page No 18.5:

#### Question 8:

What is the difference between line, a line segment and a ray?

#### Answer:

A line can be drawn to infinity in both the directions. AB is a line

A line segment has both ends fixed. EF is a line segment.

A ray has one end fixed and another end can be drawn to infinity. CD is a ray.

#### Page No 18.5:

#### Question 9:

How many rays are represented in Fig. 18.18? Name them.

#### Answer:

We know that a ray has fixed starting point and it can be drawn to infinity. If we take O as starting point, we will have a ray in every given direction.

So, our rays are, $\overrightarrow{\mathrm{OA}},\overrightarrow{\mathrm{OB}},\overrightarrow{\mathrm{OC}},\overrightarrow{\mathrm{OD}},\overrightarrow{\mathrm{OE}},\overrightarrow{\mathrm{OF}},\overrightarrow{\mathrm{OG}},\overrightarrow{\mathrm{OH}}$.

Thus, the number of rays in the figure is 8.

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