RS Aggarwal Solutions for Class 6 Chapter 18 Circles

 

Exercise 18.1

 

Question 1. Construct the following angles using set-squares:
(i) 45°
(ii) 90°
(iii) 60°
(iv) 105°
(v) 75°
(vi) 150°
Answer:
(i) 45°
Position the 45° set-square with the vertex of the 45° angle at point B. Draw the rays AB and AC using the edges from the vertex of the 45° angle on the set-square. The resulting angle measures 45°.
\( \angle BAC = 45° \)
(ii) 90°
Position the 90° set-square as shown in the figure. Draw the rays BC and BA using the edges from the vertex of the 90° angle. The resulting angle measures 90°.
\( \angle ABC = 90° \)
(iii) 60°
Position the 30° set-square as shown in the figure. Draw the rays BA and BC using the edges from the vertex of the 60° angle. The resulting angle measures 60°.
\( \angle ABC = 60° \)
(iv) 105°
Position the 30° set-square and create a 60° angle by drawing the rays BA and BC as shown in the figure. Next, position the vertex of the 45° set-square on the ray BA as shown in the figure and draw the ray BD. The resulting angle measures 105°.
\( \angle DBC = 105° \)
(v) 75°
Position the 45° set-square and create a 45° angle by drawing the rays BD and BC as shown in the figure. Next, position the vertex of the 30° set-square on the ray BD as shown in the figure and draw the ray BA. The resulting angle measures 75°.
\( \angle ABC = 75° \)
(vi) 150°
Position the 45° set-square and create a 90° angle by drawing the rays BD and BC as shown in the figure. Next, position the vertex of the 30° set-square on the ray BS as shown in the figure and draw the ray BA. The resulting angle measures 150°.
\( \angle ABC = 150° \)

Exam Tip: Always position the set-square so that the vertex of the desired angle aligns with the point where you want to construct the angle. For compound angles, combine two set-squares by placing one after marking the first angle from the second.

 

Question 2. Given a line BC and a point A on it, construct a ray AD using set-squares so that \( \angle DAC \) is:
(i) 30°
(ii) 150°
Answer:
(i) 30°
Draw the line BC and mark point A on it. Position the 30° set-square on the line BC such that the vertex of the 30° angle lies on point A and one edge coincides with the ray AB as shown in the figure. Draw the ray AD. Thus \( \angle DAC \) is the required angle of 30°.
(ii) 150°
Draw the line BC and mark point A on it. Position the 30° set-square on the line BC such that the vertex of the 30° angle lies on point A and one edge coincides with the ray AB as shown in the figure. Draw the ray AD. Thus \( \angle DAB = 30° \).
We know that angles on one side of a straight line always sum to 180°.
\( \angle DAB + \angle DAC = 180° \)
\( \angle DAC = 150° \)

Exam Tip: When constructing angles on a line, remember that supplementary angles on a straight line must add to exactly 180 degrees - use this property to verify your construction is accurate.

 

Exercise 18.2

 

Question 1. Mark the 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 find the length of AB, position the ruler with its edge along AB, such that the zero mark of the cm side of the ruler aligns 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 gives 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:
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-point 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:
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. Half of AB = 5.6÷2 = 2.8 cm. Now, we take a divider and open it so much that its end of one hand is at 0 and end of the other 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.

Exam Tip: When using a divider, always avoid changing its opening width once it is set - any disturbance to the fixed width will make the construction inaccurate. For collinear segments, ensure the ruler is perfectly aligned along the original line before extending it.

 

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 the point zero of the ruler and mark 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

Exam Tip: When a line segment does not start from the zero mark of a ruler, always subtract the smaller mark from the larger mark to find the actual length - never use the marked values directly.

 

Question 3. Draw a line segment CD. Produce it to CE such that CE = 3 CD
Answer: [Content continues on next page - answer not provided in fragment]

Exam Tip: When producing a line segment, ensure that the extension lies on the same straight line as the original segment, maintaining perfect collinearity throughout the construction.

 

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: To build segments of the required lengths, we use the compass and straightedge method described in the preliminary construction steps. Each construction involves marking points on a line and transferring segment lengths using a divider set to match specific distances.

(i) For AB - CD: Place CD backward from point B on segment AB. The remaining distance gives 7.5 - 2.5 = 5 cm.

(ii) For 2 AB: Mark point F on line l at distance AB from E. Then mark point G at distance AB from F. Segment EG = 2 AB = 15 cm.

(iii) For 3 CD: Using the divider set to CD, mark three successive equal segments on line l. Point H is placed so EH = 3 CD = 7.5 cm.

(iv) For AB + CD: Place segment AB and segment CD end to end on the same line. The total length is 7.5 + 2.5 = 10 cm.

(v) For 2 AB + 3 CD: First construct 2 AB = 15 cm, then add 3 CD = 7.5 cm. The final segment EJ has length 22.5 cm, as shown in the accompanying figure with measurements: EF = 7.5 cm, FG = 7.5 cm, GH = 2.5 cm, HI = 2.5 cm, IJ = 2.5 cm.
In simple words: Use a compass to copy lengths and build new segments by combining or dividing the given lengths AB and CD on a straight line.

Exam Tip: Always mark measurement points clearly and verify each constructed length by placing the compass back on the original segment and the new segment to confirm they match.

 

Question 5. Fill in the blanks:
(i) A part of a line with two end - points is called __________.
(ii) Segment AB is __________ segment BA
(iii) The length of a line segment is the __________ distance between two segments.
(iv) Two segments are congruent only if they have __________.
(v) Two segments of the same length are said to be __________.
Answer:
(i) A part of a line with two end-points is called **Line segment**

Explanation: A line segment is a portion of a line that has two fixed boundary points.

Example: Sides of a triangle or any polygon.

Points A and B are fixed endpoints, so AB is a line segment.

(ii) Segment AB is **equal to** segment BA

Explanation: When naming a line segment, the order does not matter. The length stays the same. If AB = 4 cm, then BA = 4 cm.

(iii) The length of a line segment is the **shortest** distance between two segments.

Explanation: We find the distance between two points by measuring the line segment connecting them. This straight line gives the shortest distance between the two endpoints.

(iv) Two segments are congruent only if they have **equal length**.

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

Explanation: Line segments are congruent when they share the same length. If AB = 4 cm and CD = 4 cm, then segment AB is congruent to segment CD.
In simple words: Line segments with matching lengths are equal in size and shape - we call this congruence.

Exam Tip: Remember that "congruent" means both the same length and same shape, while "equal" refers strictly to length; always state equal length as the defining condition for segment congruence.

 

Question 6. Match the following statements:

Answer:
(i) A line segment is a part of a line that has two distinct endpoints. It cannot extend beyond these fixed points.

(ii) Two line segments can meet at no point (if they do not touch), or they can meet at one point only. They cannot meet at more than one location.

(iii) Line segments with the same length are called congruent. If AB = 6 cm and CD = 6 cm, then AB and CD are congruent.

(iv) A line segment is a bounded section of a line. Unlike a full line (which goes infinitely in both directions), a segment has a fixed start and end.
In simple words: A line segment is a piece of a straight line with two marked endpoints, and two such pieces match when they have the same length.

Exam Tip: For matching questions, first identify key terms (endpoints, intersection, congruence, properties) in Column A, then carefully pair each with its correct definition or attribute in Column C.

 

Question 7. Tell which of the following statements are true (T) and false (F):
(i) The intersection of two segments may be 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: When two line segments cross, they can meet at one point only - a single location cannot form a line segment. Two segments cannot have an entire portion in common.

(ii) **True**

Explanation: If two line segments cross, the intersection point may fall somewhere in the middle of both segments rather than at either of their endpoints.

(iii) **True**

Explanation: A ray is a line that starts at a fixed point and extends endlessly in one direction. A segment is a bounded piece with two fixed endpoints. Every ray extends infinitely, making it fundamentally different from a segment with fixed endpoints.

(iv) **False**

Both endpoints of a line segment are fixed; however, a ray has only one endpoint fixed, with the other end extending to infinity. Therefore, a segment cannot ever be a ray.
In simple words: A ray has one fixed end that goes on forever, but a segment has both ends fixed - they are different things.

Exam Tip: For true-false questions on line geometry, draw quick diagrams to visualize intersections and check endpoint behavior - this prevents mistakes on ray vs. segment comparisons.

 

Question 8. What is the difference between a line, a line segment and a ray?:
Answer: A line extends infinitely in both directions with no fixed endpoints. Line AB can go on forever toward both A and B and beyond. A line segment has both ends fixed and bounded. Segment EF has two fixed points and cannot extend past either endpoint. A ray starts from one fixed point and extends infinitely in a single direction. Ray CD begins at C and goes on forever toward D and beyond.

The key difference is the number of fixed endpoints: a line has zero fixed endpoints (both directions are free), a segment has two fixed endpoints (both ends are stopped), and a ray has one fixed endpoint (one end is stopped, the other goes to infinity).
In simple words: A line goes forever both ways, a segment stops at both ends, and a ray starts at one point and goes forever in one direction.

Exam Tip: Always draw the arrow symbols correctly - double arrows for lines, single arrow for rays, no arrows for segments - to show which direction(s) extend infinitely.

 

Question 9. How many rays are represented in fig 18.8? Name them
Answer: A ray is defined as a portion of a line with a fixed starting point that can be drawn endlessly in one direction. If O is chosen as the starting point, each direction from O creates a distinct ray.

From the figure, the rays originating from point O are: OA, OB, OC, OD, OE, OF, OG, and OH.

Thus, there are **8 rays** in total in the figure.
In simple words: A ray has one fixed starting point and goes on forever in one direction. In this figure, point O is the starting point, and 8 different directions create 8 different rays.

Exam Tip: When counting rays from a central point, multiply the number of lines passing through that point by 2 (since each line creates two rays in opposite directions from the point).