CBSE Class 6 Mathematics Basic Geometrical Ideas Assignment Set 05

Question. From the adjoining figure (Cuboid), answer the following:
(a) Points marked
(b) Line segments
(c) Planes
Answer:
(a) Points marked: A, B, C, D, E, F, G, H
(b) Line segments: \( AB \), \( BC \), \( CD \), \( DA \), \( EF \), \( FG \), \( GH \), \( HE \), \( AE \), \( BF \), \( CG \), \( DH \)
(c) Planes: ABCD, EFGH, ABFE, DCGH, ADHE, BCGF

Question. What are concurrent lines? These lines meet in how many points? Draw figure also.
Answer: Three or more lines in a plane are said to be concurrent if they all pass through the same point. These lines meet in exactly one point (called the point of concurrence).

Question. Lines \( l, m \) and \( n \) are concurrent. Also lines \( l, n \) and \( p \) are also concurrent. Will the lines \( l, m, n \) and \( p \) be concurrent?
Answer: Yes

Question. From one point how many rays can we draw? Draw the figure also.
Answer: Infinite

Question. Give the name of that point from where a number of lines passes.
Answer: Point of concurrence

Question. Construct line segments of length:
(a) \( 5 \text{ cm} \)
(b) \( 4.7 \text{ cm} \)
Answer:
(a) Draw a line \( l \) and mark a point A on it. Place the compass pointer on the zero mark of the ruler. Open it to place the pencil point at the \( 5 \text{ cm} \) mark. Without changing the opening of the compass, place the pointer on A and swing an arc to cut \( l \) at B. \( AB \) is the required line segment of length \( 5 \text{ cm} \).
(b) Similarly, open the compass pointer to \( 4.7 \text{ cm} \) and mark an arc from a starting point on a line to construct a segment of length \( 4.7 \text{ cm} \).

Question. Construct a line segment whose length is the sum of lengths of two segments measuring \( 3.6 \text{ cm} \) & \( 2.4 \text{ cm} \)
Answer: The sum of the lengths of the two segments is \( 3.6 \text{ cm} + 2.4 \text{ cm} = 6.0 \text{ cm} \). Construct a line segment of length \( 6 \text{ cm} \) using a ruler and compass.

Question. If \( AB = 6.2 \text{ cm} \) and \( CD = 4.3 \text{ cm} \), construct the following line segments:
(a) \( 2AB \)
(b) \( 2AB - CD \)
(c) \( AB + CD \)
Answer:
(a) Length of \( 2AB = 2 \times 6.2 \text{ cm} = 12.4 \text{ cm} \)
(b) Length of \( 2AB - CD = 12.4 \text{ cm} - 4.3 \text{ cm} = 8.1 \text{ cm} \)
(c) Length of \( AB + CD = 6.2 \text{ cm} + 4.3 \text{ cm} = 10.5 \text{ cm} \)

Question. Given 3 collinear points, how many line segments do they determine? Name them.
Answer: Three: \( AB \), \( BC \), and \( AC \)

Question. From the adjoining figure, write:
(a) 5 line segments
(b) Five rays
(c) Non-intersecting line segments
Answer:
(a) Five line segments: \( AE \), \( EF \), \( FB \), \( GH \), \( CG \)
(b) Five rays: \( \overrightarrow{EB} \), \( \overrightarrow{EA} \), \( \overrightarrow{GC} \), \( \overrightarrow{HD} \), \( \overrightarrow{FE} \)
(c) Non-intersecting line segments: \( EF \) and \( GH \)