CBSE Class 6 Mathematics Practical Geometry Assignment Set 05

VI - Mathematics Assignment No - 01 - Practical Geometry

Question 1. Draw a line segment AB of 8 cm length and find its mid point by using ruler and compass
Answer: 1. Draw a line segment \(AB = 8\text{ cm}\) using a ruler.
2. With A as center and radius more than half of AB (\(> 4\text{ cm}\)), draw arcs above and below the line segment AB.
3. With B as center and the same radius, draw arcs above and below AB to cut the previous arcs at points P and Q.
4. Join PQ intersecting AB at point M. Point M is the required mid point of AB such that \(AM = MB = 4\text{ cm}\).
In simple words: The perpendicular bisector of a line segment divides it into two equal halves, giving us the exact midpoint.

Exam Tip: Keep the compass radius fixed while drawing intersecting arcs from both endpoints to ensure absolute accuracy.

Question 2. Draw a line of 5 cm length. Find a point on this line in such a way that the two parts (in length) are equal.
Answer: 1. Let the line segment be XY of length 5 cm.
2. Open the compass to a radius greater than half of 5 cm (e.g., 3 cm).
3. Taking X and Y as centers alternatively, draw intersecting arcs on both sides of the line segment XY.
4. Join the points of intersection of the arcs with a line that crosses XY at point C. Point C divides the 5 cm line into two equal parts of \(2.5\text{ cm}\) each.
In simple words: Finding a point that divides a line into two equal parts is the same as constructing its perpendicular bisector.

Exam Tip: Always verify your result with a ruler to ensure both segments measure exactly half of the total given length.

Question 3. Draw a line segment AB = 5.6 cm. Find its perpendicular bisector.
Answer: 1. Draw a line segment \(AB = 5.6\text{ cm}\) using a ruler.
2. Place the compass pointer at A and open it to a radius greater than half the length of AB (\(> 2.8\text{ cm}\)).
3. Draw arcs above and below the line segment AB.
4. With the same radius and pointer at B, draw arcs above and below AB to cut the previous arcs.
5. Join the intersection points of the arcs. This line is the required perpendicular bisector of AB.
In simple words: A perpendicular bisector meets the line segment at a right angle (\(90^\circ\)) and cuts it right in the middle.

Exam Tip: Use a protractor at the end to check if the angle formed between the bisector and the line segment is exactly \(90^\circ\).

Question 4. Draw a circle of radius 5 cm
Answer: 1. Open the compass to a distance of 5 cm using a scale.
2. Mark a point O on the paper to serve as the center.
3. Place the metal pointer of the compass on point O.
4. Rotate the pencil end smoothly around O to draw the complete circle of radius 5 cm.
In simple words: The fixed point is the center, and the distance from the center to the boundary is the radius.

Exam Tip: Ensure the pencil lead is sharp and tightly secured in the compass so the radius does not slip while drawing.

Question 5. Draw three circles with same centre and of radius 4 cm, 5 cm and 6 cm.
Answer: 1. Mark a point O on the paper as the common center.
2. With O as center, draw a circle of radius 4 cm.
3. From the same center O, draw a second circle of radius 5 cm.
4. From the same center O, draw a third circle of radius 6 cm.
5. These are concentric circles sharing the same center with increasing radii.
In simple words: Circles that share the exact same center point are called concentric circles.

Exam Tip: Do not move the compass needle position on the paper when changing the radius for the next circle.

Question 6. Draw a circle of radius 4.5 cm. Draw a line passing through the centre and meeting the circumference at two different points. Name the line so formed.
Answer: 1. Mark a point O and draw a circle of radius 4.5 cm.
2. Draw a straight line passing directly through the center O such that it touches the circle's boundary at two distinct points, say P and Q.
3. The line segment PQ passing through the center is called the diameter.
In simple words: A line passing through the center connecting two points on the circle is a diameter, which is equal to twice the radius.

Exam Tip: Remember that the diameter is always the longest chord of a circle.

Question 7. Draw a circle of radius 5.5 cm. Draw two chords not passing through the centre
Answer: 1. Mark a point O and draw a circle with a radius of 5.5 cm.
2. Choose any two pairs of points on the boundary (circumference) of the circle that do not pass through the center O.
3. Connect the first pair with a straight line segment, naming it AB.
4. Connect the second pair with another straight line segment, naming it CD.
5. AB and CD are the required chords not passing through the center.
In simple words: Any line segment whose endpoints lie on the circle is called a chord.

Exam Tip: Ensure your line segments do not touch or cross the center point O.

Question 8. Draw a circle of radius 4 cm. Draw the longest chord. Will it pass through the centre if yes, what is the name of such chord?
Answer: 1. Draw a circle with center O and radius 4 cm.
2. Draw a line segment passing through the center O with endpoints on the circle.
3. Yes, the longest chord passes through the center.
4. The name of such a chord is the Diameter.
In simple words: The diameter is the special chord that cuts right through the center and is longer than any other chord in the circle.

Exam Tip: Length of the longest chord (Diameter) = \(2 \times \text{radius} = 2 \times 4\text{ cm} = 8\text{ cm}\).

Question 9. Draw a circle of radius 3 cm. Divide the circle into two equal parts by drawing a line AB passing through the Centre. Now take a point P on the circumference of circle. Join PA and PB. Measure the angle \angle APB.
Answer: 1. Draw a circle with radius 3 cm and center O.
2. Draw a diameter AB passing through the center O, dividing the circle into two equal semicircles.
3. Take any point P on the circumference of the circle.
4. Join P to A and P to B to form triangle APB.
5. Measure the angle \angle APB using a protractor; it measures exactly \(90^\circ\).
In simple words: An angle inscribed in a semicircle is always a right angle (\(90^\circ\)).

Exam Tip: This is a fundamental geometric theorem: angle in a semicircle is a right angle.

Question 10. Draw a circle of radius 3.5 cm and mark points P, Q, R such that
(i) P is on the circle
(ii) Q is outside the circle
(iii) R is in the interior of the circle
Answer: 1. Draw a circle with center O and radius 3.5 cm.
2. Mark point P directly on the boundary (circumference) of the circle.
3. Mark point Q at a distance greater than 3.5 cm from the center O (in the exterior region outside the circle).
4. Mark point R at a distance less than 3.5 cm from the center O (inside the circular region).
In simple words: Points can lie on the boundary, outside, or inside a circle based on their distance from the center.

Exam Tip: Distance \(OP = 3.5\text{ cm}\), \(OQ > 3.5\text{ cm}\), and \(OR < 3.5\text{ cm}\).