CBSE Class 6 Mathematics Practical Geometry Worksheet Set 04

Question 1. Draw a circle of radius 5.6 cm.
Answer: Steps of construction:
1. Open the compasses for the required radius of 5.6 cm using a ruler.
2. Mark a point O on the paper to serve as the center.
3. Place the pointer of the compasses on O and rotate the pencil gently to draw the complete circle.
In simple words: Open your compass to 5.6 cm, put the needle on a center point, and draw a circle around it.

Exam Tip: Always measure the compass width carefully against a ruler starting precisely from the 0 cm mark.

Question 2. With the same centre O, draw two circles of radii 5 cm and 2.5 cm.
Answer: Steps of construction:
1. Mark a point O on the paper as the common center.
2. Open the compasses to a radius of 2.5 cm, place the pointer at O, and draw the inner circle.
3. Without disturbing the center, open the compasses to a radius of 5 cm, place the pointer at O, and draw the outer concentric circle.
In simple words: Draw two circles from the exact same center point, one smaller (2.5 cm) and one larger (5 cm).

Exam Tip: Circles sharing the same center are called concentric circles.

Question 3. Draw any circle and mark points P, Q and R such that
a) P is on the circle
b) Q is in the interior of the circle.
c) R is in the exterior of the circle.
Answer: Steps of construction:
1. Draw a circle with any convenient radius and center O.
2. Mark a point P directly on the boundary (circumference) of the circle.
3. Mark a point Q inside the circular region, closer to the center O.
4. Mark a point R outside the circular region, anywhere away from the boundary.
In simple words: Put point P on the circle line, point Q inside, and point R outside.

Exam Tip: Interior means inside, exterior means outside, and on the circle means lying on its boundary.

Question 4. Draw any line segment PQ. Mark any point B on it. Through B, draw a perpendicular to PQ.
Answer: Steps of construction:
1. Draw a line segment PQ of any convenient length.
2. Mark a point B anywhere on PQ.
3. With B as center and a convenient radius, draw an arc intersecting PQ at two points, say X and Y.
4. With X and Y as centers and a radius greater than XB, draw two arcs intersecting each other at point C on one side of PQ.
5. Join B and C. Then \(BC \perp PQ\).
In simple words: Draw a line, pick a point on it, and use compasses to construct a 90-degree upright line through that point.

Exam Tip: Ensure the radius used for intersecting arcs from X and Y is equal and greater than half the distance XY.

Question 5. Draw XY of length 8.3 cm and find its axis of symmetry.
Answer: Steps of construction:
1. Draw a line segment \(XY = 8.3\) cm.
2. With X as center and radius more than half of XY, draw two arcs (one above XY and one below XY).
3. With Y as center and the same radius, draw two arcs cutting the previous arcs at points P and Q.
4. Join PQ. Line PQ is the perpendicular bisector and acts as the axis of symmetry for segment XY.
In simple words: Draw an 8.3 cm line and construct its perpendicular bisector, which divides the line into two equal symmetrical halves.

Exam Tip: The axis of symmetry of a line segment is its perpendicular bisector passing through its exact midpoint.

Question 6. Draw a line segment of length 10.5 cm and construct its perpendicular bisector.
Answer: Steps of construction:
1. Draw a line segment AB of length 10.5 cm.
2. Open the compasses to more than half the length of AB (more than 5.25 cm).
3. Keeping the compass width fixed, place the pointer at A and draw arcs above and below AB.
4. With the pointer at B and the same radius, draw arcs cutting the previous arcs at points C and D.
5. Join CD. CD is the required perpendicular bisector dividing AB into two equal parts of 5.25 cm each.
In simple words: Draw a 10.5 cm line, then make intersecting arcs from both ends to draw a straight line right down the middle at 90 degrees.

Answer: Steps of construction:
1. Draw a line segment \(AB = 6.2\) cm.
2. Find the midpoint of AB by constructing its perpendicular bisector. Let the midpoint be O.
3. Keeping the compass pointer at O and adjusting the pencil up to A (or B), the radius will be \(\frac{6.2}{2} = 3.1\) cm.
4. Draw the circle. This circle will pass through both A and B, having AB as its diameter.
In simple words: Find the exact middle of a 6.2 cm line and use that middle point to draw a circle that touches both ends.

Exam Tip: The radius of a circle is always half of its diameter (\(r = \frac{d}{2}\)).

Question 8. Draw a circle of radius 4.5 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet.
Answer: Steps of construction:
1. Draw a circle of radius 4.5 cm with center O.
2. Draw any two non-parallel chords, say AB and CD.
3. Construct the perpendicular bisector for chord AB.
4. Construct the perpendicular bisector for chord CD.
5. Extend both perpendicular bisectors inside the circle. They meet at the center O of the circle.
In simple words: Draw a circle, make two straight lines (chords) inside it, and find their middle perpendicular lines. They will always intersect at the exact center of the circle.

Exam Tip: The perpendicular bisector of any chord of a circle always passes through the center of the circle.

Question 9. Draw a line segment of length 10.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
Answer: Steps of construction:
1. Draw a line segment AB of length 10.8 cm.
2. Construct the perpendicular bisector of AB to divide it into two equal halves at point M. Each part is \(\frac{10.8}{2} = 5.4\) cm.
3. Now construct the perpendicular bisector of AM to get point P. Length of AP = 2.7 cm.
4. Construct the perpendicular bisector of MB to get point Q. Length of QB = 2.7 cm.
5. Thus, points P, M, and Q divide AB into four equal parts: AP, PM, MQ, and QB.
6. Verification: Measuring with a scale shows each part measures exactly 2.7 cm.
In simple words: Draw a 10.8 cm line and bisect it, then bisect each half again to get four equal sections of 2.7 cm each.

Exam Tip: Dividing into four equal parts is achieved by successive bisection (\(1 \rightarrow 2 \rightarrow 4\) parts).

Question 10. Draw the perpendicular bisector of AB whose length is 8.3 cm.
a) Take any point P on the bisector drawn. Examine whether PA = PB
b) If M is the mid point of AB, what can you say about the length of MA and MB?
Answer: Steps of construction:
1. Draw line segment \(AB = 8.3\) cm and construct its perpendicular bisector XY.
a) Take any point P on the bisector line XY. Measure the distance PA and PB using a ruler or compasses. We observe that \(PA = PB\).
b) Since M is the midpoint of AB, the line segment AB is divided into two equal halves. Therefore, \(MA = MB = \frac{8.3}{2} = 4.15\) cm.
In simple words: (a) Any point on the middle perpendicular line is equidistant from both endpoints A and B. (b) The two halves MA and MB are equal in length.

Exam Tip: Any point on the perpendicular bisector of a line segment is always at an equal distance from the endpoints of the segment.

Question 11. Draw an angle of measure \(137^\circ\) and construct its bisector.
Answer: Steps of construction:
1. Use a protractor to draw an angle \(\angle ABC = 137^\circ\).
2. With B as center and a convenient radius, draw an arc that cuts ray BA at X and ray BC at Y.
3. With X as center and radius more than half of XY, draw an arc in the interior of the angle.
4. With Y as center and the same radius, draw another arc intersecting the previous arc at point Z.
5. Join BZ and extend it. Ray BZ is the required angle bisector dividing \(137^\circ\) into two equal angles of \(68.5^\circ\) each.
In simple words: Draw a \(137^\circ\) angle using a protractor, then use your compasses to draw intersecting arcs that split the angle exactly in half.

Exam Tip: Always use a sharp pencil for precise angle construction and bisection.

Question 12. Draw a right angle and construct its bisector.
Answer: Steps of construction:
1. Draw a ray OA.
2. With O as center and a convenient radius, draw an arc cutting OA at X.
3. With X as center and the same radius, cut the arc at Y, and with Y as center, cut the arc further at Z.
4. Draw perpendicular lines from Y and Z to intersect at point P. Join OP to get a right angle (\(90^\circ\)).
5. To bisect \(90^\circ\), place the compass pointer at X and draw an arc, then with the same radius place the pointer at the arc point on the perpendicular ray and cut it.
6. Join the intersection to vertex O to get an angle of \(45^\circ\).
In simple words: Construct a \(90^\circ\) angle and bisect it to get a \(45^\circ\) angle.

Answer: Steps of construction:
1. Draw an angle of \(152^\circ\) using a protractor.
2. Bisect this angle to get two equal parts of \(76^\circ\) each.
3. Bisect each of the two \(76^\circ\) angles further to get four equal parts.
4. Each of the four resulting angles measures \(\frac{152^\circ}{4} = 38^\circ\).
In simple words: Draw a \(152^\circ\) angle, bisect it into two halves, and then bisect both halves again to get four equal \(38^\circ\) sections.

Exam Tip: Dividing into four equal parts requires three successive angle bisections (\(1 \rightarrow 2 \rightarrow 4\)).

Question 14. Draw an angle of measure \(60^\circ\) and bisect it.
Answer: Steps of construction:
1. Draw a ray OA.
2. With O as center and any convenient radius, draw an arc cutting OA at point X.
3. With X as center and the same radius, draw an arc cutting the first arc at point Y.
4. Join OY and extend it to form ray OB. Angle AOB is \(60^\circ\).
5. With X and Y as centers, draw two intersecting arcs in the interior of the angle.
6. Join the point of intersection to vertex O. This bisects the \(60^\circ\) angle into two angles of \(30^\circ\) each.
In simple words: Construct a standard \(60^\circ\) angle with compasses and bisect it to create a \(30^\circ\) angle.

Exam Tip: A \(60^\circ\) angle is constructed using the first step of drawing an equilateral triangle arc on a circle.

Question 15. Draw an angle of measure \(150^\circ\) and bisect it.
Answer: Steps of construction:
1. Draw an angle of \(150^\circ\) using a protractor (or construct it as \(90^\circ + 60^\circ\)).
2. With the vertex as center, draw an arc intersecting both arms of the \(150^\circ\) angle.
3. From the two intersection points on the arms, draw intersecting arcs inside the angle using a suitable radius.
4. Join the vertex to the intersection point of the arcs. This is the angle bisector.
5. Each resulting angle measures \(\frac{150^\circ}{2} = 75^\circ\).
In simple words: Draw a \(150^\circ\) angle and use compasses to split it exactly in half into two \(75^\circ\) angles.

Exam Tip: \(150^\circ\) can also be constructed by combining a \(90^\circ\) angle and a \(60^\circ\) angle.

Question 16. Construct with ruler and compasses, angles of following measures:
a) \(60^\circ\)
b) \(120^\circ\)
c) \(90^\circ\)
d) \(45^\circ\)
e) \(15^\circ\)
f) \(30^\circ\)}
Answer:
a) \(60^\circ\): Draw a ray, take a convenient radius and draw an arc from the vertex. With the intersection point as center and same radius, cut the arc. Join the vertex to the cut point.
b) \(120^\circ\): Construct a \(60^\circ\) arc, then from that arc cut a second arc along the circle path to get \(120^\circ\).
c) \(90^\circ\): Construct a \(90^\circ\) angle by erecting a perpendicular at the endpoint of a ray using standard \(60^\circ\) and \(120^\circ\) arc intersections.
d) \(45^\circ\): Construct a \(90^\circ\) angle and bisect it to obtain \(45^\circ\).
e) \(15^\circ\): Construct a \(60^\circ\) angle, bisect it to get \(30^\circ\), and bisect the \(30^\circ\) angle to get \(15^\circ\).
f) \(30^\circ\): Construct a \(60^\circ\) angle and bisect it to get \(30^\circ\).
In simple words: Use standard compass constructions starting from a \(60^\circ\) base arc and successive bisections to draw all these standard angles.

Exam Tip: Master the construction of \(60^\circ\) and \(90^\circ\) first, as all other standard angles are derived from them by bisection.

Question 17. Draw a circle and any two of its diameters. What is the figure obtained? What figure is obtained if the diameters are perpendicular to each other?
Answer:
- If any two diameters of a circle are drawn, joining their four endpoints in cyclic order forms a rectangle.
- If the two diameters are perpendicular to each other, joining their four endpoints in cyclic order forms a square.
In simple words: Connecting the ends of two normal diameters gives a rectangle, and connecting the ends of two perpendicular diameters gives a square.

Exam Tip: The diagonals of a rectangle are equal and bisect each other; if they are also perpendicular, the rectangle becomes a square.

[Section: Symmetry]

Question 1. Can you draw a triangle which has
a) exactly one line of symmetry
b) exactly two lines of symmetry?
c) exactly three lines of symmetry?
d) no lines of symmetry?
Answer:
a) Yes, an isosceles triangle has exactly one line of symmetry.
b) No, a triangle can never have exactly two lines of symmetry.
c) Yes, an equilateral triangle has exactly three lines of symmetry.
d) Yes, a scalene triangle has no lines of symmetry.
In simple words: Isosceles triangles have 1 line of symmetry, equilateral have 3, scalene have 0, and triangles can never have exactly 2 lines of symmetry.

Exam Tip: Remember that a triangle cannot possess exactly two lines of symmetry because symmetry properties in triangles require either 0, 1, or 3 equal sides.

Question 2. Find the number of lines of symmetry in each of the following shapes.
Answer:
- First shape (Four-petaled flower/quatrefoil design): 4 lines of symmetry.
- Second shape (Diamond/Rhombus with internal lines): 2 lines of symmetry.
- Third shape (Stacked rounded capsule shapes): 2 lines of symmetry (one horizontal, one vertical).
- Fourth shape (Regular hexagon with star): 6 lines of symmetry.
- Fifth shape (Double-ended arrow): 2 lines of symmetry (one horizontal, one vertical).
In simple words: Count how many ways you can fold each shape so that both halves match perfectly.

Exam Tip: Look for both vertical, horizontal, and diagonal axes when counting lines of symmetry in complex geometric shapes.

Question 3. Consider the letters of English alphabets, A to Z. List among them the letters which have
a) Vertical lines of symmetry
b) Horizontal lines of symmetry
c) No lines of symmetry
Answer:
a) Vertical lines of symmetry: A, H, I, M, O, T, U, V, W, X, Y (11 letters).
b) Horizontal lines of symmetry: B, C, D, E, H, I, K, O, X (9 letters).
c) No lines of symmetry: F, G, J, L, N, P, Q, R, S, Z (10 letters).
In simple words: Group the capital letters based on whether they can be folded symmetrically from top-to-bottom, left-to-right, or not at all.

Exam Tip: Letters like H, I, O, and X have both vertical and horizontal lines of symmetry.

Question 4. Complete the following table:
Answer:

Exam Tip: Memorize the symmetry counts for standard geometric figures as they frequently appear in exams.

Question 5. Write some application of symmetry in everyday life.
Answer:
Applications of symmetry in everyday life include:
1. Architecture and monument design (e.g., the Taj Mahal) for aesthetic balance and structural stability.
2. Art, rangoli designs, and textile/clothing patterns.
3. Nature and biology (leaves, flowers, butterfly wings, and human/animal body structures).
4. Engineering, automobile design, and furniture manufacturing.
In simple words: Symmetry is used everywhere to make things look balanced and beautiful, from nature to buildings and clothes.

Exam Tip: Give clear, practical real-world examples like buildings, tools, or natural objects when asked about symmetry applications.