CBSE Class 6 Mathematics Practical Geometry Assignment Set 02

VI - Mathematics Assignment No - 04 - Practical Geometry

Q1. Using protractor, draw the following angles: (i) \(50^\circ\) (ii) \(133^\circ\) (iii) \(140^\circ\)
Answer: Steps of construction using a protractor:
1. Draw a base ray with an initial point (vertex).\br /> 2. Place the center point of the protractor on the vertex and align the baseline of the protractor along the ray.
3. Read the required angle degree on the protractor scale and mark a point on the paper.
4. Join the vertex to the marked point using a ruler to form the complete angle.
- For \((i) 50^\circ\): Acute angle measuring \(50^\circ\).
- For \((ii) 133^\circ\): Obtuse angle measuring \(133^\circ\).
- For \((iii) 140^\circ\): Obtuse angle measuring \(140^\circ\).
In simple words: A protractor is placed on a line to measure and draw exact angle values.

Exam Tip: Always ensure the baseline of the protractor coincides exactly with the ray.

Q2. Draw any angle \(\angle ABC\). Draw another angle \(\angle DEF\) equal to \(\angle ABC\).
Answer: Steps of construction:
1. Draw any angle \(\angle ABC\) of arbitrary measure using a ruler.
2. Draw a line segment (or ray) \(EF\).
3. With center \(B\) and any convenient radius, draw an arc intersecting the arms \(BA\) and \(BC\) at points \(P\) and \(Q\) respectively.
4. With the same radius and center \(E\), draw an arc intersecting ray \(EF\) at point \( D \.
5. Measure the distance \(PQ\) using compasses. With center \(D\) and radius equal to \(PQ\), draw an arc to intersect the previous arc at point \(R\).
6. Join \(E\) to \(R\) and extend it to form point \(\angle DEF\) equal in measure to \(\angle ABC\).
In simple words: Copy an angle using compasses by transferring the arc length without changing the radius setting.

Exam Tip: Keep the compass width fixed when measuring arc segments between the arms.

Q3. Draw an angle of \(60^\circ\). Draw its bisector.
Answer: Steps of construction:
1. Draw a ray \(OA\).
2. With center \(O\) and a convenient radius, draw an arc that cuts ray \(OA\) at point \(X\).
3. With center \(X\) and the same radius, draw an arc cutting the previous arc at point \(Y\).
4. Join \(O\) through \(Y\) and extend to form ray \(OB\). Thus, \(\angle AOB = 60^\circ\).
5. With center \(X\) and radius more than half of \(XY\), draw an arc in the interior of \(\angle AOB\).
6. With center \(Y\) and the same radius, draw another arc to intersect the previous arc at point \(Z\).
7. Join \(O\) to \(Z\). Ray \(OZ\) is the required angle bisector dividing \(60^\circ\) into two equal angles of \(30^\circ\) each.
In simple words: An angle bisector divides the angle into two equal halves.

Exam Tip: The compass radius must be kept greater than half the distance between arc intersections when bisecting.

Q4. Draw an angle of \(120^\circ\) and divide it into 4 equal parts. Measure each part.
Answer: Steps of construction:
1. Draw a ray \(OA\).
2. With center \(O\) and any radius, draw an arc intersecting \(OA\) at \(P\).
3. With center \(P\) and same radius, cut the arc at \(Q\), and with center \(Q\) cut the arc further at \(R\).
4. Join \(O\) through \(R\) to get \(\angle AOR = 120^\circ\).
5. Bisect the \(120^\circ\) angle to get two \(60^\circ\) angles.
6. Further bisect each \(60^\circ\) angle to get four equal parts.
- Measurement of each part = \(\frac{120^\circ}{4} = 30^\circ\).
In simple words: Dividing a \(120^\circ\) angle into four equal sections gives four angles of \(30^\circ\) each.

Exam Tip: Successive bisection of angles is used to divide an angle into \(2, 4, 8\) equal parts.

Q5. Using ruler and compass, construct the following angles: (i) \(30^\circ\) (ii) \(60^\circ\) (iii) \(120^\circ\)
Answer:
- (i) \(30^\circ\): Construct a \(60^\circ\) angle first, then bisect it to get \(30^\circ\).
- (ii) \(60^\circ\): Draw a ray, make an arc from the vertex, and cut it once with the same radius to form \(60^\circ\).
- (iii) \(120^\circ\): Draw a ray, make an arc from the vertex, and cut it twice along the arc to form \(120^\circ\).
In simple words: Standard angles like \(60^\circ\) and \(120^\circ\) are constructed directly using arcs, and \(30^\circ\) is half of \(60^\circ\).

Exam Tip: Maintain a uniform compass radius while marking standard arc intersections.

Q6. Using ruler and compass, construct an angle of \(90^\circ\).
Answer: Steps of construction:
1. Draw a ray \(OA\).
2. With center \(O\) and a convenient radius, draw a semi-circle cutting \(OA\) at \(P\).
3. With center \(P\) and the same radius, cut the semi-circle at \(Q\). With center \(Q\) and the same radius, cut the semi-circle further at \(R\).
4. With centers \(Q\) and \(R\) and radius greater than half of \(QR\), draw two arcs intersecting each other at \(S\).
5. Join \(O\) to \(S\). Ray \(OS\) forms a \(90^\circ\) angle with ray \(OA\).
In simple words: A \(90^\circ\) angle is constructed by bisecting the \(60^\circ\) and \(120^\circ\) marks on a semi-circular arc.

Exam Tip: A \(90^\circ\) angle lies exactly midway between \(60^\circ\) and \(120^\circ\).

Q7. Draw an angle of \(45^\circ\) and construct its line of symmetry.
Answer: Steps of construction:
1. Construct a \(90^\circ\) angle.
2. Bisect the \(90^\circ\) angle to obtain an angle of \(45^\circ\).
3. The angle bisector of the \(45^\circ\) angle acts as its line of symmetry, dividing it into two \(22.5^\circ\) angles.
In simple words: The line of symmetry for an angle is its angle bisector.

Exam Tip: Always construct the prerequisite angle (\(90^\circ\) then \(45^\circ\)) before drawing the symmetry line.

Q8. Draw an angle of \(75^\circ\) using ruler and compass. Verify your construction by measurement.
Answer: Steps of construction:
1. Construct a \(90^\circ\) angle and a \(60^\circ\) angle from the same vertex.
2. The \(75^\circ\) angle lies exactly halfway between \(60^\circ\) and \(90^\circ\).
3. Bisect the angle between the \(60^\circ\) mark and the \(90^\circ\) mark to obtain \(75^\circ\).
4. Verification: Measure the constructed angle using a protractor; it reads exactly \(75^\circ\).
In simple words: \(75^\circ\) is constructed by bisecting the space between \(60^\circ\) and \(90^\circ\).

Exam Tip: \(75^\circ = 60^\circ + \frac{1}{2}(30^\circ)\).

Q9. Draw an angle \(\angle AOB = 70^\circ\) and construct another angle \(\angle XYZ\) such that \(\angle XYZ = \frac{1}{4} \angle AOB\).
Answer: Steps of construction:
1. Draw \(\angle AOB = 70^\circ\) using a protractor.
2. To find \(\frac{1}{4} \angle AOB\), calculate \(\frac{1}{4} \times 70^\circ = 17.5^\circ\).
3. Construct the bisector of \(70^\circ\) to get \(35^\circ\).
4. Bisect the \(35^\circ\) angle to obtain \(17.5^\circ\) (or \(17^\circ 30'\)).
5. Draw ray \(YX\) and construct \(\angle XYZ = 17.5^\circ\) using a protractor or compass bisection method.
In simple words: Dividing an angle by 4 requires bisecting it twice.

Exam Tip: Successive bisection halves the angle value at each step.

Q10. Draw \(\angle PQR = 75^\circ\), bisect this angle \(\angle PQR\) and mark a point \(S\) on it. From \(S\), draw perpendiculars to arms \(QP\) and \(QR\).
Answer: Steps of construction:
1. Construct \(\angle PQR = 75^\circ\).
2. Draw the angle bisector of \(\angle PQR\).
3. Mark any point \(S\) on this angle bisector ray.
4. From point \(S\), draw perpendicular line segments to both arms \(QP\) and \(QR\) using a set-square or compass perpendicular construction method.
In simple words: Any point on the angle bisector is equidistant from the arms of the angle.

Exam Tip: Use a set-square aligned with the arm to drop an accurate perpendicular from an exterior point.

Q11. Draw a line \(AB\) of \(6.5\text{ cm}\) length. At the end points \(A, B\) draw two perpendiculars. Are these perpendiculars parallel? (yes)
Answer: Steps of construction:
1. Draw a line segment \(AB = 6.5\text{ cm}\).
2. At point \(A\), construct a perpendicular line using compasses.
3. At point \(B\), construct a perpendicular line using compasses.
4. Observe the two perpendicular lines; since both make \(90^\circ\) with the baseline \(AB\), corresponding interior angles add up to \(180^\circ\), proving the lines are parallel.
- Answer to question: Yes, the perpendiculars are parallel.
In simple words: Two lines perpendicular to the same base line are parallel to each other.

Exam Tip: Parallel lines never intersect and maintain a constant perpendicular distance.

Q12. Draw \(\angle BAC = 50^\circ\) such that \(AB = 3.5\text{ cm}\) and \(AC = 4\text{ cm}\). Through \(C\), draw a line parallel to \(AB\) and through \(B\) draw a line parallel to \(AC\), intersecting each other at \(D\). Measure \(BD\) and \(CD\). (\(BD = 4\text{ cm}\), \(CD = 3.5\text{ cm}\))
Answer: Steps of construction:
1. Draw line segment \(AB = 3.5\text{ cm}\).
2. At point \(A\), construct \(\angle BAC = 50^\circ\).
3. Along the ray, cut off \(AC = 4\text{ cm}\).
4. Through point \(C\), draw a line parallel to \(AB\).
5. Through point \(B\), draw a line parallel to \(AC\) such that it intersects the line from \(C\) at point \(D\), forming a parallelogram \(ABDC\).
6. Measure the lengths: \(BD = 4\text{ cm}\) and \(CD = 3.5\text{ cm}\).
In simple words: Opposite sides of a constructed parallelogram are equal in length.

Exam Tip: Opposite sides of a parallelogram are equal: \(BD = AC = 4\text{ cm}\) and \(CD = AB = 3.5\text{ cm}\).