GSEB Class 6 Maths Solutions Chapter 14 Practical Geometry Exercise 14.5

Gujarat Board Textbook Solutions Class 6 Chapter 14 Practical Geometry Ex 14.5

Question 1. Draw \( \overline{\mathrm{AB}} \) of length 7.3 cm and find its axis of symmetry. Note: The perpendicular bisector of a line segment is its axis of symmetry.
Answer:

Steps of construction:

Step I: Draw a line segment \( \overline{\mathrm{AB}} = 7.3 \) cm.

Step II: With centres A and B and radius more than half of AB, draw two arcs that intersect each other at P and Q.

Step III: Join P and Q. Thus, PQ is the axis of symmetry of \( \overline{\mathrm{AB}} \).

Exam Tip: Remember that the perpendicular bisector of a line segment acts as its axis of symmetry. Ensure your arcs are drawn with a radius greater than half the segment length for accurate intersection points.

Question 2. Draw a line segment of length 9.5 cm and construct its perpendicular bisector.
Answer:

Steps of construction:

Step I: Draw a line segment AB = 9.5 cm.

Step II: With A and B as centres and radius more than half of \( \overline{\mathrm{AB}} \), draw two arcs on either side of AB which intersect each other at P and Q.

Step III: Join P and Q. Thus, PQ is the required perpendicular bisector of \( \overline{\mathrm{AB}} \).

Exam Tip: A perpendicular bisector divides a line segment into two equal parts and is at a 90-degree angle to it. Make sure your construction clearly shows these two properties.

Question 3. Draw the perpendicular bisector of \( \overline{\mathrm{XY}} \) whose length is 10.3 cm.
(a) Take any point P on the bisector drawn. Examine whether PX = PY.
(b) If M is the mid-point of \( \overline{\mathrm{XY}} \), what can you say about the lengths MX and XY?
Answer:

Steps of construction:

Step I: Draw a line segment \( \overline{\mathrm{XY}} = 10.3 \) cm.

Step II: With X and Y as centres and radius more than half of XY, draw two arcs on either side of XY which intersect each other at A and B.

Step III: Join A and B. Thus, AB is perpendicular to \( \overline{\mathrm{XY}} \).

Step IV: Mark a point 'P' on AB and join PX and PY.

(a) On measuring \( \overline{\mathrm{PX}} \) and \( \overline{\mathrm{PY}} \) (using a divider), we get that \( \overline{\mathrm{PX}} = \overline{\mathrm{PY}} \).

(b) The mid-point of XY is M. On measuring, we have that \( \overline{\mathrm{XM}} = \overline{\mathrm{MY}} = \frac { 1 }{ 2 } XY \).

Exam Tip: Any point on the perpendicular bisector of a line segment is equidistant from the endpoints of that segment. The midpoint always divides the segment into two equal halves.

Question 4. Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
Answer:

Steps of construction:

Step I: Draw a line segment \( \overline{\mathrm{AB}} = 12.8 \) cm.

Step II: Draw the perpendicular bisector of AB, which meets \( \overline{\mathrm{AB}} \) at O. (i.e. O is the mid-point of \( \overline{\mathrm{AB}} \)).

Step III: Draw the perpendicular bisector of \( \overline{\mathrm{AO}} \), which meets \( \overline{\mathrm{AB}} \) at P. (i.e. P is the mid-point of \( \overline{\mathrm{AO}} \)).

Step IV: Now, draw the perpendicular bisector of \( \overline{\mathrm{BO}} \), which meets AB at Q. (i.e. Q is the mid-point of OB).

The segment is divided into 4 equal parts by the points P, O, and Q.

By actual measurement, we have: \( \overline{\mathrm{AP}} = \overline{\mathrm{PO}} = \overline{\mathrm{OQ}} = \overline{\mathrm{QB}} = 3.2 \) cm.

Exam Tip: To divide a line segment into equal parts, you repeatedly bisect segments. First bisect the entire segment, then bisect each of the resulting halves. Verification by actual measurement ensures accuracy.

Question 5. With \( \overline{\mathrm{PQ}} \) of length 6.1 cm as diameter, draw a circle.
Answer:

Steps of construction:

Step I: Draw a line segment \( \overline{\mathrm{XY}} = 6.1 \) cm.

Step II: Draw the perpendicular bisector of PQ which meets \( \overline{\mathrm{XY}} \) at O. (i.e. O is the mid-point of PQ).

Step III: With centre O and \( \overline{\mathrm{OP}} \) or \( \overline{\mathrm{OQ}} \) as radius, draw a circle passing through P and Q. The circle having \( \overline{\mathrm{XY}} \) as the diameter is the required circle.

Exam Tip: To draw a circle with a given diameter, first find the midpoint of the diameter; this will be the center of your circle. The radius will be half of the diameter's length.

Question 6. Draw a circle with centre C and radius 3.4 cm. Draw any chord \( \overline{\mathrm{AB}} \). Construct the perpendicular bisector of \( \overline{\mathrm{AB}} \) and examine if it passes through C.
Answer:

Steps of construction:

Step I: Mark a point C on a paper.

Step II: With centre 'C' and radius 3.4 cm, draw a circle.

Step III: Draw a chord \( \overline{\mathrm{AB}} \).

Step IV: Draw the perpendicular bisector of \( \overline{\mathrm{AB}} \).

We find that 'l' (the perpendicular bisector of \( \overline{\mathrm{AB}} \)) passes through the centre of the circle.

Exam Tip: A key property of circles is that the perpendicular bisector of any chord always passes through the center of the circle. This is useful for finding the center if only a chord is given.

Question 7. Repeat question 6, if \( \overline{\mathrm{AB}} \) happens to be a diameter.
Answer:

Steps of construction:

Step I: Mark a point 'C' on a paper.

Step II: With centre C and radius 3.4 cm, draw a circle.

Step III: Draw a diameter \( \overline{\mathrm{AB}} \) of the circle.

Step IV: Draw the perpendicular bisector 'l' of \( \overline{\mathrm{AB}} \). We find that 'l' passes through C and C is the mid-point of AB.

Exam Tip: The perpendicular bisector of a diameter always passes through the circle's center, which is also the midpoint of the diameter itself. This demonstrates the consistency of geometric rules.

Question 8. Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?
Answer:

Steps of construction:

Step I: Mark a point O on a paper.

Step II: With centre O and radius 4 cm, draw a circle.

Step III: Draw two chords \( \overline{\mathrm{AB}} \) and \( \overline{\mathrm{CD}} \).

Step V: Draw perpendicular bisector 'M' of \( \overline{\mathrm{CD}} \).

Step VI: Produce l and m to meet each other. We find that l and m meet at O, the centre of the circle.

Exam Tip: A crucial property of circles is that the perpendicular bisectors of any two chords will always intersect at the exact center of the circle. This is a dependable method to find the center of a circle.

Question 9. Draw any angle with vertex O. Take a point A on one of its arms and B on another such that OA = OB. Draw the perpendicular bisectors of \( \overline{\mathrm{OA}} \) and \( \overline{\mathrm{OB}} \). Let them meet at P. Is PA = PB?
Answer:

Steps of construction:

Step I: Mark a point O on a paper.

Step II: Draw an angle \( \angle \mathrm{XOY} \), having a vertex at O.

Step III: Mark a point A on \( \overrightarrow{\mathrm{OX}} \) and another point B on \( \overrightarrow{\mathrm{OY}} \), such that \( \overline{\mathrm{OA}} = \overline{\mathrm{OB}} \).

Step IV: Draw l, the perpendicular bisector of \( \overline{\mathrm{OB}} \).

Step V: Draw m, the perpendicular bisector of \( \overline{\mathrm{OA}} \).

Step VI: Mark the intersecting points of l and m as P.

Step VII: Join \( \overline{\mathrm{PA}} \) and \( \overline{\mathrm{PB}} \), measure them with the help of a divider.

Step VIII: On measuring, we find \( \overline{\mathrm{PA}} = \overline{\mathrm{PB}} \).

Exam Tip: The intersection point of the perpendicular bisectors of two sides of a triangle (or in this case, two segments from a common vertex) is equidistant from the vertices of those segments. This property confirms that PA will equal PB.