*Question 1: *

*Draw \(\overline{XY}\) of length 10 cm and What is the axis of symmetry*

*Answer:*

Axis of symmetry of line segment \(\overline{XY}\) will be the perpendicular bisector of \(\overline{XY}\). So, draw the perpendicular bisector of XY.

Steps of construction:

(i)A line segment is drawn XY = 10 cm

(ii) Assume X and Y as centers and radius more than half of XY, construct 2 arcs which intersect each other at P and Q .

(iii) Join PQ. Then PQ is the symmetry axis of the segment line XY.

*Question 2: *

*Construct a segment line of length 8 cm and draw its perpendicular bisector.*

* Answer:** *

Construction step**:**

(i) A line segment \(\overline{XY}\) is drawn= 8 cm

(ii) Taking X and Y as radius and centre more than XY half, draw 2 arcs at P they intersect each other’s and Q.

(iii) PQ is then joined. Then PQ is the perpendicular bisector of \(\overline{XY}\).

**Question 3: **

**\(\overline{AB}\) bisector is drawn perpendicular that length is 8 cm.**

**(a) A point P is taken according to the drawn bisector. Prove OA = OB.**

**(b)If X is the \(\overline{AB}\) midpoint, the lengths XA and XB explain about it?**

*Answer:*

Construction steps:

(1)A line segment is drawn \(\overline{AB}\) = 8 cm

(2)Taking A and B as radius and centre more than half of AB, draw two arcs which intersect each other at P and Q.

(3) Join PQ. Then PQ is the required perpendicular bisector of \(\overline{AB}\).

Now,

(1)Point O is taken on the drawn bisector. By using the divider and we can check that that \(\overline{OA}\)= \(\overline{OB}\)

(2) If X is the midpoint of \(\overline{AB}\), then \(\overline{XA}\) =1/2 \(\overline{XB}\)

** **

*Question 4: *

*Draw a line segment of length 10 cm. Using compasses; separate it into 4 equal parts. Verify by original measurement.*

*Answer:*

Construction steps:

(1)A line segment is drawn PQ = 10 cm

(2) \(\overline{PQ}\) of perpendicular bisector is drawn which cuts it at a point X. Thus, \(\overline{PQ}\) is the bisector of X.

(3) Draw \(\overline{PX}\) perpendicular bisector which cuts it at P. Thus P is the midpoint of.

(4) Again, \(\overline{XQ}\) perpendicular bisector is drawn which cuts it at B . Thus, B is the mid-point of \(\overline{XQ}\).

(5) Now, point X, A and B separates the line segment \(\overline{PQ}\) in the 4 equal parts.

(6) By actual measurement, we find that \(\overline{PA}\) = \(\overline{AX}\) = \(\overline{XB}\) = \(\overline{BQ}\) =2.5cm

** **

*Question 5: *

*Draw a circle with \(\overline{AB}\) of length 5 cm as diameter,.*

*Answer:*

construction steps:

(1) \(\overline{AB}\) a line segment is drawn = 5 cm.

(2) \(\overline{AB}\) perpendicular bisector is drawn which cuts, it at X. Thus X is the mid-point of AB.

(3) Assuming X as centre and XA or XB as radius a circle is drawn where diameter is the line segment \(\overline{AB}\)

* *

*Question 6: *

*Draw a circle with centre O and radius 5 cm. Draw any chord \(\overline{PQ}\). Construct the perpendicular bisector \(\overline{PQ}\) and examine if it passes through O.*

*Answer:*

Steps of construction:

(i) With centre O draw a circle and radius 5 cm.

(ii)Chord \(\overline{PQ}\) is drawn.

(iii) Taking P and Q as centers and radius more than half of \(\overline{PQ}\), Make 2 arcs which cut each other at Y and X.

(iv) XY is joined. Then XY is the perpendicular bisector of \(\overline{PQ}\).

(v) Through the centre O of the circle this perpendicular bisector of \(\overline{PQ}\) passes.

*Question 7: *

*Question 6 is repeated, if \(\overline{PQ}\) be a diameter*

*Answer:*** **

** **

Construction steps:

(1)Draw a circle with centre O and radius 5 cm.

(2) \(\overline{PQ}\) diameter is drawn

(3)P and Q is taken as centers and radius more than half of it, Make 2 arcs which intersect each other at X and Y.

(4)Join XY . Then XY is the perpendicular bisector of \(\overline{PQ}\).

(5)We observe that this perpendicular bisector of \(\overline{PQ}\) passes through the centre O of the circle.

*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:

(i) Draw the circle with O and radius 4 cm.

(ii) Draw any two chords AB and CD in this circle.

(iii) Taking A and B as centers and radius more than half AB, draw two arcs which intersect each other at E and F.

(iv) Join EF. Thus EF is the perpendicular bisector of chord CD.

(v) Similarly draw GH the perpendicular bisector of chord CD.

(vi) These two perpendicular bisectors meet at O, the centre of the 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 OA and OB. Let them meet at P. Is PA = PB?*

*Answer:*

Steps of construction:

(I)Draw any angle with vertex O.

(ii) Take a point A on one of its arms and B on another such that OA = OB.

(iii) Draw perpendicular bisector of OA and OB.

(iv) Let them meet at P. Join PA and PB.

(v) With the help of divider, we check that PA PB.