Question 1

When constructing the perpendicular bisector, is it necessary to have the same radius for the arcs above and below XY? Explore this through construction, and then justify your answer.
[Hint 1: Any point that is of the same distance from X and Y lies on the perpendicular bisector.
Hint 2: We can draw the whole line if any two of its points are known.]

Accepted Answer

Draw a line segment XY.
Choose distances k and k’ which are slightly greater than half of the distance XY.
With centres at X and Y, draw arcs of radius ‘k’ below XY.
With centres at X and Y, draw arcs of radius ‘k’’ below XY.
Let the arcs above XY intersect at A, and the arcs below XY intersect at B.
Join A and B. Let AB intersect XY at O.
Join AX, AY, BX, and BY.
∆ABX and ∆ABY are congruent because AX = AY = k, BX = BY = k’, and AB is common.
∴ ∠XAO = ∠YAO
∆AOX and ∆AOY are congruent because AX = AY = k, ∠XAO = ∠YAO, and OA is common.
∴ OX = OY and ∠AOX = ∠AOY
Also, ∠AOX + ∠AOY = 180°
∴ 2∠AOX = 180° or ∠AOX = 90°
∴ OX = OY and ∠AOX = ∠AOY = 90°.
∴ AB is the perpendicular bisector of the line XY.
Here, A and B are points that are of the same distance from X and Y.
Thus, any point that is of the same distance from X and Y lies on the perpendicular bisector.

Question 2

Is it necessary to construct the pairs of arcs above and below XY? Instead, can we construct both pairs of arcs on the same side of XY? Explore this through construction, and then justify your answer.

Accepted Answer

Draw a line segment XY.
Choose distances k and k’ which are slightly greater than half of the distance XY.
With centres at X and Y, draw arcs of radius ‘k’ above XY.
With centres at A and Y, draw arcs of radius ‘k’’ above XY.
Let the arcs intersect at the points A and B.
Join A and B and produce this line to intersect XY at O.
Join AX, AY, BX, and BY.
∆ABX and ∆ABY are congruent because AX = AY = k, BX = BY = k’, and AB is common.
∴ ∠XAO = ∠YAO
∆AOX and ∆AOY are congruent because AX = AY = k, ∠XAO = ∠YAO, and OA is common.
∴ OX = OY and ∠AOX = ∠AOY
Also, ∠AOX + ∠AOY = 180°
∴ 2∠AOX = 180° or ∠AOX = 90°
∴ OX = OY and ∠AOX = ∠AOY = 90°
∴ AB is the perpendicular bisector of the line XY.
Here, the pairs of arcs are both on the same side of XY.
∴ It is not necessary to construct the pairs of arcs above and below XY.

Question 3

While constructing one pair of intersecting arcs, is it necessary that we use the same radii for both of them? Explore this through construction, and then justify your answer.

Accepted Answer

Draw a line segment XY.
With centres at X and Y, draw arcs of unequal radii, say k and k’.
Let the arcs intersect at the point A.
Let PQ be the perpendicular bisector of XY.
Let R be any point on PQ.
Join RX and RY.
∆ROX and ∆ROY are congruent, because OX = OY, ∠ROX = ∠ROY, and OR is common.
∴ RX = RY
∴ For any point R on the perpendicular bisector, we have RX = RY.
∴ Every point on the perpendicular bisector is equidistant from X and Y.
Since AX ≠ AY, the point A is not on the perpendicular bisector.
Thus, to construct a perpendicular bisector, we must use the same radius for both arcs of a pair of arcs.

Question 4

Recreate this design using only a ruler and compass-

Accepted Answer

Let ABCD be a square. We draw perpendicular bisectors of the sides AB and BC as shown in the figure.
Let the perpendicular bisectors intersect the square at the points P, Q, R, and S.
With centres at P, Q, R, and S, draw semicircles in the square with radius equal to AP.
Colour the boundary of the design using a coloured pencil.
This will make the design stand out from the supporting lines and curve.
Figure It Out (Page 142)

Question 5

Come up with a method to construct the angle bisector using a rope.

Accepted Answer

Let ∠XOY be the given angle.
Fix a small pole at the point O.
Take a rope and make a loop at one end.
Mark a point at some distance on the rope.
Fix the loop of the rope at the pole at O and rotate the rope from OX to OY.
Mark points A and B at a fixed distance mark on the rope.
Fix small poles at A and B.
Take a rope and make loops on both ends.
Fix the loops of this piece of rope with poles at A and B.
Mark the midpoint of this rope and hold the rope at the midpoint.
Make both ends of the rope tight and mark the point at the midpoint of the rope.
Let this point be M.
Join AM, BM, and OM.
We have OA = OB and AM = BM.
In ∆OAM and ∆OBM, we have OA = OB, AM = BM, and OM is common.
∴ ∆OAM = ∆OBM (Using SSS-rule)
∴ ∠AOM = ∠BOM
∴ ∠XOM = ∠YOM
∴ OM is the bisector of the given angle ∠XOY.

Question 6

Construct the following figure:
How do we construct the petals so that they are of the maximum possible size within a given square?

Accepted Answer

Draw a line and take points A and B on it.
With A and B as centres, draw semicircles intersecting the lines at the points C, D, E, and F.
With centres at C and D, draw arcs of equal radius to intersect at the point G.
With centres at E and F, draw arcs of equal radius to intersect at the point H.
Join AG and BH.
Take point I and J on AG and BH, respectively, so that AI = BJ = AB.
Join I and J.
Here ABJI is a square.
Using a ruler and a compass, find perpendicular bisectors of the sides AB and BJ.
With centres at K, L, M, and N, draw semicircles in the square with radius equal to AK.
Now, we erase the extra lines and arcs.
The above figure looks as given below:
This is the required figure with 4 petals in a square.
Here we have drawn 4 semicircles in the square, so that the petals are of the maximum possible size.
Figure It Out (Page 147)

Constructions and Tilings