Constructions and Tilings | FIO

Question 11

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

Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • An angle bisector divides a given angle into two equal halves.
  • Using a rope, we can measure and mark equal distances along both arms from vertex OO to obtain points AA and BB.
  • Finding the midpoint MM of the segment ABAB creates two congruent triangles (DeltaOAMcongDeltaOBM\\Delta OAM \\cong \\Delta OBM by the SSS criterion), ensuring angleAOM=angleBOM\\angle AOM = \\angle BOM, so ray OMOM is the angle bisector.

Step 1 · Mark Equal Distances on Both Arms

Consider an angle XOY\angle XOY with vertex OO.Diagram 1

Fix one end of a rope at vertex OO. Stretch the rope taut along arm OXOX to a fixed length L1L_1 and mark point AA. Without changing the length, swing the rope to arm OYOY and mark point BB at the same distance L1L_1.

OA=L1OB=L1\begin{aligned} OA &= L_1 \\ OB &= L_1 \end{aligned}

OA=OB\boxed{OA = OB}

Step 2 · Find the Midpoint Between the Two Points

Fix a rope between point AA and point BB. Fold or measure the rope to find its exact middle point MM.

AM=L2BM=L2\begin{aligned} AM &= L_2 \\ BM &= L_2 \end{aligned}

AM=BM\boxed{AM = BM}

Step 3 · Draw and Verify the Angle Bisector

Join point OO to point MM with a straight line OMOM.

In ΔOAM\Delta OAM and ΔOBM\Delta OBM:

  • OA=OBOA = OB
  • AM=BMAM = BM
  • OM=OMOM = OM (common side)

By SSS congruence criterion: ΔOAMΔOBM\Delta OAM \cong \Delta OBM

Therefore, the corresponding angles are equal: AOM=BOM\angle AOM = \angle BOM

Thus, OMOM is the angle bisector of XOY\angle XOY.

OM is the angle bisector\boxed{\text{OM is the angle bisector}}

Answer

Fix a rope at vertex OO to mark points AA and BB such that OA=OBOA = OB. Then find the midpoint MM of ABAB. The straight line joining OO to MM is the angle bisector.

Common Mistakes
  • Unequal Arm Lengths: Failing to keep the rope length strictly constant when marking points AA and BB (OAOBOA \neq OB) will prevent the triangles from being congruent.
  • Slack in the Rope: The rope must be pulled straight and taut; any slack leads to incorrect distance measurements and an inaccurate bisector.

More questions in FIO

Q1

When constructing the perpendicular bisector, is it necessary to have the same radius for the arcs above and below XYXY? Explore this through construction, and then justify your answer.

[Hint 1: Any point that is of the same distance from XX and YY lies on the perpendicular bisector.

Hint 2: We can draw the whole line if any two of its points are known.]

Q2

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

Q3

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.

Q4

Recreate this design using only a ruler and compass —

Q5

Justify why AB\text{AB} in Fig. 6.4 is the perpendicular bisector.

Q6

Can you think of different methods to construct a 9090^\circ angle at a given point on a line using a rope?

Q7

Construct at least 4 different angles. Draw their bisectors.

Q8

Construct the 8-petalled figure shown in Fig. 6.5.

Q9

In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side, as shown in the figure, will the line OCOC still be an angle bisector? Explore this through construction, and then justify your answer.

Q10

What are the other angles that can be constructed using angle bisection? Can you construct 65.565.5^\circ angle?

Q11

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

Q12

Construct the following figure.

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

Q13

Construct at least 4 different angles in different orientations without taking any measurement. Make a copy of all these angles.

Q14

Construct the Fig. 6.6.

Q15

Construct 4 pairs of parallel lines in different orientations.

Q16

Construct the following figure.

Q17

Use support lines in Fig. 6.11 to construct a pointed arch. Make different arches, by changing the radius of the arcs.

Q18

Make your own arch designs.

Q19

Construct the following figures:

Q20

Optical Illusion: Do you notice anything interesting about the following figure? How does this happen? Recreate this in your notebook.

Q21

Construct this figure.

[Hint: Find the angles in this figure.]

Q22

Draw a line ll and mark a point PP anywhere outside the line. Construct a perpendicular to the given line ll through PP.

[Hint: Find a line segment on ll whose perpendicular bisector passes through PP.]

Q23

How can the tangram pieces be rearranged to form each of the following figures?

Q24

Are the following tilings possible?

← Back to Constructions and Tilings