Place two rubber bands perpendicular to each other, forming diagonals of equal length. Join the ends.

What is the quadrilateral that you get? Justify your answer.

Question

Place two rubber bands perpendicular to each other, forming diagonals of equal length. Join the ends. What is the quadrilateral that you get? Justify your answer.

Homi · Accepted Answer

Understand the Question

The two rubber bands form the diagonals of the quadrilateral, denoted as $AC$ and $BD$ .
The given conditions state:
- The diagonals are perpendicular to each other ( $AC \perp BD$ ).
- The diagonals are equal in length ( $AC = BD$ ).
When joining their endpoints to form a regular convex quadrilateral, the diagonals bisect each other at their intersection point $O$ .
We use triangle properties and angle sums to determine the specific type of quadrilateral formed.

Step 1 · Analyze Properties of the Diagonals

Let the two diagonals formed by the rubber bands be $AC$ and $BD$ , intersecting at point $O$ .Given:

Diagonals are perpendicular: $AC \perp BD \implies \angle AOB = \angle BOC = \angle COD = \angle DOA = 90^\circ$
Diagonals are equal: $AC = BD$

Since the diagonals bisect each other: $AO = OC = \dfrac{AC}{2}$ $BO = OD = \dfrac{BD}{2}$

Since $AC = BD$ , all four half-segments are equal: $AO = BO = CO = DO$

Step 2 · Evaluate Side Lengths and Interior Angles

Consider the four right-angled triangles $\triangle AOB$ , $\triangle BOC$ , $\triangle COD$ , and $\triangle DOA$ .

In $\triangle AOB$ : $AO = BO \implies \angle OAB = \angle OBA$

Using angle sum property in $\triangle AOB$ :

\begin{aligned} \angle OAB + \angle OBA + \angle AOB &= 180^\circ \\ 2 \times \angle OAB + 90^\circ &= 180^\circ \\ 2 \times \angle OAB &= 180^\circ - 90^\circ \\ 2 \times \angle OAB &= 90^\circ \\ \angle OAB &= 45^\circ \end{aligned}

Similarly, in all four congruent isosceles right triangles, the base angles are $45^\circ$ .

For each vertex angle of quadrilateral $ABCD$ :

\begin{aligned} \angle DAB &= \angle DAO + \angle OAB = 45^\circ + 45^\circ = 90^\circ \\ \angle ABC &= \angle BCD = \angle CDA = 90^\circ \end{aligned}

By SAS congruence (or Pythagoras theorem in each triangle): $AB = BC = CD = DA$

Step 3 · Identify the Quadrilateral

A quadrilateral having all four sides equal in length and all four interior angles equal to $90^\circ$ is a square.

Answer

The quadrilateral formed is a square.

Common Mistakes

Assuming Rhombus Only: Forgetting that equal diagonals with perpendicular bisectors make the quadrilateral a square, not just a rhombus.
Assuming Rectangle Only: Forgetting that perpendicular diagonals make all sides equal, ensuring it is a square rather than just a general rectangle.

Question 3

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