Question 3

A quadrilateral formed by joining the ends of two intersecting diagonals that are equal in length and perpendicular to each other is a square.

Step 1 — Understanding the given conditions

Let us consider the two rubber bands as the diagonals of a quadrilateral. Let these diagonals be AC and BD. We are told that these diagonals are perpendicular to each other. This means they intersect at a 90-degree angle. We are also told that these diagonals are of equal length. This means the length of AC is equal to the length of BD. We then join the ends of these rubber bands to form a quadrilateral ABCD.

Step 2 — Properties of the diagonals

Let O be the point where the diagonals AC and BD intersect. We are given that the diagonals are perpendicular ($AC \perp BD$). We are given that the diagonals are of equal length ($AC = BD$). When we form a quadrilateral by joining the ends of two intersecting line segments that act as diagonals, for the resulting figure to be a standard convex quadrilateral (like a a square, rectangle, or rhombus), the diagonals must bisect each other. So, point O is the midpoint of AC, which means $AO = OC$. Point O is also the midpoint of BD, which means $BO = OD$. Since $AC = BD$, and O is the midpoint of both, it means all the half-diagonals must be equal. So, $AO = OC = BO = OD$.

Step 3 — Identifying the quadrilateral

Let us look at the four triangles formed by the diagonals: $\triangle AOB$, $\triangle BOC$, $\triangle COD$, and $\triangle DOA$. Since the diagonals are perpendicular, all these triangles are right-angled triangles at O. For example, in $\triangle AOB$, $\angle AOB = 90^\circ$. We found that $AO = BO$ (from Step 2). So, $\triangle AOB$ is an isosceles right-angled triangle. This means the base angles are equal: $\angle OAB = \angle OBA$. The sum of angles in a triangle is $180^\circ$. So, $\angle OAB + \angle OBA + \angle AOB = 180^\circ$. $2 \times \angle OAB + 90^\circ = 180^\circ$ $2 \times \angle OAB = 180^\circ - 90^\circ$ $2 \times \angle OAB = 90^\circ$

$\boxed{\angle OAB = 45^\circ}$ Similarly, all four triangles ($\triangle AOB$, $\triangle BOC$, $\triangle COD$, $\triangle DOA$) are congruent isosceles right-angled triangles. This means all the angles at the vertices of the quadrilateral are $45^\circ + 45^\circ = 90^\circ$. For example, $\angle DAB = \angle DAO + \angle OAB = 45^\circ + 45^\circ = 90^\circ$. So, all four angles of the quadrilateral are right angles. Also, since the triangles are congruent, all sides of the quadrilateral are equal in length. For example, $AB = BC = CD = DA$. A quadrilateral with all four sides equal and all four angles equal to $90^\circ$ is a square.

Answer

(i) The quadrilateral formed is a square. (ii) Justification:

1. The two rubber bands act as diagonals that are of equal length and are perpendicular to each other.
2. When forming a convex quadrilateral by joining the ends of intersecting diagonals, the diagonals bisect each other.
3. Since the diagonals are equal and bisect each other, all four half-diagonals (from the center to each vertex) are equal in length.
4. Because the diagonals are perpendicular, the four triangles formed by the diagonals are congruent isosceles right-angled triangles.
5. This means all four sides of the quadrilateral are equal in length (making it a rhombus).
6. Also, all four interior angles of the quadrilateral are $90^\circ$ (making it a rectangle).
7. A quadrilateral that is both a rhombus and a rectangle is a square.