Question 19

We will check each statement about quadrilaterals to see if it is true or false, and then explain why.

Step 1 — Statement (i) analysis

This statement says a quadrilateral with equal and bisecting diagonals must be a square. Let us consider a quadrilateral where diagonals are equal. Let us also consider that these diagonals bisect each other. When diagonals bisect each other, the quadrilateral is always a parallelogram. When a parallelogram has equal diagonals, it is a rectangle. A rectangle has all angles equal to 90 degrees. However, a rectangle does not always have all sides equal. A square is a special type of rectangle where all sides are equal. Since a rectangle is not always a square, the statement is not always true.

(i) False.

Step 2 — Statement (ii) analysis

This statement says a quadrilateral with three right angles must be a rectangle. Let the four angles of the quadrilateral be $\angle A$, $\angle B$, $\angle C$, and $\angle D$. The sum of all angles in any quadrilateral is $\mathbf{360}$ degrees. We are given that three angles are right angles, meaning they are $\mathbf{90}$ degrees each. Let $\angle A = 90^\circ$, $\angle B = 90^\circ$, and $\angle C = 90^\circ$. We can find the fourth angle, $\angle D$.

$\angle A + \angle B + \angle C + \angle D = 360^\circ$

$90^\circ + 90^\circ + 90^\circ + \angle D = 360^\circ$

$270^\circ + \angle D = 360^\circ$

$\angle D = 360^\circ - 270^\circ$

$\boxed{\angle D = 90^\circ}$

Since all four angles are $\mathbf{90}$ degrees, the quadrilateral is a rectangle.

(ii) True.

Step 3 — Statement (iii) analysis

This statement says a quadrilateral whose diagonals bisect each other must be a parallelogram. Let the quadrilateral be ABCD. Let its diagonals AC and BD intersect at point O. "Diagonals bisect each other" means that O is the midpoint of both AC and BD. So, $\mathbf{AO = OC}$ and $\mathbf{BO = OD}$.

Let us look at $\triangle AOB$ and $\triangle COD$. We know $\mathbf{AO = OC}$ (given). We know $\mathbf{BO = OD}$ (given). The angles $\angle AOB$ and $\angle COD$ are vertically opposite angles. Vertically opposite angles are always equal, so $\mathbf{\angle AOB = \angle COD}$. By the SAS (Side-Angle-Side) congruence rule, $\triangle AOB \cong \triangle COD$. This means their corresponding parts are equal. So, $\mathbf{AB = CD}$. Also, $\mathbf{\angle OAB = \angle OCD}$. These are alternate interior angles. If alternate interior angles are equal, then the lines AB and CD must be parallel. So, $\mathbf{AB \parallel CD}$.

Similarly, let us look at $\triangle AOD$ and $\triangle COB$. We know $\mathbf{AO = OC}$ (given). We know $\mathbf{DO = OB}$ (given). The angles $\angle AOD$ and $\angle COB$ are vertically opposite angles. So, $\mathbf{\angle AOD = \angle COB}$. By the SAS congruence rule, $\triangle AOD \cong \triangle COB$. This means their corresponding parts are equal. So, $\mathbf{AD = CB}$. Also, $\mathbf{\angle ODA = \angle OBC}$. These are alternate interior angles. If alternate interior angles are equal, then the lines AD and CB must be parallel. So, $\mathbf{AD \parallel CB}$. Since both pairs of opposite sides are parallel ($\mathbf{AB \parallel CD}$ and $\mathbf{AD \parallel CB}$), the quadrilateral ABCD is a parallelogram.

(iii) True.

Step 4 — Statement (iv) analysis

This statement says a quadrilateral whose diagonals are perpendicular to each other must be a rhombus. A rhombus is a quadrilateral where all four sides are equal. A rhombus does have perpendicular diagonals. However, other quadrilaterals also have perpendicular diagonals. For example, a kite has diagonals that are perpendicular to each other. But a kite does not necessarily have all four sides equal. A kite only has two pairs of equal-length adjacent sides. Since a kite is not always a rhombus, the statement is not always true.

(iv) False.

Step 5 — Statement (v) analysis

This statement says a quadrilateral in which the opposite angles are equal must be a parallelogram. Let the quadrilateral be ABCD. We are given that opposite angles are equal. So, $\mathbf{\angle A = \angle C}$ and $\mathbf{\angle B = \angle D}$. The sum of all angles in a quadrilateral is $\mathbf{360}$ degrees.

$\angle A + \angle B + \angle C + \angle D = 360^\circ$

Substitute $\angle C$ with $\angle A$ and $\angle D$ with $\angle B$:

$\angle A + \angle B + \angle A + \angle B = 360^\circ$

$2\angle A + 2\angle B = 360^\circ$

Divide by 2:

$\angle A + \angle B = 180^\circ$

Angles $\angle A$ and $\angle B$ are consecutive interior angles if we consider AD and BC as parallel lines cut by transversal AB. If the sum of consecutive interior angles is $\mathbf{180}$ degrees, then the lines are parallel. So, $\mathbf{AD \parallel BC}$. Similarly, since $\angle B + \angle C = \angle B + \angle A = 180^\circ$, then $\mathbf{AB \parallel DC}$. Since both pairs of opposite sides are parallel, the quadrilateral ABCD is a parallelogram.

(v) True.

Step 6 — Statement (vi) analysis

This statement says a quadrilateral in which all the angles are equal is a rectangle. Let the quadrilateral have four equal angles. Let each angle be $x$. The sum of all angles in a quadrilateral is $\mathbf{360}$ degrees.

$x + x + x + x = 360^\circ$

$4x = 360^\circ$

$x = \frac{360^\circ}{4}$

$\boxed{x = 90^\circ}$

Since all four angles are $\mathbf{90}$ degrees, the quadrilateral is a rectangle.

(vi) True.

Step 7 — Statement (vii) analysis

This statement says isosceles trapeziums are parallelograms. An isosceles trapezium (or trapezoid) is a quadrilateral with exactly one pair of parallel sides. The non-parallel sides are equal in length. A parallelogram is a quadrilateral with two pairs of parallel sides. Since an isosceles trapezium has only one pair of parallel sides, it cannot be a parallelogram, which requires two pairs of parallel sides.

(vii) False.

Answer

(i) False. (ii) True. (iii) True. (iv) False. (v) True. (vi) True. (vii) False.