i. ABCD is a rectangle if Ac = 5 cm, let’s write the length of BD

ii. PQRS is a square, PR and QS are two diagonals intersecting at O. PR = 5cm. Let’s write the length of QO.

iii. Let’s write the measurement of in parallelogram ABCD where

iv. The diagonals AC and BD of rhombus ABCD are intersected at O. Let’s write the measurement of

v. A square is always a rhombus but a rhombus is not always a

vi. A square is always a but a rectangle is not always a

i. We are given that,

There is a rectangle named ABCD, where

AC = 5 cm.

We need to find the length of the other diagonal, BD.

We know that,

A rectangle is a quadrilateral with four right angles having equal opposite sides.

This means,

AD = BC

AB = DC

And

∠ABC = ∠BCD = ∠CDA = ∠DAB = 90°

Take ∆ADC and ∆BAD,

AD = AD [∵, common sides]

DC = AB [∵, opposite sides are equal in rectangle]

∠ADC = ∠BAD [∵, each angle measure 90° in a rectangle]

⇒ ∆ADC ≅ ∆BAD by SAS congruency.

So, by cpct (corresponding parts of congruent triangles),

AC = BD

So, if AC = 5 cm, then BD = 5 cm.

ii. We are given that,

There is a square PQRS.

PR and QS are two diagonals intersecting at O.

PR = 5 cm.

We need to find the length of QO.

We know that,

The square is a geometric shape that belongs to the quadrilateral family because it has 4 sides. The 4 sides are the same length and are parallel to each other.

This implies,

PQ = QR = RS = SP

And,

∠PQR = ∠QRS = ∠RSP = ∠SPQ = 90°

Take ∆PQR and ∆QRS,

PQ = QR [∵, every side is equal in a square]

QR = RS [∵, every side is equal in a square]

∠PQR = ∠QRS [∵, every angle is equal to 90° in a square]

⇒ ∆PQR ≅ ∆QRS by SAS congruency.

So, by cpct (corresponding parts of congruent triangles),

PR = QS

So, if PR = 5 cm, then QS = 5 cm.

Also, we know the property of the square that the diagonals bisects each other.

So, QS = QO + OS

Where, QO = OS.

⇒ QS = QO + QO

⇒ QS = 2 QO

And QS = 5 cm

⇒ QO = 2.5 cm

Thus, QO = 2.5 cm.

iii. We are given that,

There is a parallelogram ABCD.

∠ABC = 60°

We need to find the ∠ADC.

We know that,

A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

This means,

AB = DC & AB ∥ DC

BC = AD & BC ∥ AD

Also,

∠ABC = ∠ADC

∠BCD = ∠DAB

So, if ∠ABC = 60°, then ∠ADC = 60°.

iv. We are given that,

There is a rhombus ABCD.

AC and BD are diagonals of the rhombus intersecting at O.

We need to find ∠AOB.

We know that,

A rhombus is a simple quadrilateral whose four sides all have the same length, with opposite angles being equal. The diagonals bisect each other at right angles.

This means,

AB = BC = CD = DA

Also,

∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

Thus, ∠AOB = 90°.

v. A square is always a rhombus but a rhombus is not always a square.

Explanation – Recall, square and rhombus both have 4 sides that is why they belong to quadrilateral family.

A square has 4 equal sides with interior angles of 90° each.

While, a rhombus has 4 equal sides but opposite interior angles are equal.

So, we can say that, a square can also be called a rhombus since it satisfies rhombus’s properties. But a rhombus cannot be called a square always since it doesn’t have interior angles as 90°.

iv. A square is always a rectangle but a rectangle is not always a square.

Explanation – Recall, square and rectangle both have 4 sides that is why they belong to quadrilateral family.

A square has 4 equal sides with interior angles of 90° each.

While, rectangle has equal opposite sides with interior angles of 90° each. The adjacent sides are not equal.

Since, square is a quadrilateral with all interior angles of 90°, it can be called a rectangle. (Since, it satisfies every property of a rectangle)

But, a rectangle cannot be called a square since every side of the rectangle is not equal.