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Understanding Quadrilaterals - Special Parallelograms (Rhombus, Rectangle, Square)

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Rhombus is a parallelogram where all sides are of equal length. Its diagonals are perpendicular bisectors of each other, meaning they intersect at 90∘90^\circ and divide each other into two equal halves.

Rhombus ABCD with perpendicular diagonals intersecting at O
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A Rectangle is a parallelogram with four right angles (90∘90^\circ). Because it is a parallelogram, opposite sides are equal. Crucially, the diagonals of a rectangle are equal in length (AC=BDAC = BD).

Rectangle ABCD showing equal diagonals and 90 degree corners
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A Square is a special parallelogram that is both a rhombus and a rectangle. It has four equal sides and four right angles. Its diagonals are equal and bisect each other at 90∘90^\circ.

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Relationship Hierarchy: Every Square is a Rectangle and a Rhombus. Every Rectangle and Rhombus is a Parallelogram. Every Parallelogram is a Quadrilateral.

📐Formulae

Perimeter of a Rhombus or Square: P=4×sP = 4 \times s (where ss is the side length)

Area of a Rhombus: A=12×d1×d2A = \frac{1}{2} \times d_1 \times d_2 (where d1d_1 and d2d_2 are the lengths of the diagonals)

Perimeter of a Rectangle: P=2(l+b)P = 2(l + b) (where ll is length and bb is breadth)

Area of a Rectangle: A=l×bA = l \times b

Diagonal of a Rectangle: d=l2+b2d = \sqrt{l^2 + b^2} (derived from Pythagoras Theorem)

Area of a Square: A=s2A = s^2 or A=12×d2A = \frac{1}{2} \times d^2 (where dd is the diagonal)

💡Examples

Problem 1:

In a rhombus ABCDABCD, the diagonals ACAC and BDBD intersect at point OO. If OA=3OA = 3 cm and OB=4OB = 4 cm, find the length of the side ABAB.

Solution:

  1. In a rhombus, diagonals bisect each other at 90∘90^{\circ}. Therefore, △AOB\triangle AOB is a right-angled triangle with ∠AOB=90∘\angle AOB = 90^{\circ}.
  2. Using Pythagoras Theorem in △AOB\triangle AOB: AB2=OA2+OB2AB^2 = OA^2 + OB^2 AB2=32+42AB^2 = 3^2 + 4^2 AB2=9+16=25AB^2 = 9 + 16 = 25
  3. AB=25=5AB = \sqrt{25} = 5 cm.

Explanation:

Since the diagonals of a rhombus are perpendicular bisectors, they create four right-angled triangles at the intersection. We use the legs of one triangle (33 and 44) to find the hypotenuse, which is the side of the rhombus.

Problem 2:

RENTRENT is a rectangle. Its diagonals meet at OO. Find xx if OR=2x+4OR = 2x + 4 and OT=3x+1OT = 3x + 1.

Solution:

  1. In a rectangle, the diagonals are equal in length (RN=ETRN = ET).
  2. Since diagonals bisect each other, their halves are also equal. Therefore, OR=OTOR = OT.
  3. Set up the equation: 3x+1=2x+43x + 1 = 2x + 4.
  4. Subtract 2x2x from both sides: x+1=4x + 1 = 4.
  5. Subtract 11 from both sides: x=3x = 3.

Explanation:

We use the property that diagonals of a rectangle are equal and bisect each other, which implies that the distance from the center to any vertex is the same.

Problem 3:

In the given square PQRSPQRS, find the value of ∠SOT\angle SOT where OO is the intersection of diagonals and TT is a point on SRSR such that OT⊥SROT \perp SR.

Square PQRS with diagonals meeting at O and altitude OT

Solution:

1. In a square, diagonals intersect at 90∘. So, ∠SOR=90∘1. \text{ In a square, diagonals intersect at } 90^\circ \text{. So, } \angle SOR = 90^\circ 2. Since PQRS is a square, △SOR is an isosceles right triangle (OS=OR).2. \text{ Since } PQRS \text{ is a square, } \triangle SOR \text{ is an isosceles right triangle (} OS = OR \text{).} 3. OT⊥SR implies OT is the altitude from vertex O to the base SR.3. \text{ } OT \perp SR \text{ implies } OT \text{ is the altitude from vertex } O \text{ to the base } SR. 4. In an isosceles triangle, the altitude to the base bisects the vertex angle.4. \text{ In an isosceles triangle, the altitude to the base bisects the vertex angle.} ∠SOT=12×∠SOR=12×90∘=45∘\angle SOT = \frac{1}{2} \times \angle SOR = \frac{1}{2} \times 90^\circ = 45^\circ

Explanation:

The diagonals of a square are perpendicular. Since the triangle formed by the intersection point and one side is isosceles, the altitude drawn to that side bisects the 90∘90^\circ angle at the center.

Problem 4:

In rectangle ABCDABCD, the diagonals ACAC and BDBD meet at OO. If ∠BOC=70∘\angle BOC = 70^\circ, find ∠ODA\angle ODA.

Rectangle ABCD with diagonals intersecting at O and angle BOC labeled

Solution:

1. Diagonals of a rectangle are equal and bisect each other, so OA=OB=OC=OD.1. \text{ Diagonals of a rectangle are equal and bisect each other, so } OA = OB = OC = OD. 2. In △BOC, since OB=OC, it is an isosceles triangle.2. \text{ In } \triangle BOC, \text{ since } OB = OC, \text{ it is an isosceles triangle.} 3. ∠BOC=70∘. Vertically opposite angle ∠AOD=∠BOC=70∘.3. \text{ } \angle BOC = 70^\circ \text{. Vertically opposite angle } \angle AOD = \angle BOC = 70^\circ. 4. In △AOD, since OA=OD, let ∠ODA=∠OAD=x.4. \text{ In } \triangle AOD, \text{ since } OA = OD, \text{ let } \angle ODA = \angle OAD = x. 5. In △AOD:x+x+70∘=180∘5. \text{ In } \triangle AOD: x + x + 70^\circ = 180^\circ 2x=110∘  ⟹  x=55∘2x = 110^\circ \implies x = 55^\circ Thus, ∠ODA=55∘\text{Thus, } \angle ODA = 55^\circ

Explanation:

We use the property that diagonals of a rectangle are equal and bisect each other to identify isosceles triangles. Then, we apply the Angle Sum Property of a triangle.