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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 special parallelogram where all four sides are equal in length. Visually, it looks like a tilted square. Its most important property is that the diagonals bisect each other at right angles (90∘90^{\circ}), meaning they form a perpendicular '+' shape at the center.

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A Rectangle is a parallelogram where every interior angle is a right angle (90∘90^{\circ}). Visually, it consists of two pairs of equal opposite sides. A defining property of a rectangle is that its diagonals are equal in length, so if you measure from one corner to the opposite, both diagonal lines will be identical.

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A Square is the most specific special parallelogram; it is both a rectangle (four 90∘90^{\circ} angles) and a rhombus (four equal sides). Visually, it is perfectly symmetrical. Its diagonals are equal in length and bisect each other at 90∘90^{\circ}.

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Diagonal Properties: In all parallelograms, diagonals bisect each other. However, in a Rhombus, they are perpendicular. In a Rectangle, they are equal. In a Square, they are both perpendicular and equal.

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Angle Relationships: In these parallelograms, adjacent angles are supplementary, meaning they add up to 180∘180^{\circ} (e.g., in a rhombus, ∠A+∠B=180∘\angle A + \angle B = 180^{\circ}). Opposite angles are always equal.

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Hierarchical Relationship: It is helpful to visualize a hierarchy where the Square sits at the bottom as the most restricted shape. Every Square is a Rhombus and a Rectangle, but not every Rhombus or Rectangle is a Square. All of them are types of Parallelograms.

📐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.