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Understanding Quadrilaterals - Properties of Parallelogram, Rhombus, Rectangle, and Square

Grade 8ICSE

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

🔑Concepts

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A Parallelogram is a quadrilateral where both pairs of opposite sides are parallel and equal in length. Visually, the opposite angles are congruent (∠A=∠C\angle A = \angle C and ∠B=∠D\angle B = \angle D), and any two adjacent angles are supplementary, meaning they add up to 180∘180^\circ (∠A+∠B=180∘\angle A + \angle B = 180^\circ).

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The diagonals of a parallelogram bisect each other, meaning they cut each other into two equal parts at the point of intersection. Visually, if diagonals ACAC and BDBD meet at OO, then AO=OCAO = OC and BO=ODBO = OD.

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A Rectangle is a special type of parallelogram where every interior angle is a right angle (90∘90^\circ). Beyond standard parallelogram properties, a rectangle's diagonals are equal in length (AC=BDAC = BD). Visually, it looks like a stretched square with four perfectly square corners.

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A Rhombus is a parallelogram where all four sides are of equal length. Its unique visual property is that its diagonals are perpendicular bisectors of each other, intersecting at exactly 90∘90^\circ. Furthermore, the diagonals bisect the vertex angles of the rhombus.

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A Square is the most regular quadrilateral, combining all properties of a rectangle and a rhombus. It has four equal sides and four 90∘90^\circ angles. Visually, its diagonals are equal in length, bisect each other at right angles (90∘90^\circ), and bisect the corner angles into two 45∘45^\circ angles.

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The Angle Sum Property of any quadrilateral states that the sum of all four interior angles is exactly 360∘360^\circ. This can be visualized by drawing a diagonal to split the quadrilateral into two triangles, each contributing 180∘180^\circ.

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The perimeter of these shapes is the total boundary length. For a parallelogram or rectangle with sides aa and bb, it is calculated as 2(a+b)2(a + b). For a square or rhombus with side ss, it is 4s4s.

📐Formulae

Sum of interior angles of a quadrilateral =360∘= 360^\circ

Perimeter of a Parallelogram/Rectangle =2(length+breadth)= 2(length + breadth)

Perimeter of a Rhombus/Square =4×side= 4 \times side

Area of a Parallelogram =base×height= base \times height

Area of a Rectangle =length×breadth= length \times breadth

Area of a Rhombus =12×d1×d2= \frac{1}{2} \times d_1 \times d_2 (where d1,d2d_1, d_2 are diagonals)

Area of a Square =(side)2=12×(diagonal)2= (side)^2 = \frac{1}{2} \times (diagonal)^2

Diagonal of a Square =side×2= side \times \sqrt{2}

💡Examples

Problem 1:

In a parallelogram ABCDABCD, the measure of ∠A\angle A is (3x−20)∘(3x - 20)^\circ and the measure of ∠B\angle B is (x+40)∘(x + 40)^\circ. Find the measure of all the angles of the parallelogram.

Solution:

Step 1: Identify the relationship between ∠A\angle A and ∠B\angle B. In a parallelogram, adjacent angles are supplementary. ∠A+∠B=180∘\angle A + \angle B = 180^\circ Step 2: Substitute the given expressions: (3x−20)+(x+40)=180(3x - 20) + (x + 40) = 180 4x+20=1804x + 20 = 180 Step 3: Solve for xx: 4x=180−204x = 180 - 20 4x=1604x = 160 x=40x = 40 Step 4: Calculate individual angles: ∠A=3(40)−20=120−20=100∘\angle A = 3(40) - 20 = 120 - 20 = 100^\circ ∠B=40+40=80∘\angle B = 40 + 40 = 80^\circ Step 5: Use the property that opposite angles are equal: ∠C=∠A=100∘\angle C = \angle A = 100^\circ ∠D=∠B=80∘\angle D = \angle B = 80^\circ

Explanation:

This solution uses the property that adjacent angles in a parallelogram add up to 180∘180^\circ to form a linear equation. Once xx is found, we apply the property that opposite angles are congruent.

Problem 2:

The diagonals of a rhombus are 16 cm16 \text{ cm} and 12 cm12 \text{ cm}. Find the length of the side of the rhombus.

Solution:

Step 1: Recall that diagonals of a rhombus bisect each other at right angles (90∘90^\circ). Let the diagonals be d1=16 cmd_1 = 16 \text{ cm} and d2=12 cmd_2 = 12 \text{ cm}. Step 2: The segments of the diagonals from the center to the vertices are half the total length: Half-diagonal 1=162=8 cm\text{Half-diagonal 1} = \frac{16}{2} = 8 \text{ cm} Half-diagonal 2=122=6 cm\text{Half-diagonal 2} = \frac{12}{2} = 6 \text{ cm} Step 3: These segments form a right-angled triangle with the side of the rhombus (ss) as the hypotenuse. Apply Pythagoras Theorem: s2=82+62s^2 = 8^2 + 6^2 s2=64+36s^2 = 64 + 36 s2=100s^2 = 100 Step 4: Take the square root: s=100=10 cms = \sqrt{100} = 10 \text{ cm}

Explanation:

This problem uses the property that rhombus diagonals are perpendicular bisectors. By considering one of the four small right-angled triangles formed inside the rhombus, we use the Pythagorean theorem to find the side length.