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Quadrilaterals - Conditions for a Quadrilateral to be a Parallelogram

Grade 9CBSE

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

🔑Concepts

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Definition of a Parallelogram: A quadrilateral is defined as a parallelogram if both pairs of its opposite sides are parallel. Visually, this creates a closed figure where side ABAB is parallel to CDCD and side BCBC is parallel to DADA.

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Opposite Sides Property (Theorem 8.2): If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram. In a visual representation of quadrilateral ABCDABCD, if length AB=CDAB = CD and BC=DABC = DA, the shape is guaranteed to be a parallelogram.

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Opposite Angles Property (Theorem 8.4): A quadrilateral is a parallelogram if each pair of its opposite angles is equal. For example, if ∠A=∠C\angle A = \angle C and ∠B=∠D\angle B = \angle D, the quadrilateral ABCDABCD must be a parallelogram.

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Diagonals Bisecting Property (Theorem 8.6): If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. This means if the diagonals ACAC and BDBD intersect at point OO such that AO=OCAO = OC and BO=ODBO = OD, then ABCDABCD is a parallelogram.

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The One-Pair Condition (Theorem 8.8): A quadrilateral is a parallelogram if one pair of opposite sides is both equal and parallel. This is a shortcut condition: if you can prove AB=CDAB = CD and AB∥CDAB \parallel CD, you do not need to check the other pair of sides.

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Consecutive Interior Angles: In any parallelogram, the sum of any two consecutive (adjacent) angles is supplementary, meaning they add up to 180∘180^\circ. Visually, if you extend the sides, these angles behave like interior angles between parallel lines: ∠A+∠B=180∘\angle A + \angle B = 180^\circ, ∠B+∠C=180∘\angle B + \angle C = 180^\circ, etc.

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Angle Sum Property: Like all quadrilaterals, the sum of all four interior angles in a parallelogram is always 360∘360^\circ, expressed as ∠A+∠B+∠C+∠D=360∘\angle A + \angle B + \angle C + \angle D = 360^\circ.

📐Formulae

Opposite sides: AB=CDAB = CD and BC=DABC = DA

Opposite angles: ∠A=∠C\angle A = \angle C and ∠B=∠D\angle B = \angle D

Sum of adjacent angles: ∠A+∠B=180∘\angle A + \angle B = 180^\circ or ∠B+∠C=180∘\angle B + \angle C = 180^\circ

Diagonal bisection: AO=OCAO = OC and BO=ODBO = OD (where OO is the intersection point)

Perimeter of a parallelogram: P=2(a+b)P = 2(a + b) where aa and bb are adjacent sides

Area of a parallelogram: A=base×heightA = \text{base} \times \text{height}

💡Examples

Problem 1:

In a quadrilateral ABCDABCD, ∠A=115∘\angle A = 115^\circ, ∠B=65∘\angle B = 65^\circ, and ∠C=115∘\angle C = 115^\circ. Determine if ABCDABCD is a parallelogram.

Solution:

Step 1: Use the angle sum property of a quadrilateral to find the fourth angle ∠D\angle D. ∠A+∠B+∠C+∠D=360∘\angle A + \angle B + \angle C + \angle D = 360^\circ Step 2: Substitute the known values into the equation: 115∘+65∘+115∘+∠D=360∘115^\circ + 65^\circ + 115^\circ + \angle D = 360^\circ 295∘+∠D=360∘295^\circ + \angle D = 360^\circ Step 3: Solve for ∠D\angle D: ∠D=360∘−295∘=65∘\angle D = 360^\circ - 295^\circ = 65^\circ Step 4: Check the pairs of opposite angles. ∠A=115∘\angle A = 115^\circ and ∠C=115∘\angle C = 115^\circ (Equal). ∠B=65∘\angle B = 65^\circ and ∠D=65∘\angle D = 65^\circ (Equal). Since both pairs of opposite angles are equal, ABCDABCD is a parallelogram.

Explanation:

This solution applies the theorem that a quadrilateral is a parallelogram if its opposite angles are equal. We first calculate the missing angle to verify the condition for both pairs.

Problem 2:

In quadrilateral PQRSPQRS, diagonals PRPR and QSQS intersect at OO. Given PO=x+2PO = x + 2, OR=10OR = 10, QO=y−3QO = y - 3, and OS=12OS = 12. Find the values of xx and yy that make PQRSPQRS a parallelogram.

Solution:

Step 1: For PQRSPQRS to be a parallelogram, the diagonals must bisect each other. This means PO=ORPO = OR and QO=OSQO = OS. Step 2: Set up the equation for diagonal PRPR: x+2=10x + 2 = 10 x=10−2=8x = 10 - 2 = 8 Step 3: Set up the equation for diagonal QSQS: y−3=12y - 3 = 12 y=12+3=15y = 12 + 3 = 15 Step 4: Therefore, for x=8x = 8 and y=15y = 15, the diagonals bisect each other, making PQRSPQRS a parallelogram.

Explanation:

This example uses the diagonal bisection property. By equating the two segments of each diagonal, we ensure the intersection point is the midpoint for both, which is a necessary and sufficient condition for a parallelogram.