1
Area(parallelogram) = Area(△ABC)
2
bₚ × hₚ = ½ × bₜ × hₜ
FORMULA CHAPTER
Chapter 13 – Construction of an Equal-Area Parallelogram
WBBSEClass 09Mathematics2 formulas
Complete WBBSE Madhyamik Class 9 equal-area parallelogram construction guide with conditions, notation and step-by-step examples for exam preparation.
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Equal-Area Construction Condition
Area(parallelogram) = Area(△ABC)
SymbolMeaningUnit
△ABCGiven triangleNot applicable
θSpecified angle of the parallelogramdegrees
AreaMeasure of the plane regionsquare units
Worked example
Question: A triangle ABC has area 36 cm². A parallelogram is constructed with one angle equal to 60° and equal in area to the triangle. What must be the area of the parallelogram?
Calculation:
Step 1: Area(△ABC) = 36 cm².
Step 2: By the construction condition, Area(parallelogram) = Area(△ABC).
Step 3: Area(parallelogram) = 36 cm².
Answer: The parallelogram must have area 36 cm² and the specified 60° angle.
Base–Height Relation for the Equal-Area Parallelogram
bₚ × hₚ = ½ × bₜ × hₜ
SymbolMeaningUnit
bₚBase of the parallelogramlength units
hₚPerpendicular height of the parallelogramlength units
bₜBase of the trianglelength units
hₜPerpendicular height of the trianglelength units
Worked example
Question: A triangle has base 10 cm and height 8 cm. An equal-area parallelogram is to have base 8 cm. Find its required perpendicular height.
Calculation:
Step 1: Area of triangle = ½ × 10 × 8 = 40 cm².
Step 2: For equal areas, 8 × hₚ = 40.
Step 3: hₚ = 40 ÷ 8 = 5 cm.
Answer: The required parallelogram height is 5 cm.