The area of a triangle ABC is 63 sq.units. Two parallel lines DE,FG are drawn such that they divide the line segments AB and AC into three segments AB and AC into three equal parts. What is the area of the quadrilateral DEFG?
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Two parallel lines DE,FG divide the line segments AB and AC of ΔABC with 63 sq units area into three equal parts then Area of Quadrilateral DEFG = 21 sq units
Step-by-step explanation:
as DE ║ FG ║ BC
=> Δ ADE ≈ Δ AFG ≈ ΔABC
AD = AB/3 , AF = 2AB/3
AE = AC/3 AG = 2AC/3
=> Area of ΔADE = (1/3)² Area of ΔABC
=> Area of ΔADE = 63/9
=> Area of ΔADE = 7 sq units
=> Area of ΔAFG = (2/3)² Area of ΔABC
=> Area of ΔAFG = 63 * 4 / 9
=> Area of ΔAFG = 28 sq units
Area of Quadrilateral DEFG = Area of ΔAFG - Area of ΔADE
=> Area of Quadrilateral DEFG = 28 - 7
=> Area of Quadrilateral DEFG = 21 sq units
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