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3. In the picture below, triangles ABC and CDE have the same areas. Let Fbe the point of intersection of ACand DE. It is known that AB is parallelto DE. AB = 9 cm and DF = 7.5 cm.Find the length of EF in cm. |
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Answer» Given : triangles ABC and CDE have the same areas. F is the point of intersection of AC and DE. AB is parallel to DE. AB = 9 cm and DF = 7.5 cm. To Find : the length of EF in cm. Solution: Let say AREA of Δ ABC & ΔCDE = A AB ║ DE => AB ║ EF => Δ ABC ≈ Δ FEC => Area of ΔABC / Area of ΔFEC = ( AB/ EF)² => A / Area of ΔFEC = ( 9/ x)² => Area of ΔFEC = Ax² / 81 in Δ CDE Area of Δ FEC = ( FE / DE ) Area of Δ CDE => Area of Δ FEC = ( x / (x +7.5) ) * A => Ax² / 81= ( x / (x +7.5) ) * A => x(x + 7.5) = 81 => x² + 7.5x - 81 = 0 => x² + 13.5x - 6x - 81 = 0 => x(x + 13.5) - 6(x + 13.5) = 0 => (x - 6)(x + 13.5) = 0 => x = 6 ( x = - 13.5 not possible ) Length of FE = 6 cm Learn more: perimeters of two similar triangles are 30cm and 40CM respectively ... If ∆ABC is similar to ∆DEF web that BC = 4 cm, EF = 7 cm sod area of ... |
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