3. In the picture below, triangles ABC
and CDE have the same areas. Let F
be the point of intersection of AC
and DE. It is known that AB is parallel
to DE. AB = 9 cm and DF = 7.5 cm.
Find the length of EF in cm.
Answers
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
Let say FE = x 6 cm
=> 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
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