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Question
It is given that ∆ABC ~ ∆EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm. Find the lengths of the remaining sides of the triangles.
Solution
Given,
∆ABC ∼ ∆EDF,
So the corresponding sides of ∆ABC and ∆EDF are in the same ratio.
i.e., `("AB")/("ED") = ("AC")/("EF") = ("BC")/("DF")` ...(i)
Also,
AB = 5 cm,
AC = 7 cm,
DF = 15 cm
And DE = 12 cm
On putting these values in equation (i), we get
`5/12 = 7/("EF") = ("BC")/15`
On taking first and second terms, we get
`5/12 = 7/("EF")`
⇒ EF = `(7 xx 12)/5` = 16.8 cm
On taking first and third terms, we get
`5/12 = ("BC")/15`
⇒ BC = `(5 xx 15)/12` = 6.25 cm
Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 6.25 cm.
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