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Question
In the given fig, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to Find the value of AC.
Activity: 2AX = 3BX
∴ `"AX"/"BX" = square/square`
`("AX" +"BX")/"BX" = (square + square)/square` ...(by componendo)
`"AB"/"BX" = square/square` ...(I)
ΔBCA ~ ΔBYX ...`square` test of similarity,
∴ `"BA"/"BX" = "AC"/"XY"` ...(corresponding sides of similar triangles)
∴ `square/square = "AC"/9`
∴ AC = `square` ...[From(I)]
Solution
Given: XY || seg AC, 2AX = 3BX and XY = 9
To Find: Seg AC.
Solution:
Consider, 2AX = 3BX
∴ `"AX"/"BX"=bb(3/2)`
∴ `"AX + BX"/"BX" = bb((3+2)/2)` ...(By Componendo)
∴ `"AB"/"BX" = bb(5/2)` ...(I)
ΔBCA ~ ΔBYX ...(AA test of similarity)
∴ `"BA"/"BX" = "AC"/"XY"` ...(corresponding sides of similar triangles)
∴ `bb(5/2) = "AC"/9`
∴ AC = `(5 × 9)/2`
∴ AC = 22.5 ...[From(I)]
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In the above figure, line AB || line CD || line EF, line l, and line m are its transversals. If AC = 6, CE = 9. BD = 8, then complete the following activity to find DF.
Activity :
`"AC"/"" = ""/"DF"` (Property of three parallel lines and their transversal)
∴ `6/9 = ""/"DF"`
∴ `"DF" = "___"`
