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प्रश्न
In the following figure, seg XY || side AC. If 2AX = 3BX and XY = 9, complete the activity to find the value of AC.
Activity:
2AX = 3BX
∴ `"AX"/"BX" = 3/2`
∴ `(AX + BX)/"BX"` = `(square + square)/2` ...[By componendo]
∴ `"AX"/"BX" = square/2`
ΔBCA ~ ΔBYX ...[`square` test of similarity]
∴ `"BA"/"BX" = "AC"/"XY"` ....[corresponding sides of similar triangles]
∴ `5/square = "AC"/9`
∴ AC = `square`
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उत्तर
2AX = 3BX
∴ `"AX"/"BX" = 3/2`
∴ `(AX + BX)/"BX"` = \[\frac{\boxed{3} + \boxed{2}}{2}\] ...[By componendo]
∴ `"AB"/"BX"` = \[\frac{\boxed5}{{2}}\]
Δ BCA ~ Δ BYX ...[\[\boxed{AA}\] test of similarity]
∴ `"BA"/"BX" = "AC"/"XY"` ....[corresponding sides of similar triangles]
∴ \[\frac{5}{\boxed{2}}\] = `"AC"/9`
∴ AC = `45/2`
∴ AC = \[\boxed{22.5}\]
