Question

In the following figure, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC. 

Activity:

2AX = 3BX     ...(Given)

∴ `(AX)/(BX) = 3/square`

`(AX + BX)/(BX) = (3 + 2)/2`  ...(By componendo)

`square/(BX) = 5/2`    ...(I)

Now ΔBCA ~ ΔBYX    ...`(square" test of similarity")`

∴ `(BA)/(BX) = (AC)/(XY)`    ...(Corresponding sides of similar triangles)

∴ `square/square = "AC"/9`    ...[From (I)]

∴ AC = `square`

Answer 1

2AX = 3BX     ...(Given)

∴ \[\frac{{AX}}{BX} = \frac{3}{\boxed{2}}\]

`(AX + BX)/(BX) = (3 + 2)/2`  ...(by componendo)

\[\frac{\boxed{AB}}{BX} = \frac{5}{2}\]    ...(I)

Now ΔBCA ~ ΔBYX  ...\[(\boxed{SS}\text{ test of similarity}\])

∴ `(BA)/(BX) = (AC)/(XY)` ...(Corresponding sides of similar triangles)

∴ \[\frac{\boxed{5}}{\boxed{2}} = \frac{AC}{9}\]    ...[From (I)]

∴ AC = \[\boxed{22.5}\]