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  ...(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)

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

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

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

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

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

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

∴ AC = \[\boxed{22.5}\]