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In the given fig, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to Find the value of AC. 

Problem Set 1 | Q 12 | Page 29
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)]

In fig 1.80, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity
Solution Given: XY || seg AC, 2AX = 3BX and XY = 9

Solution:

Consider, 2AX = 3BX 

∴ `"AX"/"BX"=3/2`

∴ `"AX + BX"/"BX" = (3+2)/2`   ...(By Componendo)

∴ `"AB"/"BX" = 5/2`                   ...(I)

ΔBCA ~ ΔBYX                           ...(AA test of similarity)

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

∴ `5/2 = "AC"/9`      

∴ AC = `(5 × 9)/2`        

∴ AC = 22.5         ...[From(I)]

Answer:-

Given: XY || segment AC, 2AX = 3BX and XY = 9

To Find: The length of AC

Solution:

From the given information, we have:

  • XY || AC
  • 2AX = 3BX
  • XY = 9

We need to find the length of AC.

Let us first find the length of BX in terms of AX:

Since 2AX = 3BX, we can write BX as:

BX = (2/3)AX

Now, let us use the similarity of triangles BCA and BYX (using AA test of similarity):

  • ∠BCA = ∠BYX (alternate angles)
  • ∠CAB = ∠YXB (alternate angles)
  • Therefore, ΔBCA ~ ΔBYX (by AA test of similarity)

Using corresponding sides of similar triangles BCA and BYX, we have:

  • BA/BX = AC/XY

Substituting the values of BA and BX in terms of AX, we get:

  • (3/2)AX/[(2/3)AX] = AC/9

Simplifying, we get:

  • AC = (5/2) * 9
  • AC = 22.5

Therefore, the length of AC is 22.5 units.

Problem Set 1 | Q 12 | Page 29

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