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)]
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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