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In the following figure, seg XY || side AC. If 2AX = 3BX and XY = 9, complete the activity to find the value of AC.[Figure: Δ ABC with A at the top, B at the lower left and C at the lower right; X lies on side AB and Y lies on side BC, and seg XY is…

Q.3 (A)(1) In the following figure, seg XY || side AC. If 2AX = 3BX and XY = 9, complete the activity to find the value of AC.

[Figure: Δ ABC with A at the top, B at the lower left and C at the lower right; X lies on side AB and Y lies on side BC, and seg XY is parallel to side AC.]

Activity :

2AX = 3BX

∴ \(\dfrac{AX}{BX} = \dfrac{3}{2}\)

∴ \(\dfrac{AX + BX}{BX} = \dfrac{____ + ____}{2}\) .......... (By componendo)

\(\dfrac{AB}{BX} = \dfrac{____}{2}\)

Δ BCA ~ Δ BYX ............. (____ Test of similarity)

∴ \(\dfrac{BA}{BX} = \dfrac{AC}{XY}\) .............. (Corresponding sides of similar triangles)

∴ \(\dfrac{5}{____} = \dfrac{AC}{9}\)

∴ AC = ____

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Answer: AC = 22.5

2AX = 3BX gives \(\dfrac{AX}{BX} = \dfrac{3}{2}\), so by componendo \(\dfrac{AX + BX}{BX} = \dfrac{3 + 2}{2}\), i.e. \(\dfrac{AB}{BX} = \dfrac{5}{2}\).

In Δ BCA and Δ BYX: ∠B is common and ∠BCA ≅ ∠BYX (corresponding angles, XY || AC), so Δ BCA ~ Δ BYX by the AA test.

∴ \(\dfrac{BA}{BX} = \dfrac{AC}{XY}\), i.e. \(\dfrac{5}{2} = \dfrac{AC}{9}\).

∴ AC = \(\dfrac{5 \times 9}{2}\) = 22.5.

Blanks in order: 3, 2, 5, AA, 2, 22.5.

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report
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