From the following diagram the line segment XY is parallel to side AC of $△$ ABC and it divides the triangle into two parts of equal areas then AX / AB =

Question

Question

This question was previously asked in

TSPSC FBO (Forest Beat Officer) 2017 Official Paper 2

Answer 1

Calculation:

F3 Savita State G 8-2-23 D10

Area ( $△$ ABC) = 2 × Area( $△$ BXY)

In $△$ ABC and $△$ BXY

∠ABC = ∠XBY ( common angle)

∠ACB = ∠XYB (since XY||AC, angles are equal)

ΔABC ∼ ΔXBY (AA similarity)

(AB/BX)² = Area(

△

ABC) / Area(

△

XBY) = 2

AB² = 2× (BX)²

⇒ AB = √ 2 BX

AX = AB - BX = AB - AB/√ 2

⇒ AX = (√2 -1 / √ 2) AB

⇒ AX/AB = (√ 2 -1 / √ 2) =

\frac{2 - \sqrt{2}}{2}

∴ The correct answer is $\frac{2 - \sqrt{2}}{2}$ .