$In the figure, XY ∥ AC and XY divides the triangular region ABC into two equal areas . Determine AX : AB .$

Open in App

Solution

$It is given that XY divides the triangular region ABC into two equal areas . That means A (△ BXY) = A (□ AXYC) \Rightarrow \frac{A (△ BXY)}{A (□ AXYC)} = \frac{1}{1} \Rightarrow \frac{A (△ BXY)}{A (△ BAC) - A (△ BXY)} = \frac{1}{1} \Rightarrow 2 A (△ BXY) = A (△ BAC) \Rightarrow \frac{A (△ BXY)}{A (△ BAC)} = \frac{1}{2} . . . (1) It is given that XY ∥ AC . In △ BXY and △ BAC, ∠ BXY = ∠ BAC [Corresponding angles] ∠ BYX = ∠ BCA [Corresponding angles] Therefore, by AA similarity test △ BXY ~ △ BAC Now, we can say that \frac{A (△ BXY)}{A (△ BAC)} = {(\frac{BX}{AB})}^{2} \Rightarrow \frac{1}{2} = {(\frac{BX}{AB})}^{2} [From (1)] \Rightarrow {(\frac{AB - AX}{AB})}^{2} = \frac{1}{2} \Rightarrow \frac{{AB}^{2} {(1 - \frac{AX}{AB})}^{2}}{{AB}^{2}} = \frac{1}{2}$ ^{$\Rightarrow (1 - \frac{AX}{AB}) = \frac{1}{\sqrt{2}} \Rightarrow \frac{AX}{AB} = 1 - \frac{1}{\sqrt{2}} \Rightarrow \frac{AX}{AB} = \frac{\sqrt{2} - 1}{\sqrt{2}}$}

Suggest Corrections

Similar questions

\frac{A X}{A B}

∥

A B C, X Y | | A C

\frac{A X}{A B}

Δ

¯ ¯¯¯¯¯¯¯ ¯ X Y

¯ ¯¯¯¯¯¯ ¯ A C

\frac{A X}{A B}

Δ A B C

\frac{A X}{A B}

Join BYJU'S Learning Program