In $Δ$ ABC, $¯ ¯¯¯¯¯¯¯ ¯ X Y$ is paralled to $¯ ¯¯¯¯¯¯ ¯ A C$ and divides the triangle into two parts of equal area. Then the $\frac{A X}{A B}$ equals

\frac{\sqrt{2} + 1}{2}

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

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\frac{2 + \sqrt{2}}{2}

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\frac{\sqrt{2} - 1}{2}

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Solution

The correct option is B $\frac{2 - \sqrt{2}}{2}$
${(\frac{B X}{A B})}^{2} = \frac{1}{2}$
$\frac{B X}{A B} = \frac{1}{\sqrt{2}}$
$1 - \frac{B X}{A B} = 1 - \frac{1}{\sqrt{2}}$
$\frac{A B - B X}{A B} = \frac{\sqrt{2} - 1}{\sqrt{2}}$
$\frac{A X}{A B} = \frac{2 - \sqrt{2}}{2}$

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Similar questions

Q. In the given figure, the line segment

X Y

is parallel in

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Δ A B C

and is divided the triangle into two parts of equal areas then

\frac{A X}{A B} = \frac{\sqrt{2} - 1}{\sqrt{2}}

In $Δ P Q R$ , MN||QR and MN divides the triangle into two parts of equal areas. $\frac{Q M}{P Q} = ? ?$

The line segment XY is parallel to side AC of $△$ ABC and it divides the triangle into two parts of equal areas. Find the ratio $\frac{B X}{A B}$ .

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x y

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and its dividend the triangle into two points of equal areas. Prove that

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In ΔABC, ¯¯¯¯¯¯¯¯¯XY is paralled to ¯¯¯¯¯¯¯¯AC and divides the triangle into two parts of equal area. Then the AXAB equals

The correct option is B 2−√22(BXAB)2=12BXAB=1√21−BXAB=1−1√2AB−BXAB=√2−1√2AXAB=2−√22

In $Δ$ ABC, $¯ ¯¯¯¯¯¯¯ ¯ X Y$ is paralled to $¯ ¯¯¯¯¯¯ ¯ A C$ and divides the triangle into two parts of equal area. Then the $\frac{A X}{A B}$ equals

The correct option is B $\frac{2 - \sqrt{2}}{2}$
${(\frac{B X}{A B})}^{2} = \frac{1}{2}$
$\frac{B X}{A B} = \frac{1}{\sqrt{2}}$
$1 - \frac{B X}{A B} = 1 - \frac{1}{\sqrt{2}}$
$\frac{A B - B X}{A B} = \frac{\sqrt{2} - 1}{\sqrt{2}}$
$\frac{A X}{A B} = \frac{2 - \sqrt{2}}{2}$