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  • In the given figure, the line segment XY is parallel to the side AC of the triangle ABC and it divides the triangle into two parts of equal area. Then AX:AB is

In the given figure, the line segment XY is parallel to the side AC of the triangle ABC and it divides the triangle into two parts of equal area. Then AX:AB is

Option: 1

1:2


Option: 2

1:4


Option: 3

1:\sqrt2


Option: 4

None of these


Answers (1)

best_answer

Since AX is parallel to AC

\begin{array}{l} \angle A=\angle B X Y \text { and } \angle C=\angle B Y X[\text { corresponding } \angle] \\ \therefore \quad \triangle A B C \sim \triangle X B Y \end{array}

\Rightarrow \quad \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\Delta X B Y)}=\frac{A B^{2}}{X B^{2}}\;\;\;\;\;\;\;\;\;\ldots(i)

\text { But, } \operatorname{ar}(\triangle A B C)=2 \times \operatorname{ar}(\triangle X B Y)[\text { given }]

\Rightarrow \quad \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle X B Y)}=2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(ii)

From (i) and (ii)

\begin{aligned} & \frac{A B^{2}}{X B^{2}}=2 \Rightarrow\left(\frac{A B}{X B}\right)^{2}=2 \\ \Rightarrow \quad & \frac{A B}{X B}=\sqrt{2} \Rightarrow A B=\sqrt{2}(X B) \\ \Rightarrow \quad & A B=\sqrt{2}(A B-A X) \\ \Rightarrow \quad & \sqrt{2} A X=(\sqrt{2}-1) A B \\ \Rightarrow \quad & \frac{A X}{A B}=\frac{(\sqrt{2}-1)}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{(2-\sqrt{2})}{2} \end{aligned}

\text { Hence, } A X: A B=(2-\sqrt{2}): 2

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Sayak

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