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In the fig. XY || AC and XY divides triangular region ABC into two part equal in area.
Then AX is equal to : AB 
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1 √2 -
√2 + 2 √2 -
1 2 -
√2 - 2 √2 - None of these
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Correct Option: D
From above given figure , we can see that
ar △XBY = ar trap. XYCA(Given)
ar(ΔABC) = 2 ar(ΔXBY)
| ar(∆XBY) | = | 1 |
| ar(∆ABC) | 2 |
But ΔXBY ∼ ΔABC (∴ XY || AC)
| ∴ | ar(∆XBY) | = | XB2 | ( Areas of similar triangle ) |
| ar(∆ABC) | AB2 |
| ∴ | 1 | = | XB2 | |
| 2 | AB2 |
| ∴ | XB | = | 1 |
| AB | √2 |
| AB - AX | = | 1 | { ∴ AB - AX = XB } |
| AB | √2 |
| ∴ 1 - | AX | = | 1 |
| AB | √2 |
| ⇒ | AX | = 1 - | 1 |
| AB | √2 |
| ⇒ | AX | = | √2 - 1 |
| AB | √2 |
