I know Trigonometry is out of the syllabi of GMAT and no GMAT coaching teaches Trigonometry( I Think?)
But for those of us who do know Trigonometry, it it does come in handy on multiple occasions in such problems.
So I would recommend learning the basics of Trigonometry to all students. Just the basics and you are good to go. It will barely take an hour of your time.
Anyway, coming back to the problem.
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Screenshot 2021-05-20 at 00.35.31.png [ 17.27 KiB | Viewed 1530 times ]
We know from the above question, AB=AC=2 (since the question mentions that the triangle is isosceles).
We need to find BE.
And BE=AB-AE=2-AE.
Hence we need to find AE first.
Since we know the triangle is Right angled (at angle A) and isosceles, angle B & C must be 45 degree (as 90+45+45=180)
sin ang B = sin 45 = AC/BC
\(\frac{1}{\sqrt{2}}=\frac{2}{BC}\)
\(BC=2\sqrt{2}\)
Now perpendicular from A to BC at point X (lets call the point X (forgot to mark it on the image
)) divides BC into two equal parts.
Hence \(BX=CX=2\sqrt{2}\) -----> (1)
Using triangle ABX, we take tan of angle B
tan B = tan 45 = \(\frac{AX}{BX}\)
\(1 = \frac{AX}{BX}\)
AX=BX=\(\sqrt{2}\) {From (1) }
Since AX and AE are both radii of the arc AFE, we know AE=AX=\(\sqrt{2}\)
AE= \(\sqrt{2}\)
BE=AB-AE
= \(2-\sqrt{2}\)
=\(\sqrt{2}(\sqrt{2}-1)\)
Hence option B is the answer.