Arc FE is inscribed within isosceles right triangle ABC (∠A=90°), as

Question

https://gmatclub.com/forum/download/file.php?id=61013 Arc FE is inscribed within isosceles right triangle ABC (∠A=90°), as shown above. The arc is centered at A. If AC=2, what is EB? A. \sqrt{2}-1 B. \sqrt{2}*(\sqrt{2}-1) C. 2*(\sqrt{2}-1) D. 1 E. \sqrt{2} Project PS Butler Subscribe to get Daily Email - Click Here | Subscribe via RSS - RSS 1.png

sumitkrocks · Answer

Arc EF and side BC will intersect at a point which is equidistant from B, c and A

Radius= half of side BC i.e 1/2 *2 \sqrt{2} 
and EB = 2- radius
=2 - \sqrt{2} 
= \sqrt{2}  ( \sqrt{2} -1)
Option B is the right answer

TarunKumar1234 · Answer

Let's say circle meets at BC side point D.

AD will be perpendicular bisector at BC, as ABC is isosceles triangle.
BD = DC = 2√2 /2 = √2

AD^2 = (2^2 - √2^2) or, AD = √2

EB= 2-√2.

So, I think B.

sahibr · Answer

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.

Attachment:

Screenshot 2021-05-20 at 00.35.31.png [ 17.27 KiB | Viewed 2061 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.

)) 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.

Arc FE is inscribed within isosceles right triangle ABC (∠A=90°), as

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GMAT Problem Solving (PS) Questions

Arc FE is inscribed within isosceles right triangle ABC (∠A=90°), as

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