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

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Arc FE is inscribed within isosceles right triangle ABC (∠A=90°), as s  [#permalink]

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06 Jun 2015, 02:22
2
1
00:00

Difficulty:

45% (medium)

Question Stats:

57% (02:17) correct 43% (02:51) wrong based on 21 sessions

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T8900.png [ 4.84 KiB | Viewed 1208 times ]

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. $$\sqrt{2}$$

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Re: Arc FE is inscribed within isosceles right triangle ABC (∠A=90°), as s  [#permalink]

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06 Jun 2015, 03:01
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reto wrote:
Attachment:
The attachment T8900.png is no longer available

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. $$\sqrt{2}$$

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Sol 9.jpg [ 190.68 KiB | Viewed 1187 times ]

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Re: Arc FE is inscribed within isosceles right triangle ABC (∠A=90°), as s  [#permalink]

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06 Jun 2015, 03:05
reto wrote:
Attachment:
T8900.png

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. $$\sqrt{2}$$

Since ABC is an Isoceles right triangle Radius of circle inscribed(line segment from vertex containing right angle to the mid point of hypotenuse) =$$\frac{1}{2}$$ the hypotenuse BC =$$2\sqrt{2}$$

Hence EB = AB - AE = 2 - $$\sqrt{2}$$ . ie B
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Re: Arc FE is inscribed within isosceles right triangle ABC (∠A=90°), as s  [#permalink]

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08 Feb 2019, 04:19
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Re: Arc FE is inscribed within isosceles right triangle ABC (∠A=90°), as s   [#permalink] 08 Feb 2019, 04:19
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# Arc FE is inscribed within isosceles right triangle ABC (∠A=90°), as s

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