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If the triangle ABC is inscribed in semi-circle BAC as above figure an

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If the triangle ABC is inscribed in semi-circle BAC as above figure an  [#permalink]

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New post 06 Mar 2016, 17:00
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Triangle ABC.jpg
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If the triangle ABC is inscribed in semi-circle BAC as above figure and BC is a diameter, the length of AB is 6 and the length of AC is 8, what is the length of arc BAC?

A. 5π(pi)
B. 6π(pi)
C. 7π(pi)
D. 8π(pi)
E. 10 π(pi)


* A solution will be posted in two days.

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Re: If the triangle ABC is inscribed in semi-circle BAC as above figure an  [#permalink]

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New post 07 Mar 2016, 05:40
BC is the diameter. Triangle ABC is right angled at A. If AB = 6 and AC = 8 then BC = 10

Radius = 10/2 = 5.

Length of arc BAC = Circumference/2 = pi * r = 5 * pi

Answer: A
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Re: If the triangle ABC is inscribed in semi-circle BAC as above figure an  [#permalink]

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New post 07 Mar 2016, 06:52
Diameter BC = 10 (Pythagoras theorem)
Radius of the semicircle = 5

the length of arc BAC = circumference of the circle with radius 5 / 2
= (2x \(\pi\) x 5) / 2
= 5\(\pi\)

Answer = A
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If the triangle ABC is inscribed in semi-circle BAC as above figure an  [#permalink]

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New post Updated on: 30 Apr 2018, 10:04
MathRevolution wrote:
Attachment:
Triangle ABC.jpg

If the triangle ABC is inscribed in semi-circle BAC as above figure and BC is a diameter, the length of AB is 6 and the length of AC is 8, what is the length of arc BAC?

A. 5π(pi)
B. 6π(pi)
C. 7π(pi)
D. 8π(pi)
E. 10 π(pi)


Since BC is the diameter of the semi-circle, we know that ∠BAC is 90º

In other words, we can conclude that BAC is a RIGHT TRIANGLE and side BC is the HYPOTENUSE.
This means we can apply the Pythagorean Theorem to get: 6² + 8² = (side BC)²
Simplify: 36 + 64 = (side BC)²
Simplify: 100 = (side BC)²
So, side BC = 10
In other words, the DIAMETER = 10

Circumference of COMPLETE circle = (DIAMETER)(π)
So, circumference of SEMIcircle = (DIAMETER)(π)/2
= (10)(π)/2
= 5π
= A

Cheers,
Brent
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Originally posted by GMATPrepNow on 07 Mar 2016, 12:24.
Last edited by GMATPrepNow on 30 Apr 2018, 10:04, edited 1 time in total.
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Re: If the triangle ABC is inscribed in semi-circle BAC as above figure an  [#permalink]

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New post 08 Mar 2016, 17:03
If the triangle ABC is inscribed in semi-circle BAC as above figure and BC is a diameter, the length of AB is 6 and the length of AC is 8, what is the length of arc BAC?

A. 5π(pi)
B. 6π(pi)
C. 7π(pi)
D. 8π(pi)
E. 10 π(pi)


->Since BC is a diameter, angle A is 90 degrees. According to Pythagoras' theorem, AB^2+AC^2=BC^2 -> 6^2+8^2=10^2 and BC=10. Then, a circumference with a diameter 10 is 10π(pi). The question asks arc BAC, which is 10π/2=5π. Thus, A is the answer.
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Re: If the triangle ABC is inscribed in semi-circle BAC as above figure an  [#permalink]

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Re: If the triangle ABC is inscribed in semi-circle BAC as above figure an   [#permalink] 09 Apr 2019, 10:56
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