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In the figure shown, the triangle is inscribed in the

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In the figure shown, the triangle is inscribed in the [#permalink]

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25 Jan 2006, 09:29
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In the figure shown, the triangle is inscribed in the semicircle. If the length of line segment AB is 8 and the length of line segment BC is 6, what is the length of arc ABC?
A. 15 TT
B. 12 TT
C. 10 TT
D. 7 TT
E. 5 TT
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Director
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25 Jan 2006, 09:49
as the sides of the triangle are 8 & 6, it is a right angled triangle. Therefore, the diameter of the semi circle(also the third side) is 10.

Answer would be (pie)*R (as it is a semicircle, half the perimeter will be the length of the semi-circle) = 5*pie (E)

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Vithal
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25 Jan 2006, 11:37
E.

Another approach:

The third side ( diameter ) should be between (8-6) and (8+6).

Thus 2 < d < 14.

From the answer choices, 2TTr/2 ( half of the cirumference of circle ) gives
r = 15,12,10,7 and 5 and the corresponding diameter as 30,24,20,14,10.

Only d = 10, satisfies , 2 < d < 14. Thus, the answer is E.
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26 Jan 2006, 05:48
oa is 5TT

thanks guys, i made a silly error
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26 Jan 2006, 12:52
cj,

there is a rule:

If a triangle is inscribed in a semi-circle, it will always be a right angle triangle with right angle at the corner which touches the arc portion of semi-circle (Like angle ABC here).
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26 Jan 2006, 23:45
The angle ABC is a right angle (triangle inscribed in semicircle). So lenght of AC = sqrt(64+36) = 10.

Length of the arc ABC = (2* pi * r)/2 = 5pi
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11 Sep 2011, 02:48
Quote:
cj,

there is a rule:

If a triangle is inscribed in a semi-circle, it will always be a right angle triangle with right angle at the corner which touches the arc portion of semi-circle (Like angle ABC here).

Thank you for this. E is the answer.
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11 Sep 2011, 03:48
Its E i.e 5*pi

Solution

Angle in a semi circle is a right angle and ABC forms a right angle triangle with right angled at B.

so AB=8 AND BC=6 so AC=10 as 8-6-10 forms a right angled triangle with Hypotenuse AC

so Now we got the diameter =10 and radius =5

Now length of the Segment ABC=circumference/2 i.e 2*pi*r/2 as it is a semicircle
so Now length of ABC=Pi*r and from above we know radius=5 so
Lenght of segment ABC= 5*pi
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11 Sep 2011, 08:36
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In a circle, whenever one of the sides of the triangle is diameter, that forms the hypotenuse of the triangle.

AB = 8, BC=6 => AC = 10

Length of arc ABC = (central angle/360)*(2*pi*r)

= (180/360)*(2*pi*r)

= pi*r

AC = 2r = 10 => r = 5

=> Length of arc ABC = 5pi

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12 Sep 2011, 04:07
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Re: In the figure shown, the triangle is inscribed in the [#permalink]

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17 Nov 2016, 08:18
joemama142000 wrote:
In the figure shown, the triangle is inscribed in the semicircle. If the length of line segment AB is 8 and the length of line segment BC is 6, what is the length of arc ABC?
A. 15 TT
B. 12 TT
C. 10 TT
D. 7 TT
E. 5 TT

AC must be between 2 and 14. Because, AB+BC=8+6=14, and AB-BC=8-6=2. So, 2<AC<14
Circumference of ABC=2πr/2--->πr
If r=highest 7, then length of ABC=7π
But it is actually less than 7π. So, E is correct answer.
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Re: In the figure shown, the triangle is inscribed in the   [#permalink] 17 Nov 2016, 08:18
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