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In the figure shown, the triangle is inscribed in the semicircle. If

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

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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) 15π
(B) 12π
(C) 10π
(D) 7π
(E) 5π

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
2015-10-19_0005.png
2015-10-19_0005.png [ 4.5 KiB | Viewed 6235 times ]
[Reveal] Spoiler: OA

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Re: In the figure shown, the triangle is inscribed in the semicircle. If [#permalink]

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New post 18 Oct 2015, 13:17
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Bunuel wrote:
Image
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) 15π
(B) 12π
(C) 10π
(D) 7π
(E) 5π

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
2015-10-19_0005.png


the calculated value = half of the perimeter of circle O = pi*R

The length of AC = 10 => R = 5 => the calculated value = Pi * 5

Ans E

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Re: In the figure shown, the triangle is inscribed in the semicircle. If [#permalink]

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New post 18 Oct 2015, 13:28
Bunuel wrote:
Image
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) 15π
(B) 12π
(C) 10π
(D) 7π
(E) 5π

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
2015-10-19_0005.png



IMO: 5π .Diameter is 10. So (180/360)* 2π5 = 5π

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Re: In the figure shown, the triangle is inscribed in the semicircle. If [#permalink]

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New post 18 Oct 2015, 13:45
without any calculations, we can see that AC is the diagonal of a circle. Knowing that AB=8 and AC=6 we can spot the Pythagorean triplet 3-4-5, and can deduce that AC, the diagonal is 10. Circumference of the circle must then be 10pi. Since we have only a semicircle, the length of the arc then must be 1/2*10pi = 5pi. Answer choice E.

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

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New post 18 Oct 2015, 22:31
Bunuel wrote:
Image
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) 15π
(B) 12π
(C) 10π
(D) 7π
(E) 5π

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
2015-10-19_0005.png


Triangle ABC is right angled at B,
by Pythagoras theorem we have diameter of semicircle as 10

=> arc ABC = π * radius = 5π

Answer Choice E

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Re: In the figure shown, the triangle is inscribed in the semicircle. If [#permalink]

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New post 02 May 2016, 14:23
I am looking for the rule that states that any angle like <B in this problem will always be right angle, can't find it in my pocket reference. Theoretical explanation from anyone? Thanks

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Re: In the figure shown, the triangle is inscribed in the semicircle. If [#permalink]

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glt13 wrote:
I am looking for the rule that states that any angle like <B in this problem will always be right angle, can't find it in my pocket reference. Theoretical explanation from anyone? Thanks


A right triangle's hypotenuse is a diameter of its circumcircle (circumscribed circle).

The reverse is also true: if one of the sides of an inscribed triangle is a diameter of the circle, then the triangle is a right angled (right angel being the angle opposite the diameter/hypotenuse).
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Re: In the figure shown, the triangle is inscribed in the semicircle. If [#permalink]

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Bunuel wrote:
Image
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) 15π
(B) 12π
(C) 10π
(D) 7π
(E) 5π

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
2015-10-19_0005.png


Image

We are given that triangle ABC is inscribed in the semicircle. Since a triangle inscribed in a semicircle is always a right triangle, triangle ABC is a right triangle. Therefore, AC is both the hypotenuse of triangle ABC and the diameter of the semicircle. We are also given that side AB = 8 and that side BC = 6. With that information, we can conclude that triangle ABC is a 6-8-10 right triangle, in which side AC = 10.

We need to determine the length of arc ABC, or, in other words, the length of half of the circumference of the circle. Since the diameter is 10, the radius is 5. We use the circumference formula C = 2∏r to obtain 10∏ as the circumference of the entire circle. The length of arc ABC is half of the circumference; therefore, its value is 5∏.

Answer: E
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Re: In the figure shown, the triangle is inscribed in the semicircle. If [#permalink]

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New post 17 May 2016, 19:23
Attached is a visual that should help.
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Screen Shot 2016-05-17 at 7.12.20 PM.png
Screen Shot 2016-05-17 at 7.12.20 PM.png [ 151.44 KiB | Viewed 4650 times ]


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Re: In the figure shown, the triangle is inscribed in the semicircle. If [#permalink]

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New post 22 Mar 2017, 13:10
Bunuel wrote:
Image
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) 15π
(B) 12π
(C) 10π
(D) 7π
(E) 5π

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
2015-10-19_0005.png


A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

So as ABC is inscribed in semicircle then ABC is a right triangle and AC=diameter=hypotenuse. Also note that the length of arc ABC is half of the circumference.

\(AC=\sqrt{AB^2+BC^2}=\sqrt{64+36}=10\) --> \(radius=\frac{diameter}{2}=\frac{AC}{2}=5\) --> \(circumference=2\pi{r}=10\pi\) --> \(arc_{ABC}=\frac{circumference}{2}=5\pi\).

Answer: E.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: In the figure shown, the triangle is inscribed in the semicircle. If [#permalink]

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New post 08 Sep 2017, 06:20
Two methods:

1) Formula to find length of arc= Central angle/360 *2*pi*radius

AC (diameter):
AC^2= 8^2+6^2= 10

Putting the values in the above formula:
Length= 180/360 *2*pi*5
= 5*pi



2) 3-4-5 triangle
6-8-10

Length of arc= 2*pi*r/2= 5*pi
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Re: In the figure shown, the triangle is inscribed in the semicircle. If   [#permalink] 08 Sep 2017, 06:20
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