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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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18 Oct 2015, 12:07
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77% (01:02) correct 23% (01:13) wrong based on 1037 sessions

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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.

Attachment:

2015-10-19_0005.png [ 4.5 KiB | Viewed 12431 times ]

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

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18 Oct 2015, 12:17
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Bunuel 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) 15π
(B) 12π
(C) 10π
(D) 7π
(E) 5π

Kudos for a correct solution.

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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18 Oct 2015, 12:28
Bunuel 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) 15π
(B) 12π
(C) 10π
(D) 7π
(E) 5π

Kudos for a correct solution.

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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18 Oct 2015, 12:45
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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18 Oct 2015, 21:31
Bunuel 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) 15π
(B) 12π
(C) 10π
(D) 7π
(E) 5π

Kudos for a correct solution.

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

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

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02 May 2016, 13:23
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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Posts: 50585
Re: In the figure shown, the triangle is inscribed in the semicircle. If  [#permalink]

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03 May 2016, 00:07
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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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03 May 2016, 04:25
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Bunuel 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) 15π
(B) 12π
(C) 10π
(D) 7π
(E) 5π

Kudos for a correct solution.

Attachment:
2015-10-19_0005.png

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∏.

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

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17 May 2016, 18:23
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Attached is a visual that should help.
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Screen Shot 2016-05-17 at 7.12.20 PM.png [ 151.44 KiB | Viewed 9867 times ]

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

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22 Mar 2017, 12:10
Bunuel 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) 15π
(B) 12π
(C) 10π
(D) 7π
(E) 5π

Kudos for a correct solution.

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.

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

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

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08 Sep 2017, 05:20
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]

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03 Mar 2018, 07:19
Can you explain how we can conclude that length of ABC is half the circle?

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

Kudos for a correct solution.

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.

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

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Posts: 50585
Re: In the figure shown, the triangle is inscribed in the semicircle. If  [#permalink]

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03 Mar 2018, 07:58
AMARAAZUNWIE wrote:
Can you explain how we can conclude that length of ABC is half the circle?

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

Kudos for a correct solution.

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.

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

Semicircle is half of the circle.
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Re: In the figure shown, the triangle is inscribed in the semicircle. If &nbs [#permalink] 03 Mar 2018, 07:58
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