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In the figure shown below, the triangle PQR is inscribed in a semicirc

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In the figure shown below, the triangle PQR is inscribed in a semicirc  [#permalink]

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New post 05 Oct 2018, 00:43
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In the figure shown below, the triangle PQR is inscribed in a semicircle. If the length of line segment PQ is 4 and the length of line segment QR is 3, what is the length of arc PQR?


(A) 3π

(B) 5π

(C) 7π/2

(D) 25π/2

(E) 5π/2

Attachment:
2018-10-05_1142.png
2018-10-05_1142.png [ 9.53 KiB | Viewed 369 times ]

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Re: In the figure shown below, the triangle PQR is inscribed in a semicirc  [#permalink]

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New post 05 Oct 2018, 01:15
Since this is a semicircle. Longest radius is the diameter and forms a right angle with the semicircle. So diameter is 5 or radius is 2.5

Since, this is a semicircle, half the circumference is the required length of Arc = pi*2.5
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Re: In the figure shown below, the triangle PQR is inscribed in a semicirc  [#permalink]

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New post 05 Oct 2018, 07:30
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Bunuel wrote:
Image
In the figure shown below, the triangle PQR is inscribed in a semicircle. If the length of line segment PQ is 4 and the length of line segment QR is 3, what is the length of arc PQR?

(A) 3π
(B) 5π
(C) 7π/2
(D) 25π/2
(E) 5π/2

Attachment:
2018-10-05_1142.png


Since the figure is a semicircle, we know that PR is the diameter of the circle
If PR is the diameter of the circle, then ∠Q is an inscribed angle "holding" (aka containing) the diameter.
One of our circle properties tells us that ∠Q must equal 90° (see video below for more on this)
This means ∆PQR is a RIGHT triangle, and its legs have lengths 3 and 4.
We can EITHER apply the Pythagorean Theorem to determine the length of the hypotenuse OR we can recognize that lengths 3 and 4 are parts of a "Pythagorean triplet" 3-4-5
Either way, we can determine that PR has length 5

What is the length of arc PQR?
Arc PQR is a half of the circle's circumference

Formula: circumference = (diameter)(π)
= (5)(π)

So, the length of arc PQR = (1/2)(5)(π) = 5π/2

Answer: E

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In the figure shown below, the triangle PQR is inscribed in a semicirc  [#permalink]

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New post 06 Oct 2018, 07:18
Bunuel wrote:
Image
In the figure shown below, the triangle PQR is inscribed in a semicircle. If the length of line segment PQ is 4 and the length of line segment QR is 3, what is the length of arc PQR?


(A) 3π

(B) 5π

(C) 7π/2

(D) 25π/2

(E) 5π/2

Attachment:
2018-10-05_1142.png



Fomulas and rules to remember :)

    Any triangle inscribed in semicircle is right traingle of 90°

    Arc Length Formula = \(\frac{θ°}{360} *2piR\) where θ° is used for angle degree measure (in our case it is 180 degrees)


\(4^2+3^2 = x^2\)

\(16+9=x^2\)

\(\sqrt{25} =\sqrt{x^2}\)

\(x =5\)

if PR is 5, then radius of circle is 2.5

Arc length = \(\frac{180°}{360} *2pi *2.5\) = \(\frac{5pi}{2}\)


this is awesome :) :lol:
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Re: In the figure shown below, the triangle PQR is inscribed in a semicirc  [#permalink]

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New post 09 Oct 2018, 10:37
Bunuel wrote:
Image
In the figure shown below, the triangle PQR is inscribed in a semicircle. If the length of line segment PQ is 4 and the length of line segment QR is 3, what is the length of arc PQR?


(A) 3π

(B) 5π

(C) 7π/2

(D) 25π/2

(E) 5π/2

Attachment:
2018-10-05_1142.png

We see that arc PQR is half of the circle, so if we can determine the value of line PR, we can determine the arclength of PQR.

Any triangle inscribed in a semicircle where the diameter is one of the sides of the triangle is a right triangle. We recognize triangle PQR as a 3 - 4 - 5 right triangle, or we can determine the length of PR by the Pythagorean theorem. Since the length of PR is 5, we see that the circumference of the entire circle is 5π, and, thus, half of the circumference of the circle is 5π/2. This is the length of arc PQR.

Answer: E
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Re: In the figure shown below, the triangle PQR is inscribed in a semicirc &nbs [#permalink] 09 Oct 2018, 10:37
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