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Math Expert V
Joined: 02 Sep 2009
Posts: 59730
In the figure shown below, the triangle PQR is inscribed in a semicirc  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 72% (00:59) correct 28% (00:50) wrong based on 87 sessions

### HideShow timer Statistics 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 [ 9.53 KiB | Viewed 1813 times ]

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

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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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Bunuel wrote: 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

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

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Bunuel wrote: 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 = $$\frac{180°}{360} *2pi *2.5$$ = $$\frac{5pi}{2}$$

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

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Bunuel wrote: 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.

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If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: In the figure shown below, the triangle PQR is inscribed in a semicirc   [#permalink] 09 Oct 2018, 10:37
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