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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]
Bunuel

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]
Bunuel

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.