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A dart is thrown at a random point on a circular dartboard with a radius of 8 cm. What is the probability that the dart lands in the shaded region? Answer in terms of π.

 Apr 28, 2020
 #1
avatar+631 
+1

Angle A subtends to arc BC, which is equal to 180. Therefore angle A is 180/2 = 90. Knowing that it is a 45-45-90 triangle, can you figure out the area of the triangle? After that, just subtract it from the area of the semi-circle.

 Apr 28, 2020
 #2
avatar+485 
+1

By Thales' Theorem: angle BAC = 90 degrees(it subtends an arc of 180 degrees)

Because we know that, we know that triangle BAC is a right triangle with angle measurements of 45, 45, 90

Next, realize that the hypotenuse of said triangle is the diameter BC, which has length 16( which is 8 *2).

Now that we have our hypotenuse, realize that the ratio of a 45-45-90 triangle is 

x, x, \(x\sqrt{2}\)

That means that our leg of our right triangle is equal to:

\(16\over{\sqrt{2}}\)\(16\sqrt{2}\over2\) by rationalizing the denominator(multiplying it by \(\sqrt{2}\over\sqrt{2}\))

This equals 

\(8\sqrt{2}\)

Triangle area formula is 1/2 * b * h

Since the base and height are the same in this scenario, we have that the area of triangle BAC is:

\((8\sqrt{2})^2/2 = 64*2/2 = 64\)

The area of the entire circle is 64pi, so half of it is 32pi. Our area of the shaded region is then:

\(32\pi-64\)

The probability is then this part divided by the entire area of the dart board, which is 64pi, so our answer is:

\(32\pi-64\over64\pi\)

.
 Apr 28, 2020
 #3
avatar+23646 
+1

Area of circle = pi r^2    =    64 pi   

      subtract 1/2 (unshaded area) = 32 pi

         subtract area of triangle  1/2 b h = 1/2 16 8 = 64

  

(32 pi - 64) / (64 pi)

 Apr 28, 2020

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