We can use geometric models to solve certain types of exercises. In , on a or in a region of a represent . The geometric probability of an is a that involves geometric measures such as length or . Consider the given diagram.
We are told that a point in the figure is chosen at random, and want to find the probability that the point lies in the shaded region. The probability that the point is in the shaded region is the ratio of the area of the shaded region to the area of the figure.
P(The point is in the shaded region)= [0.8em]
Area of the shaded region/Area of the figure
We will find the area of the shaded region and the area of the entire figure one at a time. Then, we will find their ratio.
The figure is a with of 16 units. This means that its is 16÷ 2= 8 units. We can now substitute this value in the formula for the .
The area of the circle is 64Ï€ square units.
Area of the Figure: 64Ï€ square units
Area of the Shaded Region
The area of the shaded region consists of two parts.
- The difference between the and the .
- Two sixths of the difference between the area of a circle and the area of a .
We can calculate the area of each of these regions one at a time. Let's first focus on the hexagon!
Shaded Area of the Hexagon
A can be divided into six and , with side length 8 units. Let's now draw an a of the hexagon. Since the angles of an equilateral triangle measure 60^(∘) and the apothem an angle along with its opposite side, we know that the angle between the apothem and the radius measures 60^(∘)÷2= 30^(∘).
Let's write and solve an equation for a using the .
cos 30^(∘)=a/8
8cos30^(∘)=a
8( sqrt(3)/2)=a
8sqrt(3)/2=a
4sqrt(3)/1=a
4sqrt(3)=a
a=4sqrt(3)
The apothem of the hexagon is equal to 4sqrt(3) units. Now, we can recall the formula for the area of a regular polygon.
A=1/2ans
In this formula a is the apothem, n is the number of sides, and s is the side length. Finally, we can substitute a= 4sqrt(3), n= 6 and s= 8 in the above formula.
A_H=1/2ans
A_H=1/2( 4sqrt(3))( 6)( 8)
A_H=1/2(4)(6)(8)sqrt(3)
A_H=1/2(192)sqrt(3)
A_H=192/2sqrt(3)
A_H=96sqrt(3)
The area of the hexagon is equal to 96sqrt(3) units. Let's now focus on the rectangle. The apothem of the triangle 4sqrt(3) is equal to half the length of the rectangle. Therefore, the length of the rectangle is 2* 4sqrt(3)= 8sqrt(3) units.
Now, we can substitute l= 8sqrt(3) and w= 8 into the formula for the area of a rectangle.
A_R=l w
A_R=( 8sqrt(3))( 8)
A_R=64sqrt(3)
The area of the rectangle is equal to 64sqrt(3) units. Finally, we can calculate the area of the shaded part of the hexagon by subtracting A_R from A_H.
A_(SH)=A_H-A_R
A_(SH)= 96sqrt(3)-64sqrt(3)
A_(SH)=32sqrt(3)
The area of the shaded hexagon is 32sqrt(3) units.
Shaded Area of the Circle
Now, let's calculate the area of the shaded part of the circle. The region that is created by subtracting the area of a hexagon from the area of a circle consists of six . Out of these six circular segments, two are shaded.
Therefore, we can calculate the shaded part of the circle by multiplying 26 by the difference between A_C and A_H.
A_(SC)=2/6(A_C-A_H)
A_(SC)=2/6(64Ï€- 96sqrt(3))
A_(SC)=1/3(64Ï€-96sqrt(3))
A_(SC)=1/3(64Ï€)-1/3(96sqrt(3))
A_(SC)=64/3Ï€-96/3sqrt(3)
A_(SC)=64/3Ï€-32sqrt(3)
The area of the shaded part of the circle is equal to 643Ï€-32sqrt(3) units.
Sum of the Shaded Regions
To find the area of the entire shaded region, we need to add the shaded area of the circle A_(SC) and the shaded area of the hexagon A_(SH).
A=A_(SC)+A_(SH)
A=(64/3Ï€-32sqrt(3))+32sqrt(3)
A=64/3Ï€-32sqrt(3)+32sqrt(3)
A=64/3Ï€
The shaded area is equal to 643Ï€ square units.
Shaded Area: 643Ï€ square units
Probability
As previously mentioned, the probability that the point is in the shaded region is the ratio of the area of the shaded region to the area of the figure. Since we already know both areas, we can find their ratio.
P=Area of the shaded region/Area of the figure
P=64Ï€3/64Ï€
P≈0.33
P≈ 33 %
The probability that a point from the figure chosen at random is in the shaded region is about 0.33, which can be also written as 33 %.
The hexagon consists of six congruent triangles. Also, there are six congruent circular segments. Let's consider two consecutive triangles and two consecutive segments.
Now, we can highlight part of the circle that has the same area as the shaded region.
Out of 3 of the circle, only 1 is shaded. Now, we have enough information to calculate the probability P that a randomly chosen point is in the shaded region.
The probability is about 0.33, which can be also written as 33 %.
