We are asked to consider an isosceles triangle with a perimeter of 32 centimeters. We also know that the lengths of the sides of the triangle are integers. We want to find the probability that the area of the triangle is exactly 48 square centimeters.
In this case, the sample space is the set of all isosceles triangles with a perimeter of 32 centimetres and sides of integer lengths. The favorable outcomes are all such triangles that have an area of exactly 48 centimeters. Let's call the length of the legs of the triangle a and the length of its base b.
The legs of the triangle must have lengths greater than 0 centimeters, since it does not make sense for a side of the triangle to have a non-positive length. The perimeter of our triangle is 32 centimeters, so the length b of the base is 32 centimeters minus 2 times the length of a leg.
b = 32 - 2aSince b is the length of the base, it must also be greater than 0.
The value of a must be an integer that is at least 1 and at most 15. To find the valid values of a for our triangle, we will use the Triangle Inequality Theorem.
Triangle Inequality Theorem
The sum of lengths of any two sides of a triangle must be greater than the length of the third side.
From the Triangle Inequality Theorem, we know that the sum of the lengths of the legs 2a must be greater than the length of the base b. We also know that the sum of the length of the base and the length of a leg b+a must be greater than the length of a leg a, but since b>0, this is always true. Let's solve our inequality!
The possible values of a are 9, 10, 11, 12, 13, 14, and 15. This means that the number of all possible outcomes is 7. Let's list the corresponding values of b.
a
b = 32 - 2a
b
9
b = 32 - 2( 9)
14
10
b = 32 - 2( 10)
12
11
b = 32 - 2( 11)
10
12
b = 32 - 2( 10)
8
13
b = 32 - 2( 10)
6
14
b = 32 - 2( 10)
4
15
b = 32 - 2( 10)
2
Now, let's find the areas of these triangles. We will start with the triangle with legs of length a = 9 and base of length b = 14. Recall the formula for the area A of a triangle with base b and height h.
A = 1/2bh
Let's take a closer look at our triangle. We will call the height created using the vertex between the legs of the triangle h.
As we can see, half of the base, one of the legs, and the height form a right triangle. We can use the Pythagorean Theorem to write an equation for h. Then we will solve the equation.
We only considered the principal root, since the height must be positive. Now we can substitute 4sqrt(2) for h and 14 for b into the formula for the area of a triangle to find the area of our triangle.
Now let's repeat this process for the other possible triangles, first finding their heights.
a
b
Height
10
12
8
11
10
4sqrt(6)
12
8
8sqrt(2)
13
6
4sqrt(10)
14
4
8sqrt(3)
15
2
4sqrt(14)
Now that we know the heights of the triangles, we can calculate their areas.
b
h
Area
12
8
48
10
4sqrt(6)
≈ 49
8
8sqrt(2)
≈ 45.3
6
4sqrt(10)
≈ 48
4
8sqrt(3)
≈ 27.7
2
4sqrt(14)
≈ 15
Only one of the triangles has an area of exactly 48 square centimeters, so there is 1 favorable outcome. Remember, there are 7 possible outcomes. The probability that the triangle has an area of exactly 48 square centimeters is the ratio of the number of favorable outcomes to the number of possible outcomes, 17.
