Chapter 13.3, Problem 36HP

To determine

To calculate:

The probability that the area of the triangle is exactly $48c{m}^2$ .

Answer to Problem 36HP

The probability that the area of the triangle is exactly $48c{m}^2$ is equal to $\frac{1}{7}$ .

Explanation of Solution

Given information:

An isosceles triangle has side lengths that are all integers, and the sides of the triangle sum to $32cm$ .

Calculation:

An isosceles triangle has side lengths that are all integers, and the sides of the triangle sum to $32cm$ .

The triangle inequality states that the largest side length must be less than the sum of the other side lengths.

Hence, there are $7$ sets of side lengths that satisfy this condition:

1. $15cm$ , $15cm$ , $2cm$ .
2. $14cm$ , $14cm$ , $4cm$ .
3. $13cm$ , $13cm$ , $6cm$ .
4. $12cm$ , $12cm$ , $8cm$ .
5. $11cm$ , $11cm$ , $10cm$ .
6. $10cm$ , $10cm$ , $12cm$ .
7. $9cm$ , $9cm$ , $14cm$ .

To find the probability that the area of the triangle is exactly $48c{m}^2$ , determine which of the $7$ sets of side lengths yield an area of exactly $48c{m}^2$ .

Now let’s find out area for each of the $7$ sets of side lengths:

1. For side lengths: $15cm$ , $15cm$ , $2cm$:

The area of the triangle is given by:

  $\begin{array}{l}A=\sqrt{16\left(16-15\right)\left(16-15\right)\left(16-2\right)}\\ A=\sqrt{16\left(1\right)\left(1\right)\left(14\right)}\\ A=\sqrt{224}\\ A=14.96\\ A=15c{m}^2\end{array}$

According to Heron’s formula for the area of a triangle with three known sides with a perimeter of $32cm$ (half of $32cm$ is $16cm$ ).

Therefore, for side lengths: $15cm$ , $15cm$ , $2cm$ , $A=15c{m}^2$

2. For side lengths: $14cm$ , $14cm$ , $4cm$:

The area of the triangle is given by:

  $\begin{array}{l}A=\sqrt{16\left(16-14\right)\left(16-14\right)\left(16-4\right)}\\ A=\sqrt{16\left(2\right)\left(2\right)\left(12\right)}\\ A=\sqrt{278}\\ A=27.7c{m}^2\end{array}$

Therefore, for side lengths: 14 cm, 14 cm, 4 cm, $A=27.7c{m}^2$ .

3. For side lengths: $13cm$ , $13cm$ , $6cm$:

The area of the triangle is given by:

  $\begin{array}{l}A=\sqrt{16\left(16-13\right)\left(16-13\right)\left(16-6\right)}\\ A=\sqrt{16\left(3\right)\left(3\right)\left(10\right)}\\ A=\sqrt{1440}\\ A=37.9c{m}^2\end{array}$

Therefore, for side lengths: $13cm$ , $13cm$ , $6cm$ , $A=37.9c{m}^2$ .

4. For side lengths: $12cm$ , $12cm$ , $8cm$:

The area of the triangle is given by:

  $\begin{array}{l}A=\sqrt{16\left(16-12\right)\left(16-12\right)\left(16-8\right)}\\ A=\sqrt{16\left(4\right)\left(4\right)\left(8\right)}\\ A=\sqrt{2048}\\ A=45.3c{m}^2\end{array}$

Therefore, for side lengths: $12cm$ , $12cm$ , $8cm$ , $A=45.3c{m}^2$ .

5. For side lengths: $11cm$ , $11cm$ , $10cm$:

The area of the triangle is given by:

  $\begin{array}{l}A=\sqrt{16\left(16-11\right)\left(16-11\right)\left(16-10\right)}\\ A=\sqrt{16\left(5\right)\left(5\right)\left(6\right)}\\ A=\sqrt{2400}\\ A=49c{m}^2\end{array}$

Therefore, for side lengths: $11cm$ , $11cm$ , $10cm$ , $A=49c{m}^2$ .

6. For side lengths: $10cm$ , $10cm$ , $12cm$:

The area of the triangle is given by:

  $\begin{array}{l}A=\sqrt{16\left(16-10\right)\left(16-10\right)\left(16-12\right)}\\ A=\sqrt{16\left(6\right)\left(6\right)\left(4\right)}\\ A=\sqrt{2304}\\ A=48c{m}^2\end{array}$

Therefore, for side lengths: $10cm$ , $10cm$ , $12cm$ , $A=48c{m}^2$ .

7. For side lengths: $9cm$ , $9cm$ , $14cm$:

The area of the triangle is given by:

  $\begin{array}{l}A=\sqrt{16\left(16-9\right)\left(16-9\right)\left(16-14\right)}\\ A=\sqrt{16\left(7\right)\left(7\right)\left(2\right)}\\ A=\sqrt{1568}\\ A=39.6c{m}^2\end{array}$

Therefore, for side lengths: $9cm$ , $9cm$ , $14cm$ , $A=39.6c{m}^2$ .

Since only one set of side lengths out of $7$ yielded an area of $48c{m}^2$ , the probability that the area of the triangle is exactly $48c{m}^2$ is equal to $\frac{1}{7}$ .