Brenton Dixon

2022-07-21

Probability That the Sides of an Isosceles Triangle is an Equilateral Triangle

The sides of an isosceles triangle are whole numbers, and its perimeter is 30 units. What is the probability that the triangle is equilateral?

The sides of an isosceles triangle are whole numbers, and its perimeter is 30 units. What is the probability that the triangle is equilateral?

Kali Galloway

Beginner2022-07-22Added 16 answers

Step 1

There aren't a lot of options. There is only one equilateral triangle, which has side length (10,10,10).

In order to form a triangle, the sum of any two sides needs to be larger than the third side, so only these combinations are possible: (14,14,2),(13,13,4),(12,12,6),(11,11,8),(10,10,10),(9,9,12),(8,8,14)

Step 2

So the answer is $\frac{1}{7}$

There aren't a lot of options. There is only one equilateral triangle, which has side length (10,10,10).

In order to form a triangle, the sum of any two sides needs to be larger than the third side, so only these combinations are possible: (14,14,2),(13,13,4),(12,12,6),(11,11,8),(10,10,10),(9,9,12),(8,8,14)

Step 2

So the answer is $\frac{1}{7}$

Deromediqm

Beginner2022-07-23Added 2 answers

Step 1

Let a,b,c denote the lengths of the sides of our triangle. Without loss of generality, since the triangle is isoceles, let $a=c$, so that (a,a,b) describes our triangle, where $a,b\in \mathbb{N}$. Now we know additionally $2a+b=30$, and by the triangle inequality, $b\le 2a$, So $30-2a\le 2a$, i.e. $a\ge 8$, but of course $a\le 15$. Thus the possible triangles are described by $\{(8,8,14),(9,9,12),(10,10,10),(11,11,8),(12,12,6),(13,13,4),(14,14,2),(15,15,0)\}$.

Step 2

Only one of these is equilateral. Now assuming uniform distribution over these possibilities, and depending whether you include the degenerate (15,15,0), the probability is either $\frac{1}{7}$ or $\frac{1}{8}$.

Let a,b,c denote the lengths of the sides of our triangle. Without loss of generality, since the triangle is isoceles, let $a=c$, so that (a,a,b) describes our triangle, where $a,b\in \mathbb{N}$. Now we know additionally $2a+b=30$, and by the triangle inequality, $b\le 2a$, So $30-2a\le 2a$, i.e. $a\ge 8$, but of course $a\le 15$. Thus the possible triangles are described by $\{(8,8,14),(9,9,12),(10,10,10),(11,11,8),(12,12,6),(13,13,4),(14,14,2),(15,15,0)\}$.

Step 2

Only one of these is equilateral. Now assuming uniform distribution over these possibilities, and depending whether you include the degenerate (15,15,0), the probability is either $\frac{1}{7}$ or $\frac{1}{8}$.

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