Question

If the sides of a triangle are distinct positive integers in an arithmetic progression. If the smallest side is 10, then what will be the number of such triangles?
A. 8
B. 9
C. 10
D. Infinitely many

Answer 1

Hint: We have been given that the sides of the triangle are in an arithmetic progression and we know that the smallest side is 10 so we know that the first term of our arithmetic progression is our smallest side. Now we know that the sum of any two sides of a triangle must be greater than the third side. We will use that property and get to know about the range of the common difference of the A.P., we also know that common difference must be greater than 0 because the sides of the triangle are distinct. So, we will get to know about the different possible values of common difference to calculate the different combinations of sides possible, thereby get to know about the number of different triangles possible.

Complete step-by-step answer:
It has been given that the sides of the triangle are in an arithmetic progression.
Let us assume the sides to be $a,a + d,a + 2d$ where a is the first term of the arithmetic progression or the smallest side of the triangle and d is the common difference.
It is given that a = 10, so the sides of the triangle are 10, 10+d, 10+2d.
We know about the property that in a triangle the sum of any two sides must be greater than the third side.
$ \Rightarrow 10 + 10 + d > 10 + 2d$
$ \Rightarrow 20 + d > 10 + 2d$
$ \Rightarrow 20 - 10 > 2d - d$
$ \Rightarrow 10 > d$
$ \Rightarrow d < 10$
We also know that d > 0 because the sides of the triangle are distinct i.e., they must be different from each other because the condition of the smaller and larger side is there so they cannot be the same.
$ \Rightarrow 0 < d < 10$
So, the possible values of d are 1, 2, 3, 4, 5, 6, 7, 8, 9 i.e. 9 different values on the basis of which we will get 9 different possible combinations of the sides of the triangle which are listed below.
For d =1, sides are 10, 11, 12
For d=2, sides are 10, 12, 14
For d=3, sides are 10, 13, 16
For d=4, sides are 10, 14, 18
For d=5, sides are 10, 15, 20
For d=6, sides are 10, 16, 22
For d=7, sides are 10, 17, 24
For d=8, sides are 10, 18, 26
For d=9, sides are 10, 19, 28
Hence, we can say that 9 distinct triangles are possible.
$\therefore $ Option B. 9 is our correct answer.

Note: For such types of questions, just keep in mind the basic concept of arithmetic progression i.e. if three terms are in arithmetic progression they can be listed as a, a+d, a+2d where a is our first term and d is the common difference. Also, remember the property that in a triangle sum of any two sides is always greater than the third side i.e. triangle can only be formed if the sum of any two sides is always greater than the third side.