Question

how many isosceles triangle with integer sides are possible such that sum of two the sides is 10

Answer 1

Answer:

hope this will make u understand

Step-by-step explanation:

Two possibilities: Two equal sides could add up to 12 or sum of 2 unequal sides = 12.

i.e. Sum of 2 equal sides = 12

Sum of 2 unequal sides = 12

=> If sum of two equal sides were 12, sides of the triangle should be 6, 6, x.

What are the values x can take?

x could range from 1 to 11.

11 integer values exist.

Now 2 unequal sides adding to 12. This could be 1 + 11, 2 + 10, 3 + 9, 4 + 8 or 5 + 7.

How many isosceles triangles are possible with the above combinations?

Isosceles triangles with the above combination:

1, 11, 11 ☑ 1, 1, 11 ☒

2, 10, 10 ☑ 2, 2, 10 ☒

3, 9, 9 ☑ 3, 3, 9 ☒

4, 8, 8 ☑ 4, 4, 8 ☒

5, 7, 7 ☑ 5, 5, 7 ☑

6 possibilities. Triplets such as (1, 1, 11), (2, 2, 10), etc are eliminated as sum of the two smaller values is less than the largest value. These cannot form a triangle.

The question is " How many isosceles triangles with integer sides are possible such that sum of two of the side is 12?"

11 + 6 = 17 possibilities totally.

Hence, the answer is 17.

Choice C is the correct answer.