Question

The possible values of a, the base of an isosceles triangle, and b, one of the legs of the triangle, are as follows: 26≤a≤28 and 41≤b≤43 (in millimeters). Find the possible values of the perimeter of the triangle.

Answer 1

First, you should try to find the sum of the two legs. Because the triangle is isosceles so the legs (b) are equal, than the sum of the two legs:

41*2 ≤ 2b ≤ 43*2 simplified is: 82 ≤ 2b ≤ 86

But we aren't done, for the perimeter also includes the base of the triangle, which is a. Since we already know the sum of the legs, we simply have to add the base (a) to the legs.

Sum of Legs:         82 ≤ 2b ≤ 86

Base:                      26 ≤ a ≤ 28

Perimeter (a+2b):   82+26 ≤ P ≤ 86+28     Which simplified:   108 ≤ P ≤ 114

Answer 2

The perimeter of a shape is the sum of its sides

The possible values of the perimeter are: $\mathbf{108 \le P \le 114}$

The given parameters are:

$\mathbf{Base: 26 \le a \le 28}$

$\mathbf{Leg: 41 \le b \le 43}$

The perimeter (P) of an isosceles triangle is:

$\mathbf{P = a + 2b}$

So, we simply add the endpoints of the inequalities.

This gives:

$\mathbf{26 + 2 \times 41 \le P \le 28 + 2 \times 43}$

$\mathbf{26 + 82 \le P \le 28 + 86}$

$\mathbf{108 \le P \le 114}$

Hence, the possible values of the perimeter are:

$\mathbf{108 \le P \le 114}$

Read more about perimeters at:

https://brainly.com/question/6465134