The question is as follows:

Two sides of an isosceles triangle with perimeter 112 cm are in the ratio 3:2. Find ALL possible lengths for the base.

Please help. Thanks!

6 answers

I did it. Check down two

Down two where?

http://www.jiskha.com/display.cgi?id=1473719167#1473719167.1473720184

thanks!

You are welcome.

To find all possible lengths for the base of the isosceles triangle, we need to understand two key facts about isosceles triangles:

1. In an isosceles triangle, the two equal sides are opposite equal angles.
2. The sum of the lengths of two sides of a triangle must be greater than the length of the third side.

Let's solve the problem step by step:

Step 1: Determine the lengths of the two equal sides.
Given that the ratio of the lengths of the two sides is 3:2, we can set up the following equation:

Let x be the common ratio.
Length of one side = 3x
Length of the other side = 2x

Since the perimeter of the triangle is 112 cm, we can write the equation:

3x + 3x + 2x = 112
8x = 112
x = 14

So, the length of one side is 3x = 3 * 14 = 42 cm, and the length of the other side is 2x = 2 * 14 = 28 cm.

Step 2: Determine the possible lengths for the base.
The sum of the two equal sides must be greater than the length of the base. That is:

42 + 42 > base
84 > base

Therefore, any length for the base less than 84 cm is not possible for this isosceles triangle.

So, the base of the isosceles triangle can have any length greater than 84 cm.

In summary, all possible lengths for the base of the isosceles triangle are greater than 84 cm.