how do you find the height of an isosceles triangle only knowing the lengths of all three bases?

EX: There's isosceles triangle ABC with vertex angle B. AB= 5x-28. BC= 2x+11. AC= x+5

The triangle is isoceles, so there will be 1 base and 2 sides of equal length. A perpendicular line segment from the base to the vertex of the other 2 sides (aka the height) will also be the midpoint of the base. You then have a right triangle, with the hypotenuse being the side length, and the base being 1/2 the base of the isoceles triangle.

To find the height of the isosceles triangle, we can use the formula for the area of a triangle.

The formula for the area of a triangle is:
Area = (base x height) / 2

In this case, we have the lengths of all three bases, AB, BC, and AC. Since it is an isosceles triangle, we know that AB = AC, and the altitude (height) will drop from the vertex angle B and intersect the base AC at a right angle.

Let's find the length of the base AC first. We have AC = x + 5.

Now, let's calculate the area of the triangle using the given bases AB and BC:
Area = (AB x BC) / 2

Substituting the given values:
Area = [(5x - 28) x (2x + 11)] / 2

Next, we need to express the height in terms of x. We can then solve for x and substitute it back to find the height.

To find the height, we can use the formula. Rearranging the area formula, we get:
Height = (2 x Area) / Base

Substituting the values we found earlier, we have:
Height = (2 x [(5x - 28) x (2x + 11)] / 2) / (x + 5)

Simplifying this equation will give you the height of the isosceles triangle in terms of x.