Two corners of an isosceles triangle are at #(9 ,4 )# and #(1 ,8 )#. If the triangle's area is #48 #, what are the lengths of the triangle's sides?

To find the lengths of the sides of the isosceles triangle, we first need to determine the length of the base and the height of the triangle.

Given the coordinates of two corners of the triangle, we can calculate the length of the base using the distance formula:

Distance = √((x2 - x1)^2 + (y2 - y1)^2)

Using the coordinates (9, 4) and (1, 8), we find the length of the base:

Base = √((9 - 1)^2 + (4 - 8)^2) = √((8)^2 + (-4)^2) = √(64 + 16) = √80

Next, we need to find the height of the triangle. Since the triangle is isosceles, the height will be the perpendicular distance from the midpoint of the base to the opposite vertex.

To find the midpoint of the base, we average the x-coordinates and y-coordinates of the given points:

Midpoint = ((x1 + x2) / 2, (y1 + y2) / 2) = ((9 + 1) / 2, (4 + 8) / 2) = (5, 6)

Now, we calculate the distance from the midpoint to the third vertex (which is the height of the triangle):

Height = Distance from midpoint to third vertex = √((x3 - 5)^2 + (y3 - 6)^2)

Given that the area of the triangle is 48, we can use the formula for the area of a triangle:

Area = (1/2) * Base * Height

Substituting the given values:

48 = (1/2) * √80 * Height

Solving for Height:

Height = (48 * 2) / √80 ≈ 13.42

Now that we have the height, we can find the lengths of the other two sides of the triangle using the Pythagorean theorem. Since the triangle is isosceles, these two sides will have the same length.

Let's denote the length of each of these sides as S:

S^2 = (Base/2)^2 + Height^2

Substituting the given values:

S^2 = (√80 / 2)^2 + (13.42)^2 = (√80 / 2)^2 + (13.42)^2 ≈ 66.37

S ≈ √66.37

Therefore, the lengths of the sides of the isosceles triangle are approximately √80 and √66.37.