Question

Two corners of an isosceles triangle are at `(9 ,6 )` and `(7 ,2 )`. If the triangle's area is `64`, what are the lengths of the triangle's sides?

Answer 1

`"sides "a = c = 28.7" units"` and `"side " b = 2sqrt5" units"`

Explanation:

let `b =` the distance between the two points:

`b = sqrt((9-7)^2+(6-2)^2)`

`b = 2sqrt5" units"`

We are given that the `"Area" = 64" units"^2`

Let "a" and "c" be the other two sides.

For a triangle, `"Area " = 1/2bh`

Substituting in the values for "b" and the Area:

`64" units"^2 = 1/2(2sqrt5" units")h`

Solve for the height:

`h = 64/sqrt5 = 64/5sqrt5" units"`

Let `C =` the angle between side "a" and side "b", then we may use the right triangle formed by side "b" and the height to write the following equation:

`tan(C) = h/(1/2b)`

`tan(C) = (64/5sqrt5" units")/(1/2(2sqrt5" units"))`

`C = tan^-1(64/5)`

We can find the length of side "a", using the following equation:

`h = (a)sin(C)`

`a = h/sin(C)`

Substitute in the values for "h" and "C":

`a = (64/5sqrt5" units")/sin(tan^-1(64/5))`

`a = 28.7" units"`

Intuition tells me that side "c" is the same length as side "a" but we can prove this using the Law of Cosines:

`c^2 = a^2 + b^2 - 2(a)(b)cos(C)`

Substitute in the values for a, b, and C:

`c^2 = (28.7" units")^2 + (2sqrt5" units")^2 - 2(28.7" units")(2sqrt5" units")cos(tan^-1(64/5))`

`c = 28.7" units"`