Question

Two corners of an isosceles triangle are at `(1 ,3 )` and `(5 ,3 )`. If the triangle's area is `6`, what are the lengths of the triangle's sides?

Answer 1

The sides of the isosceles triangle: 4, `sqrt13,sqrt13`

Explanation:

We're being asked about the area of an isosceles triangle with two corners at (1,3) and (5,3) and area 6. What are the lengths of the sides.

We know the length of this first side: `5-1=4` and I'm going to assume this is the base of the triangle.

The area of a triangle is `A=1/2bh`. We know `b=4` and `A=6`, so we can figure out `h`:

`A=1/2bh`

`6=1/2(4)h`

`h=3`

We can now construct a right triangle with `h` as one side, `1/2b=1/2(4)=2` as the second side, and the hypotenuse being the "slanty side" of the triangle (with the triangle being isosceles, so the 2 slanty sides being of equal length, we can do this one right triangle and get both missing sides). The Pythagorean Theorem is what is called for here - but I don't like `a` and `b` and `c` - I prefer `s` for short side, `m` for medium side and `h` for hypotenuse or simply `l` for long side:

`s^2+m^2=l^2`

`2^2+3^2=l^2`

`4+9=l^2`

`13=l^2`

`l=sqrt13`

And now we have all the sides of the isosceles triangle: 4, `sqrt13,sqrt13`