Question

Two corners of an isosceles triangle are at `(5 ,4 )` and `(9 ,2 )`. If the triangle's area is `36`, what are the lengths of the triangle's sides?

Answer 1

The length of the sides are both: `s~~16.254` to 3 dp

Explanation:

It usually helps to draw a diagram:

`color(blue)("Method")`
Find base width `w`
Use in conjunction with area to find `h`
Using `h` and `w/2` in Pythagoras find `s`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("To determine the value of "w)`

Consider the green line in the diagram (base as would be plotted)

Using Pythagoras:
`w=sqrt((9-5)^2+(2-4)^2)`

`color(blue)(w=sqrt(4^2+(-2)^2) =sqrt(20) =2sqrt(5))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("To determine the value of "h)`

`"Area = w/2xxh`

`36=(2sqrt(5))/2xxh`

`36=2/2xxsqrt(5)xxh`

`color(blue)(h=36/sqrt(5))`
'~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("To determine the value of "s)`

Using Pythagoras

`(w/2)^2+h^2=s^2`

`=>s=sqrt( ( (2sqrt(5))/2)^2 +(36/sqrt(5))^2`

`=>s=sqrt( ( 5 +(36^2)/5)`

`s=sqrt( (25+36^2)/5) = sqrt(1321/5)`

`s~~16.254`

Answer 2

Support of the decision that the given points are for the base of the triangle.

Explanation:

Suppose the coordinates give were not for the base of an Isosceles triangle but for one of the other two sides. Then we would have:

Where

`x=2sqrt(5)xxsin(theta)`
`h=2sqrt(5)xxcos(theta)`

Given that Area`= 36 = x xx h`

Thus we have:

`" "color(blue)(36=(2sqrt(5)color(white)(.))^2( sin(theta)cos(theta)))`

`color(brown)("Using the Trig Identity of "sin(2theta)=2sin(theta)cos(theta))`

`" "color(brown)(36=20( sin(theta)cos(theta)) ->)color(blue)(36=20/2sin(2theta))`

`=> sin(2theta)=72/20`

But `" "-1<=sin(2theta)<=+1`

and `72/20>+1` so there is a`" "color(red)(underline("contradiction"))`

Implying that this scenario of `2sqrt(5)` not being the base is false.

`color(magenta)("The length of "2sqrt(5)" applies to the base of the triangle")`