Question

Two corners of an isosceles triangle are at `(5 ,8 )` and `(4 ,1 )`. If the triangle's area is `36`, what are the lengths of the triangle's sides?

Answer 1

side b = `sqrt(50)=5sqrt(2)~~7.07` to 2 decimal places

sides a and c =`1/10sqrt(11618) ~~10.78` to 2 decimal places

Explanation:

In geometry it is always wise to draw a diagram. It comes under good communication and gets you extra marks.

`color(brown)("As long as you label all the relevant points and include")` `color(brown)("the pertinent data you do not always need to draw the")` `color(brown)("orientation exactly as it would appear for the given points")`

Let `(x_1,y_1)->(5,8)`
Let `(x_2,y_2)->(4,1)`

Note that it does not matter that vertex C should be on the left and vertex A on the right. It will work out. I did it this way round as it is the order you used.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Method plan")`

Step 1: Determine length of side b.
Step 2: Area known so use to determine h.
Step 3: Use Pythagoras to determine length side c and a
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step1")`

`b=sqrt( (x_2-x_1)^2+(y_2-y_1)^2)`

`b=sqrt((4-5)^2+(1-8)^2)`

`color(green)(b=sqrt(50))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step2")`

Area given as 36`" units"^2`

So `" "36=sqrt(50)/2xxh`

So `color(green)(h= (2xx36)/sqrt(50) = 72/(sqrt(50))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Step3")`

`"side c "=" side a"= sqrt( (b/2)^2+h^2)`

`c=sqrt( (sqrt(50)/2)^2 +(72/(sqrt(50)))^2)`

`c= sqrt(50/4+5184/50)`

`c=sqrt((1250+10368)/100)`

`c=sqrt(11618/100)`

`c=1/10sqrt(11618)`

`=>c~~10.78` to 2 decimal places