Question

Two corners of an isosceles triangle are at `(6 ,3 )` and `(5 ,8 )`. If the triangle's area is `8`, what are the lengths of the triangle's sides?

Answer 1

case 1. Base`=sqrt26 and` leg`=sqrt(425/26)`
case 2. Leg `=sqrt26 and` base`=sqrt(52+-sqrt1680)`

Explanation:

Given Two corners of an isosceles triangle are at `(6,3) and (5,8)`.
Distance between the corners is given by the expression
`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`, inserting given values
`d=sqrt((5-6)^2+(8-3)^2)`
`d=sqrt((-1)^2+(5)^2)`
`d=sqrt26`

Now area of triangle is given by
`"Area"=1/2"base"xx"height"`

Case 1. The corners are base angles.
`:."base"=sqrt26`
`"height"=2xx"Area"/"base"` .....(1)
`=2xx8/sqrt26=16/sqrt26`
Now using the Pythagoras theorem
`"leg"=sqrt("height"^2+("base"/2)^2)`
`"leg"=sqrt((16/sqrt26)^2+(sqrt26/2)^2)`
`=sqrt(256/26+26/4`
`=sqrt(128/13+13/2)`
`=sqrt(425/26)`

Case 2. The corners are base angle and the vertex.
`"Leg"=sqrt26`
Let `"base"=b`
Also from (1) `"height"=2xx"Area"/"base"`
`"height"=2xx8/"base"`
`"height"=16/"base"`
Now using the Pythagoras theorem
`"leg"=sqrt("height"^2+("base"/2)^2)`
`sqrt26=sqrt("256/b^2+b^2/4)`, squaring both sides
`26="256/b^2+b^2/4`
`104b^2=1024+b^4`
`b^4-104b^2+1024=0`, solving for `b^2` using the quadratic formula
`b^2=(104+-sqrt((-104)^2-4xx1024xx1))/2`
`b^2=52+-sqrt1680`, taking square root
`b=sqrt(52+-sqrt1680)`, we have ignored the negative sign as length can not be negative.