Question

Two corners of an isosceles triangle are at `(4 ,8 )` and `(1 ,3 )`. If the triangle's area is `2`, what are the lengths of the triangle's sides?

Answer 1

Lengths of the triangle's sides are `AC = BC =3.0, AB=5.83`

Explanation:

Let ABC be the isocelles triangle of which AB is base and AC=BC and the corners are A`(4,8)` and B `(1,3)`. Base `AB=sqrt((3-8)^2+(1-4)^2) = sqrt 34` Let CD be the altitude(h) drawn from corner C on AB at point D , which is the mid point of AB. We know `area = 1/2*AB * h` or `2 = sqrt34 * h/2 or h = 4/sqrt34` Hence side `AC^2= (sqrt34/2)^2 + (4/sqrt34)^2 or AC =3.0 = BC` since `AC^2=AD^2+CD^2` `:.AC = BC =3.0, AB=sqrt 34=5.83` [Ans]