Show that the points A(-1,2), B(5,2) & C(2,5) are the vertices of an isosceles triangle. Find the area of triangle ABC.

Expert Answers
kjcdb8er eNotes educator| Certified Educator

By definition, an isosceles triangle is one that has two congruent (equal) angles. To prove two angles equal on the triangle ABC, let us consider the definition of tangent:

tanx = opposite/adjacent = h/a

Noting that the y values for points A and B are equal, the height is the distance from point C to the line y = 2:

h = 5 - 2 = 3

The distance AB = 5 - -1 = 6

Let D be the point through C perpendicular to AB; i.e. the line x = 2.

AD = 2 - -1 = 3   ; DB = 5 - 2 = 3

(note that a triangle with perpendicular bisector is isosceles. But proof by definition follows)

The tangent of angle CAB = h/AD = 3/3 = 1

The tangent of angle CBA = h/DB = 3/3 = 1

Therefore the angles are congruent, and the triangle is isosceles.

 

The definition of the Area of a triangle is half base times height. We know both from above:

h = 3, AB = 6

0.5*h*AB = 9

hala718 eNotes educator| Certified Educator

Let us calulate the length of the triangle sides:

AB = sqrt[(2-2)^2 + (5+1)^2]= sqrt(36) = 6

BC = sqrt[(5-2)^2 + (2-5)^2]= sqrt18 = 3sqrt2

AC = sqrt[(5-2)^2 + (2+1)^2 = sqrt18 = 3sqrt2

We notice that BC = AC = 3sqrt2

Then ABC is an isoscale triangle.

Now The area:

We know that the area a= (1/2)base * height

height is CD where D in the midpoint of AB

==> CD^2 = BC^2 - (AB/2)^2

==> CD = sqrt(18 - 9) = sqrt9 = 3

  ==> a = (1/2)AB * CD

             = (1/2)*6 * 3

              = 3*3

==> area (a) = 9

neela | Student

A )(-1,2), B(5,2) and C(2,5)

Therfore AB^2 = (5- -1)^2+(2-2)^2 = (5+2)^2 = 49. Or AB = sqrt(49) = 7.

BC^2 = = (2-5)^2+(5-2)^2 = 9+9=18. BC = sqrt(18) = 3sqrt2.

AC^2 = (2- -1)^2+(5-2)^2= 9+9= 18. AC = sqrt18 = 3sqrt2.

Therefore BC =AC  = 3sqrt2. Triangle ABC is an isosceles .

Therefore square on altitude from AB to C = AD^2 =   AC^2 - { (1/2) AB}^2 =  18 - 49/4 = 18- 12.25 = 5.75.

Therefore the area of the triangle = (1/2) *BC*AD = (1/2)*7* sqrt5.75 = 8.39271 nearly.