Two corners of an isosceles triangle are at (8 ,3 ) and (5 ,4 ). If the triangle's area is 4 , what are the lengths of the triangle's sides?

1 Answer
Jun 27, 2016

The length of the sides are sqrt 10, sqrt 10, sqrt 8 and the points are (8,3), (5,4) and (6,1)

Explanation:

Let the points of the triangle be (x_1,y_1),( x_2, y_2),( x_3, y_3).

Area of triangle is A = ((x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2))/2)

Given A = 4, (x_1,y_1) = (8,3), (x_2, y_2) = (5,4)

Substituting we have the below Area equation:

((8(4 – y_3) + 5(y_3 – 3) + x_3(3 – 4))/2 )= 4

((8(4 – y_3) + 5(y_3 – 3) + x_3(3 – 4)) = 8

(32 – 8y_3) + (5y_3 – 15) + (-1x_3) = 8

17 – 3y_3 -x_3 = 8

– 3y_3 -x_3 = (8-17)

– 3y_3 -x_3 = -9

3y_3 + x_3 = 9 ----> Equation 1

Distance between points (8,3), (5,4) using distance formula is

sqrt ( (8-5)^2+(3-4)^2) = sqrt ( 3^2+(-1)^2) = sqrt 10

Distance between points (x_3,y_3), (5,4) using distance formula is
sqrt ((x_3 -5)^2 + (y_3 - 4)^2) = sqrt 10

Squaring both sides and subsituting x_3 = 9 - 3y_3 from equation 1, we get a quadratic equation.

(9-3y_3 - 5)^2 + (y_3 - 4 )^ 2 = 0

(4-3y_3)^2 + (y_3 - 4 )^ 2 = 0

Factorizing this, we get (y-1)(10y-22) = 0

y = 1 or y = 2.2 . y=2.2 can be discarded. Hence, the third point has to be (6,1).

By calculating the distances for points (8,3), (5,4) and (6,1), we get sqrt 8 for the length of the base.