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?

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Answer 1 · 2016-06-27T06:47:00

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

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.