Two corners of an isosceles triangle are at #(8 ,3 )# and #(6 ,2 )#. If the triangle's area is #4 #, what are the lengths of the triangle's sides?

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Answer 1 · 2018-01-19T04:30:55

Three sides of the he isosceles triangle are #color(blue)(2.2361, 2, 2)#

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#a = sqrt((6-8)^2 + (2-3)^2) = 2.2361#

#h =(2* Area) / a = (2*4)/2.2361 = 3.5777#

Slope of base BC #m_a = (2-3) / (6-8) = 1/2#

Slope of altitude AD is #-(1/m_a) = -2#

Midpoint of BC #D = (8+6)/2, (3+2)/2 = (7, 2.5)#

Equation of AD is

#y - 2.5 = -2 * (x - 7)#

#y + 2x = 11.5# Eqn (1)

Slope of BA #=m_b =tan theta = h / (a/2) = (2 * 3.5777) / 2.2361 = 3.1991#

Equation of AB is

#y - 3 = 3.1991 * (x - 8)#

#y - 3.1991x = - 22.5928# Eqn (2)

Solving Eqns (1), (2) we get the coordinates of A

#A (6.5574, 1.6149)#

Length AB #=c = sqrt((8-6.5574)^2 + (3-1.6149)^2) = 2#

Three sides of the he isosceles triangle are #color(blue)(2.2361, 2, 2)#