Two corners of an isosceles triangle are at #(2 ,9 )# and #(4 ,3 )#. If the triangle's area is #9 #, what are the lengths of the triangle's sides?

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Answer 1 · 2016-12-03T04:02:58

The sides are #a = 4.25, b = sqrt(40), c = 4.25#

Let side #b = sqrt((4 - 2)^2 + (3 - 9)^2)#

#b = sqrt((2)^2 + (-6)^2)#

#b = sqrt(4 + 36)#

#b = sqrt(40)#

We can find the height of the triangle, using #A = 1/2bh#

#9 = 1/2sqrt(40)h#

#h = 18/sqrt(40)#

We do not know whether b is one of the sides that are equal.

If b is NOT one of the sides that are equal, then the height bisects the base and the following equation is true:

#a^2 = c^2 = h^2 + (b/2)^2#

#a^2 = c^2 = 324/40 + 40/4#

#a^2 = c^2 = 8.1 + 10#

#a^2 = c^2 = 18.1#

#a = c ~~ 4.25#

#s = (sqrt(40) + 2(4.25))/2#

#s ~~ 7.4#

#A = sqrt(s(s - a)(s - b)(s - c))#

#A = sqrt(7.4(3.2)(1.07)(3.2))#

#A ~~ 9#

This is consistent with the given area, therefore, side b is NOT one of the equal sides.

The sides are #a = 4.25, b = sqrt(40), c = 4.25#