Two corners of an isosceles triangle are at #(1 ,3 )# and #(9 ,7 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

1 Answer
Douglas K.
Dec 10, 2016

The sides of the triangle are #a = c = 15 and b = sqrt(80)#

Explanation:

Let the length of side b equal the distance between the two given points:

#b = sqrt((9 - 1)^2 + (7 - 3)^2)#

#b = sqrt((8)^2 + (4)^2)#

#b = sqrt(80)#

#Area = 1/2bh#

#2Area = bh#

#h = (2Area)/b#

#h = (2(64))/sqrt(80)#

#h = 128/sqrt(80)#

If side b is NOT one of the equal sides then the height is one of the legs of a right triangle and half of the length side b, #sqrt(80)/2# is the other leg. Therefore, we can use the Pythagorean Theorem to find the length of hypotenuse and this will be one of the equal sides:

#c = sqrt((128/sqrt(80))^2 + (sqrt(80)/2)^2)#

#c ~~ 15#

We need to find whether a triangle with sides, #a = c = 15 and b = sqrt(80)# has an area of 64.

I used a Heron's Formula Calculator and discovered that the area is 64.

The sides of the triangle are #a = c = 15 and b = sqrt(80)#

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