Two corners of an isosceles triangle are at #(9 ,4 )# and #(1 ,8 )#. If the triangle's area is #48 #, what are the lengths of the triangle's sides?

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Answer 1 · 2017-12-08T10:48:53

Measure of the three sides are (8.9443, 11.6294, 11.6294)

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Length #a = sqrt((9-1)^2 + (4-8)^2) = sqrt 80 = 8.9443#

Area of #Delta = 48#
#:. h = (Area) / (a/2) = 48 / (8.9443/2) = 48 / 4.4772 = 10.733#

#side b = sqrt((a/2)^2 + h^2) = sqrt((4.4772)^2 + (10.733)^2)#
#b = 11.6294#

Since the triangle is isosceles, third side is also #= b = 11.6294#

Measure of the three sides are (8.9443, 11.6294, 11.6294)