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Two corners of an isosceles triangle are at #(9 ,4 )# and #(3 ,2 )#. If the triangle's area is #48 #, what are the lengths of the triangle's sides?

Answer 1

#2sqrt(10),sqrt(1202/5),sqrt(1202/5)~=6.325,15.505,15.505#

The length of the given side is
#s=sqrt((9-3)^2+(4-2)^2)=sqrt(40)=2sqrt(10)~=6.325#

From the formula of the triangle's area:
#S=(b*h)/2# => #48=(cancel(2)sqrt(10)*h)/cancel(2)# => #h=48/sqrt(10)~=15.179#

Since the figure is an isosceles triangle we could have Case 1 , where the base is the singular side, ilustrated by Fig. (a) below

Or we could have Case 2 , where the base is one of the equal sides, ilustrated by Figs. (b) and (c) below

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Answer 2

James Atkins

The lengths of the triangle's sides are 8 and 8√5.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.