Two corners of an isosceles triangle are at #(4 ,3 )# and #(9 ,3 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

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Answer 1 · 2017-08-04T10:26:32

Length of sides of triangle are #5, 25.72(2dp) ,25.72(2dp) # unit

The base of the isosceles triangle,

#b=sqrt( (x_1-x_2)^2+(y_1-y_2)^2) = sqrt( (4-9)^2+(3-3)^2) #

# = sqrt25 = 5 # unit .

The area of the isosceles triangle is #A_t = 1/2 * b *h =1/2*5 *h #

# A_t=64 :. h = (2* A_t )/b = (2*64)/5=128/5= 25.6 # unit.

Where #h# is the altitude of triangle.

The legs of the isosceles triangle are #l_1=l_2= sqrt(h^2+(b/2)^2)=sqrt(25.6^2+(5/2)^2) ~~25.72(2dp) #unit

Hence the length of three sides of triangle are

#5, 25.72(2dp) ,25.72(2dp) # unit [Ans]