Two corners of an isosceles triangle are at $(7 ,6 )$ and $(4 ,9 )$ . If the triangle's area is $24$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-06-13T06:43:52

The length of the other sides is $=11.5$

The length of the base is

$b=sqrt((7-4)^2+(6-9)^2)=sqrt(3^2+3^2)=3sqrt2$

Let the altitude of the triangle be $=h$

Then,

The area is $A=1/2bh$

$1/2*3sqrt2*h=24$

$h=(2*24)/(3sqrt2)=8sqrt2$

The other sides of the triangle are

$a=c=sqrt(h^2+(b/2)^2)$

$=sqrt((8sqrt2)^2+(3/2sqrt2)^2)$

$=sqrt(128+9/2)$

$=sqrt(265/2)$

$=11.5$

Two corners of an isosceles triangle are at (7 ,6 )(7 ,6 ) and (4 ,9 )(4 ,9 ). If the triangle's area is 24 24 , what are the lengths of the triangle's sides?