Two corners of an isosceles triangle are at $(4 ,2 )$ and $(6 ,1 )$ . If the triangle's area is $3$ , what are the lengths of the triangle's sides?

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Answer 1 · 2016-04-27T13:22:47

Length of the sides are $(13sqrt(5))/10$

Using Pythagoras the distance between points is

$sqrt ((x_2-x_1)^2+(y_2-y_1)^2) -> sqrt((6-4)^2+(1-2)^2)$

$=sqrt(5)->"triangle base"$
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Tony B

Area = half the base $xx$ hight $->3 = sqrt(5)/2xxh$

$=>h=6/sqrt(5)$

Using Pythagoras $(sqrt(5)/2)^2 +h^2=x^2$

$=>x=sqrt((sqrt(5)/2)^2+(6/sqrt(5))^2)$

$=>x=sqrt(5/4+36/5)$

$=>x=sqrt((25+144)/20) = sqrt(169/20)$

$x=sqrt( 13^3/(2^2xx5))$

$=>x=13/2sqrt(1/5) = 12/(2sqrt(5))$

It is frowned upon to have a root in the denominator

Multiply by 1 but in the form of $1=sqrt(5)/sqrt(5)$

$x=13/(2sqrt(5))xxsqrt(5)/sqrt(5) = (13sqrt(5))/10$

Two corners of an isosceles triangle are at (4 ,2 )(4 ,2 ) and (6 ,1 )(6 ,1 ). If the triangle's area is 3 3 , what are the lengths of the triangle's sides?