Two corners of an isosceles triangle are at $(1, 3)$ $(1 ,3 )$ and $(5, 8)$ $(5 ,8 )$ . If the triangle's area is $6$ $6$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(1, 3)$ $(1 ,3 )$ and $(5, 8)$ $(5 ,8 )$ . If the triangle's area is $6$ $6$ , what are the lengths of the triangle's sides?

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Answer 1 · 2016-03-13T13:40:22

possible triangle with sides: $\sqrt{92537} = 3.70974$ $sqrt(92537)=3.70974$ , $\sqrt{92537} = 3.70974$ $sqrt(92537)=3.70974$ , $\sqrt{41} = 6.40312$ $sqrt41=6.40312$
possible triangle with sides: $\sqrt{413}$ $sqrt413$ , $\sqrt{41}$ $" "sqrt41$ , $1.89495$ $" "1.89495$

Explanation:

Distance between the two points
$= \sqrt{{(5 - 1)}^{2} + {(8 - 3)}^{2}} = \sqrt{41}$ $=sqrt((5-1)^2+(8-3)^2)=sqrt41$
Let this be the base $b = \sqrt{41}$ $b=sqrt41$
Formula for Area of triangle:

Area $= \frac{1}{2} b h$ $=1/2bh$
$6 = \frac{1}{2} \sqrt{41} \cdot h$ $6=1/2sqrt41*h$

$h = \frac{12}{\sqrt{41}}$ $h=12/sqrt41$

The two equal sides can now be computed:
Let x be one of the sides
$x = \sqrt{h^{2} + {(\frac{b}{2})}^{2}}$ $x=sqrt(h^2+(b/2)^2)$
$x = \sqrt{{(\frac{12}{\sqrt{41}})}^{2} + {(\frac{\sqrt{41}}{2})}^{2}}$ $x=sqrt((12/sqrt41)^2+((sqrt41)/2)^2)$
$x = \frac{1}{82} \sqrt{92537}$ $x=1/82sqrt(92537)$
$x = 3.70974$ $x=3.70974$
First triangle with sides:
$x = 3.70974$ $x=3.70974$
$x = 3.70974$ $x=3.70974$
$b = \sqrt{41} = 6.40312$ $b=sqrt41=6.40312$

For the second triangle:
the two equal sides are:
$x = \sqrt{41}$ $x=sqrt41$
$x = \sqrt{41}$ $x=sqrt41$
and the 3rd side $b$ $b$
compute the vertex angle of the isosceles triangle first:
Area $= \frac{1}{2} \sqrt{41} \cdot \sqrt{41} \cdot sin θ$ $=1/2sqrt41*sqrt41*sin theta$
$6 = \frac{1}{2} \sqrt{41} \cdot \sqrt{41} \cdot sin θ$ $6=1/2sqrt41*sqrt41*sin theta$
$sin θ = \frac{12}{41}$ $sin theta=12/41$
$θ = {17.0186}^{\circ}$ $theta=17.0186^@$
Using cosine law
$b = \sqrt{x^{2} + x^{2} - 2 \cdot x \cdot x \cdot cos θ}$ $b=sqrt(x^2+x^2-2*x*x*cos theta)$
$b = \sqrt{{\sqrt{41}}^{2} + {\sqrt{41}}^{2} - 2 \cdot \sqrt{41} \cdot \sqrt{41} \cdot cos ({17.0186}^{\circ})}$ $b=sqrt(sqrt41^2+sqrt41^2-2*sqrt41*sqrt41*cos(17.0186^@))$
$b = 1.89495$ $b=1.89495" "$ units

God bless....I hope the explanation is useful.

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Two corners of an isosceles triangle are at $(1, 3)$ $(1 ,3 )$ and $(5, 8)$ $(5 ,8 )$ . If the triangle's area is $6$ $6$ , what are the lengths of the triangle's sides?

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Explanation:

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Two corners of an isosceles triangle are at $(1, 3)$ $(1 ,3 )$ and $(5, 8)$ $(5 ,8 )$ . If the triangle's area is $6$ $6$ , what are the lengths of the triangle's sides?