Two corners of an isosceles triangle are at $(8, 5)$ $(8 ,5 )$ and $(6, 2)$ $(6 ,2 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(8, 5)$ $(8 ,5 )$ and $(6, 2)$ $(6 ,2 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-02-07T11:04:32

Lengths of triangle's sides are $3.61 (2 d p), 2.86 (d p), 2.86 (d p)$ $3.61(2dp) , 2.86(dp), 2.86(dp)$ unit.

Explanation:

Length of base of isoceles triangle is $b = \sqrt{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}} = \sqrt{{(8 - 6)}^{2} + {(5 - 2)}^{2}} = \sqrt{4 + 9} = \sqrt{13} = 3.61 (2 d p)$ $b=sqrt((x_1-x_2)^2+(y_1-y_2)^2) = sqrt((8-6)^2+(5-2)^2)= sqrt(4+9)=sqrt 13=3.61(2dp)$

Area of isoceles triangle is $A_{t} = \frac{1}{2} \cdot b \cdot h or 4 = \frac{1}{2} \cdot \sqrt{13} \cdot h or h = \frac{8}{\sqrt{13}} = 2.22 (2 d p)$ $A_t = 1/2 *b*h or 4 = 1/2*sqrt13*h or h = 8/sqrt 13 =2.22(2dp)$ . Where $h$ $h$ be the altitude of triangle.

Legs of isoceles triangle are $l_{1} = l_{2} = \sqrt{h^{2} + {(\frac{b}{2})}^{2}} = \sqrt{{2.22}^{2} + {(\frac{3.61}{2})}^{2}} = 2.86 (2 d p)$ $l_1=l_2= sqrt( h^2 +(b/2)^2) =sqrt( 2.22^2 +(3.61/2)^2) = 2.86(2dp)$ unit

Lengths of triangle's sides are $3.61 (2 d p), 2.86 (d p), 2.86 (d p)$ $3.61(2dp) , 2.86(dp),2.86(dp)$ unit. [Ans]

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Two corners of an isosceles triangle are at $(8, 5)$ $(8 ,5 )$ and $(6, 2)$ $(6 ,2 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Two corners of an isosceles triangle are at (8,5)(8 ,5 ) and (6,2)(6 ,2 ). If the triangle's area is 44 , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(8, 5)$ $(8 ,5 )$ and $(6, 2)$ $(6 ,2 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?