Two corners of an isosceles triangle are at $(8 ,3 )$ and $(6 ,2 )$ . If the triangle's area is $4$ , what are the lengths of the triangle's sides?

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Answer 1 · 2018-01-19T04:30:55

Three sides of the he isosceles triangle are $color(blue)(2.2361, 2, 2)$

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$a = sqrt((6-8)^2 + (2-3)^2) = 2.2361$

$h =(2* Area) / a = (2*4)/2.2361 = 3.5777$

Slope of base BC $m_a = (2-3) / (6-8) = 1/2$

Slope of altitude AD is $-(1/m_a) = -2$

Midpoint of BC $D = (8+6)/2, (3+2)/2 = (7, 2.5)$

Equation of AD is

$y - 2.5 = -2 * (x - 7)$

$y + 2x = 11.5$ Eqn (1)

Slope of BA $=m_b =tan theta = h / (a/2) = (2 * 3.5777) / 2.2361 = 3.1991$

Equation of AB is

$y - 3 = 3.1991 * (x - 8)$

$y - 3.1991x = - 22.5928$ Eqn (2)

Solving Eqns (1), (2) we get the coordinates of A

$A (6.5574, 1.6149)$

Length AB $=c = sqrt((8-6.5574)^2 + (3-1.6149)^2) = 2$

Three sides of the he isosceles triangle are $color(blue)(2.2361, 2, 2)$

Two corners of an isosceles triangle are at (8 ,3 )(8 ,3 ) and (6 ,2 )(6 ,2 ). If the triangle's area is 4 4 , what are the lengths of the triangle's sides?