Two corners of an isosceles triangle are at $(2, 9)$ $(2 ,9 )$ and $(8, 5)$ $(8 ,5 )$ . If the triangle's area is $9$ $9$ , what are the lengths of the triangle's sides?

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Answer 1 · 2018-05-03T11:42:15

Length of three sides of the triangle are $7.21, 4.39, 4.39$ $7.21 ,4.39 , 4.39$ unit

Base of the isosceles triangle is

$B = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}})$ $B= sqrt((x_2-x_1)^2+(y_2-y_1)^2))$

$= \sqrt{{(8 - 2)}^{2} + {(5 - 9)}^{2}}) = \sqrt{36 + 16} = \sqrt{52} = 7.21 (2 d p)$ $= sqrt((8-2)^2+(5-9)^2)) =sqrt(36+16)=sqrt 52 =7.21 (2dp)$

unit .We know area of triangle is $A_{t} = \frac{1}{2} \cdot B \cdot H$ $A_t =1/2*B*H$ , where

$H$ $H$ is altitude. $:. 9=1/2*7.21*H or H= 18/7.21= 2.50 (2dp)$

unit. Legs are $L = sqrt(H^2+(B/2)^2)$

$= sqrt( 2.50^2+(7.21/2)^2)= 4.39 (2dp)$ unit

Length of three sides of triangle are $7.21 ,4.39 , 4.39$ unit [Ans]

Two corners of an isosceles triangle are at (2 ,9 )(2,9)(2 ,9 ) and (8 ,5 )(8,5)(8 ,5 ). If the triangle's area is 9 99 , what are the lengths of the triangle's sides?