Two corners of an isosceles triangle are at $(5 ,8 )$ and $(9 ,1 )$ . If the triangle's area is $36$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-01-06T11:09:11

The length of three sides of triangle are $8.06 ,9.8 , 9.8$ unit

Base of the isocelles triangle is $B= sqrt((x_2-x_1)^2+(y_2-y_1)^2)) = sqrt((9-5)^2+(1-8)^2)) =sqrt(16+49)=sqrt65 =8.06(2dp)$ unit

We know area of triangle is $A_t =1/2*B*H$ Where $H$ is altitude.
$:. 36=1/2*8.06*H or H= 72/8.06=8.93(2dp)$ unit

Legs are $L = sqrt(H^2+(B/2)^2)= sqrt( 8.93^2+(8.06/2)^2)=9.80(2dp)$ unit

The length of three sides of triangle are $8.06 ,9.8 , 9.8$ unit [Ans]

Two corners of an isosceles triangle are at (5 ,8 )(5 ,8 ) and (9 ,1 )(9 ,1 ). If the triangle's area is 36 36 , what are the lengths of the triangle's sides?