Two corners of an isosceles triangle are at $(2, 4)$ $(2 ,4 )$ and $(4, 7)$ $(4 ,7 )$ . If the triangle's area is $9$ $9$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-07-21T14:45:42

The lengths of the sides of the triangle are $3.61 u, 5.30 u, 5.30 u$ $3.61u, 5.30u,5.30u$

The length of the base is

$b = \sqrt{{(4 - 2)}^{2} + {(7 - 4)}^{2}} = \sqrt{4 + 9} = \sqrt{13} = 3.61$ $b=sqrt((4-2)^2+(7-4)^2)=sqrt(4+9)=sqrt13=3.61$

Let the altitude of the triangle be $= h$ $=h$

Then

The area of the triangle is $A = \frac{1}{2} \cdot b \cdot h$ $A=1/2*b*h$

$h = 2 \frac{A}{b} = 2 \cdot \frac{9}{\sqrt{13}} = \frac{18}{\sqrt{13}} = 4.99$ $h=2A/b=2*9/(sqrt13)=18/sqrt13=4.99$

The sides of the triangle are

$= \sqrt{h^{2} + {(\frac{b}{2})}^{2}}$ $=sqrt(h^2+(b/2)^2)$

$= \sqrt{\frac{18^{2}}{13} + \frac{13}{4}}$ $=sqrt(18^2/13+13/4)$

$= 5.30$ $=5.30$

Two corners of an isosceles triangle are at (2,4)(2 ,4 ) and (4,7)(4 ,7 ). If the triangle's area is 99 , what are the lengths of the triangle's sides?