Two corners of an isosceles triangle are at $(7, 2)$ $(7 ,2 )$ and $(3, 9)$ $(3 ,9 )$ . If the triangle's area is $24$ $24$ , what are the lengths of the triangle's sides?

Question

1363 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-28T08:04:03

The lengths of the sides of the isoceles triangle are $8.1 u$ $8.1u$ , $7.2 u$ $7.2u$ and $7.2 u$ $7.2u$

The length of the base is

$b = \sqrt{{(3 - 7)}^{2} + {(9 - 2)}^{2}} = \sqrt{16 + 49} = \sqrt{65} = 8.1 u$ $b=sqrt((3-7)^2+(9-2)^2)=sqrt(16+49)=sqrt65=8.1u$

The area of the isoceles triangle is

$a r e a = a = \frac{1}{2} \cdot b \cdot h$ $area=a=1/2*b*h$

$a = 24$ $a=24$

Therefore,

$h = \frac{2 a}{b} = \frac{2 \cdot 24}{\sqrt{65}} = \frac{48}{\sqrt{65}}$ $h=(2a)/b=(2*24)/sqrt65=48/sqrt65$

Let the length of the sides be $= l$ $=l$

Then, by Pythagoras

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

$l^{2} = {(\frac{\sqrt{65}}{2})}^{2} + {(\frac{48}{\sqrt{65}})}^{2}$ $l^2=(sqrt65/2)^2+(48/sqrt65)^2$

$= \frac{65}{4} + \frac{48^{2}}{65}$ $=65/4+48^2/65$

$= 51.7$ $=51.7$

$l = \sqrt{51.7} = 7.2 u$ $l=sqrt51.7=7.2u$

Two corners of an isosceles triangle are at (7,2)(7 ,2 ) and (3,9)(3 ,9 ). If the triangle's area is 2424 , what are the lengths of the triangle's sides?