# Thread: Area of triangle inscribed in a rectangular prism

1. ## Area of triangle inscribed in a rectangular prism

i dont know where to start

*Finding the volume of the rectangular prism

2. ## Re: Area of triangle inscribed in a rectangular prism

Use the Pythagorean theorem to develop a system of equations for $a,~b,~c$ given the lengths you are given.

Solve that system.

for example

$HJ^2 = a^2+b^2 = 4^2 = 16$

2 other similar equations can be developed.

3. ## Re: Area of triangle inscribed in a rectangular prism

Originally Posted by Ilikebugs
i dont know where to start
*Finding the volume of the rectangular prism

The coordinates are: $O: (0,0,0),~A: (0,0 ,2a ),~B: (0,2b ,2a ),~C: (0,2b ,0 ),~D: (-2c, 2b,0 ),~E: (-2c,2b ,2a )$
Now the coordinates of $H,~J,~K$ are the midpoints of the diagonals of their faces.

4. ## Re: Area of triangle inscribed in a rectangular prism

b^2+c^2=(JK)^2=36
a^2+c^2=HK^2=25

5. ## Re: Area of triangle inscribed in a rectangular prism

ok ... you've got three equations, and the volume of the prism is $V=8abc$

$a^2+b^2=4^2$

$a^2+c^2=5^2$

$b^2+c^2=6^2$

... what next?

6. ## Re: Area of triangle inscribed in a rectangular prism

2-1=c^2-b^2=9
3-2=b^2-a^2=11
3-1=c^2-a^2=20

c=sqr(20+a^2)
b=sqr(11+a^2)

V=8*a*sqr(20+a^2)*sqr(11+a^2)

a^2+11+a^2=16
a=sqr(5/2)

V=8*5/2*sqr(20+5/2)*sqr(11+5/2)?

(I think I messed up

7. ## Re: Area of triangle inscribed in a rectangular prism

Originally Posted by skeeter
[IMG]http://mathhelpforum.com/geometry/

ok ... you've got three equations, and the volume of the prism is $V=8abc$

$a^2+b^2=4^2$

$a^2+c^2=5^2$

$b^2+c^2=6^2$

... what next?
subtract (1) from (3)

$c^2 - a^2 = 20$

$2c^2 = 45$

$c = \sqrt{\dfrac{45}{2}} = 3\sqrt{\dfrac 5 2}$

plug this into (2)

$a^2 + \dfrac{45}{2} = 25$

$a^2 = \dfrac{5}{2}$

$a = \sqrt{\dfrac{5}{2}}$

plug this into (1)

$\dfrac 5 2 + b^2 = 16$

$b^2 = \dfrac{27}{2}$

$b = \sqrt{\dfrac{27}{2}}=3\sqrt{\dfrac 3 2}$

$V = 8abc = 8\left(\sqrt{\dfrac{5}{2}}\right)\left(3\sqrt{ \dfrac 3 2}\right)\left(3\sqrt{ \dfrac 5 2}\right)=90\sqrt{6}$