Let the width of the barn be = x feet
So length of the barn = 2x
Height of the barn = x-8
As the stalls are 6 feet longer from both ends hence, we have to find the area with width as x-12 feet
Volume of the space is = 3840 cubic feet
Hence, equation is :
[tex](x-12)(2x)(x-8)=3840[/tex]
Solving this we get
[tex]2x^{3}-40x^{2}+192x=3840[/tex]
[tex]2x^{3}-40x^{2}+192x-3840=0[/tex]
[tex]2(x-20)(x^{2}+96)=0[/tex]
This gives x=20 and x= ±[tex]4\sqrt{6}i[/tex]
Hence, neglecting the square root value we get x = 20 feet
Hence, the width is = x-12 = 20 - 12 = 8 feet
Length is = 2x = 2*20 = 40 feet
Height is = x-8 = 20 - 8 =12 feet
And we can cross check this by multiplying all the three dimensions to get 3840 cubic feet
[tex]8\times40\times12=3840[/tex] cubic feet.
The dimension of the cube is 8, 40, and 12.
It deals with the size of geometry, region, and density of the different forms both 2D and 3D.
Given
The lower floor will have horse stalls that extend 6 feet from both of the longer walls.
The barn's length is twice the barn's width,
And the lower floor's ceiling height is 8 feet less than the barn's width.
The dimensions of the lower floor are.
Let x be the width,
The lower floor will have horse stalls that extend 6 feet from both of the longer walls is (x - 12)
The barn's length is twice the barn's width is ( 2x )
And the lower floor's ceiling height is 8 feet less than the barn's width is ( x - 8 )
The volume of 3,840 ft³
We know the formula for the volume.
Volume = langth x width x height
3800 = ( x - 12) ( 2x ) ( x - 8 )
3800 = 2x³ - 40x² + 192x
0 = 2x³ - 40x² + 192x - 3800
It is a cubic equation that can be solved by the hit and trial method.
x = 20, ±4√6 i
The value of x is 20.
Then
length = x - 12 = 20 - 12 = 8
Width = 2x = 2 (20) = 40
Height = x - 8 = 20 - 8 = 12
So, the dimension of the cube are 8, 40, and 12.
More about the geometry link is given below.
https://brainly.com/question/7558603
