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a building contractor needs to compute the volume.

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Tags: Mathematik

RASTA aktiv_icon

00:47 Uhr, 14.02.2022

Tag!
a building contractor needs to compute the volume of air contained in a building with a rectangular floor, as shown in the figure below. the building is

20

feet long, 4 yards wide,

912

feet high on two sides 9 and a half feet high on two sides and

15

feet high from the floor to the peak. what is the volume of air in the building in cubic feet?

the great FIND = Volume of air in the building in cubic ft.
measurements of the building that have been given

l = 20

w = 4

yards (

I

am converting to a single unit so 1f=3yds

= 4 \cdot 3 = 12

ft)
height on the sides=9

\frac{1}{2}

ft
height from the floor to the top of the ceiling

= 15

ft

what to do when you have to find the volume (of a structure) and you are giving two heights.
My guess: find two volumes: the volume for the floor to the top of the sides of the structure
then, the volume from the floor to the top of the ceiling then add them together.
Is this assumption correct?

i

found the area of the rectangular prism
V(rectangular prism)=l ×

w

h

= 20

12

9.5

=2280ft^3
how can I find the aea of the trianglular prism a triangular right prism (TP) with a triangular base?
thanks

supporter aktiv_icon

08:08 Uhr, 14.02.2022

The figure is missing!

RASTA aktiv_icon

13:00 Uhr, 14.02.2022

I'm sorry about that. Here's the pic.

supporter aktiv_icon

13:52 Uhr, 14.02.2022

You just have to calculate the volume of the cuboid and of the pyramid, whose basic area
a an isosceles triangle. You know its basis a and height

h

.
The area A equals

\frac{a \cdot h}{2}

The volume of the pyramid is

A \cdot 20

.

Turn the unit yard into the unit feet while calculating!

RASTA aktiv_icon

21:41 Uhr, 14.02.2022

okay, thanks for the tip,

i

am posting my entire work. I found the area first of the rectangular prism
and then I found the area of the tirangular prism, both solids that make up the whole bulding, top and bottom,
add them together and I found how much air in cubic feet the building may have

V(rectangular prism)=l ×

w

h

= 20

12

9.5

=2280ft^3

Vtriangular prism= Bh ,
where

B

represents the area of the base.
The base of this triangular prism is made up of two triangles.
The area of a triangle is equal to

\frac{1}{2}

the base times the height.
B(triangle)=1/2bh
So, I can substitute

B

on the formula for a triangular prism for

\frac{1}{2} \cdot b \cdot h

So

V(tp)=(1/2bh)h

But now I see that I have the same variable for two different things because I have two different heights..
I will distinguish them by naming them

h 1

and

h 2

.
V(tp)=(1/2bh1)h2

h 1 =

the height of our triangle, the height of our base.
h2=is the height of our original formula. This

h

represents the height of our prism.

Let's go thru our triangular prism and see if I can identify these variables.
V(tp)=(1/2bh)h
Let's find the base of our triangle
b=12ft

h 1 =

this is the height of our triangle.
To find it, I am subtracting the height from floor to peak from the height from both sides of our rectangle prism.
So,15ft - 9.5ft=5.5 ft
So

h 1 = 5.5

ft

Now to

h 2

which is the height of our prism.

To find the height of our prism I want to find the distance between our two bases.
The distance between the two bases of our triangular prism is

20

ft.
So,h2= 20ft,
Now I have values for each one of the variables in my formula.

Let's plug'em in.
V(tp)=(1/2bh1)h2
Vtp=(1/2(12ft*5.5ft))20ft
Simplifying right hand side,
Vtp=(1/2(66ft))20ft
=33ft*20ft

= 660 f^{3}

Now adding the two volumes together,
Vrp

+

Vtp
2280ft^3 +660ft^3

= 2940

ft^3$$
The volume of air in cubic feet in the building is 2940ft^3.

I be looking forward for your confirmation.
thanks for helping

Roman-22

01:20 Uhr, 15.02.2022

The result

2940 {f t}^{3}

you arrived at for the volume is correct.
But I guess it could be done a bit easier.
The volume of any prism is the area of the "base" times its height (or length

20 f t

as in your case.

So you need to find the area of the figure on the left in my drawing (yellow plus red) and multiply it by

20 f t

to get the volume.

If you now notice that this figure on the left and the rectangle on the right have the very same area, then it should be easy for you to specify this area, since the dimensions of the rectangle should be clear.

But you got the correct result by dividing the object into two different prisms with just a little more effort anyway.

RASTA aktiv_icon

01:24 Uhr, 15.02.2022

Thank you. I like your approach because 'easy'is a word in Math. So I will do it again to practice your approach.

RASTA aktiv_icon

01:44 Uhr, 15.02.2022

What do you mean by its base a?. What's that?

RASTA aktiv_icon

01:48 Uhr, 15.02.2022

You just have to calculate the volume of the cuboid and of the pyramid, whose basic area
a an isosceles triangle

You said that: whose basic area a ????. What do you mean by that?.

Roman-22

03:29 Uhr, 15.02.2022

A prism is

a 3 D

geometrical object where two identical polygons in parallel planes are connected with straight lines of equal length.

Or as its often found on the net

(e . g

. www.splashlearn.com/math-vocabulary/geometry/prism

) :

"A prism is a 3-dimensional shape with two identical shapes facing each other. These identical shapes are called “bases”."

So these shapes are called "bases" no matter which way the prism is positioned - its not necessarily standing on its base!
So if you look at the first figure here www.mathsisfun.com/geometry/prisms.html
you see a prism with a pentagon as its base.

In your case the base of the prism is the "house shaped" pentagon which can be seen in my drawing on the left side
You may see it as a combination of two trapezoids. Whatever way you look at it, its should be not much effort to calculate its area. My thought was that the easiest way would be to combine the two trapezoids in your mind to form a rectangle whose area obviously is

\frac{12}{2} f t \cdot (15 f t + 9.5 f t)

. And so the volume of the whole object is

V = \frac{12}{2} f t \cdot (15 f t + 9.5 f t) \cdot 20 f t

RASTA aktiv_icon

03:49 Uhr, 15.02.2022

Ach du meine Güte!. You guys are the best!.

RASTA aktiv_icon

10:32 Uhr, 15.02.2022

To supporter :
I have been studying what you have written but I am not totally getting what you mean here:

Supporter:
You just have to calculate the volume of the cuboid and of the pyramid, whose basic area
a an isosceles triangle. You know its basis a and height

What is that a?. What does that a you keep mentioning mean

RASTA aktiv_icon

10:40 Uhr, 15.02.2022

So, Roman; one last questions about your formula.
So, I can say that, to calculate the volume of a shape like the one given,
I can use this formula, according to your analysis of it.
Volume(total)=

\frac{w}{2} \cdot

(height1

+

height2)* width.

Is this correct?

Roman-22

16:21 Uhr, 15.02.2022

Looks OK, given that you mean the correct lengths with "W" and "width".
But it does not make much sense in my opinion to learn and remember a formula for any shape possible (which actually is impossible anyway).
Its better to know that the volume of a prism is calculated by multiplying its base area by its height (or length or width, whatever you may call it).
And to calculate the area of a polygon you should resort to disassemble it in "known" shapes like the two congruent trapezoids in case of your "house shaped" polygon.
Of course its perfectly OK, too, to dissect the shape in a rectangle and a triangle which is basically what supporter suggested and what you used in your first calculation.
There are many ways to skin a cat and you should chose he method you feel most comfortable with.

RASTA aktiv_icon

18:49 Uhr, 15.02.2022

Yes, you're right about that. There are many ways to skin a cat. Your English us very good.

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