volume of a trapezoid

**mat** · October 17th 2008, 08:43 PM

hello how to figure out the volume of a traezoid or trapezium?

if the height is 5.6 metres another height is 4 metres
the side is 4 metres and the other side is 3.8 metres thanks.

**Jhevon** · October 17th 2008, 08:50 PM

Originally Posted by mat

hello how to figure out the volume of a traezoid or trapezium?

if the height is 5.6 metres another height is 4 metres
the side is 4 metres and the other side is 3.8 metres thanks.

google is your friend. see here. that should help

**mat** · October 17th 2008, 09:35 PM

i dont know where to put the numbers though

**earboth** · October 18th 2008, 12:10 AM

Originally Posted by mat

hello how to figure out the volume of a traezoid or trapezium?

if the height is 5.6 metres another height is 4 metres
the side is 4 metres and the other side is 3.8 metres thanks.

I am guessing only, because the text of your question isn't clear.

Let H denote the height of the solid;
let h denote the height of the trapezium;
let a, b denote the parallel sides of the trapezium

then the volume is calculated by:

$V_T = \underbrace{\underbrace{\underbrace{\dfrac{a+b}2}_ {mid-parallel} \cdot h}_{area\ A} \cdot H}_{volume\ of\ a\ prism}$

With your values I've got $V_T = 87.36\ m^3$

**mat** · October 27th 2008, 06:38 PM

thanks ear but i have to build a structure
and the slant is on the roof from right to left is going downhill
so the 2 sides of the structure will be equal and 5.6 metres height with the slant tallest part and 4 metres in length and 4 metres in height.
the front is 3.8 metres length by 4 metres height.
and the back of the structure the tallest part is 3.8 metres length and 5.6 metres height.
the volume i got was 72.96 metres cubed.
how to find the surface area of this structure thankyou.

**earboth** · October 28th 2008, 01:48 AM

Originally Posted by mat

thanks ear but i have to build a structure
and the slant is on the roof from right to left is going downhill
so the 2 sides of the structure will be equal and 5.6 metres height with the slant tallest part and 4 metres in length and 4 metres in height.
the front is 3.8 metres length by 4 metres height.
and the back of the structure the tallest part is 3.8 metres length and 5.6 metres height.
the volume i got was 72.96 metres cubed.
how to find the surface area of this structure thankyou.

Honestly I find it very difficult for me to imagine how this structure looks like.
Could you provide us with a rough sketch? That would be very helpful (for me at least)

**earboth** · October 29th 2008, 11:35 AM

Originally Posted by mat

Attachment 8451
hope you can see it
the roof is at the top and
is 22 degrees downwards from right to left.
what is the volume and surface area of this trapzium.
Would you be able to solve this 12 hours from now thanks.

If I understand your sketch correctly your structure is a prism. The base area seems to be a rectangle and it looks like as if the "roof" of the prism has a sharp bend.

Divide this prism into 3 different solids:
1. A prism with a rectangle as the base area and the height of 4 m
2. A prism with a right triangle as the base area and the height of 1.6 m
3. A pyramid with the top T and the vertical rectangle as base area.

According to these settings (merely guessing!) the volume is:

$V_{baseprism} = 4\cdot 3.8 \cdot 4 = 60.8\ m^3$

$V_{topprism} = \dfrac12\cdot 4\cdot 3.8 \cdot 1.6 = 12.16\ m^3$

$V_{pyramid} =\dfrac13\cdot \underbrace{\sqrt{4^2+3.8^2}\cdot 1.6}_{base\ area}\cdot \underbrace{\dfrac12\cdot \sqrt{4^2+3.8^2} }_{height}\approx 8.117\ m^3$

The surface of the complete structure is:
1. base area
2. two rectangles
3. two trapezii
4. one right triangle, one acute triangle

$a_{base}=4\cdot 3.8=15.2\ m^2$

$a_{rectangles}= 4\cdot 5.6+3.8\cdot 5.6=43.68\ m^2$

$a_{trapezii}=\dfrac{4+5.6}2\cdot 4+\dfrac{4+5.6}2\cdot 3.8=37.44\ m^2$

$a_{right\ triangle}=\dfrac12 a_{base} = 7.6\ m^2$

To calculate the area of the last triangle I calculated the lengths of the sides first (use Pythagorean theorem)
$\sqrt{18.56}, \sqrt{17}, \sqrt{30.44}$
Calculate the angle at T:

$\cos(T)=\dfrac{30.44-17-18.56}{-2\cdot \sqrt{18.56}\cdot \sqrt{17}}\approx 0.14412...~\implies~\angle(T)\approx 81.71^\circ$

$a_{acute\ triangle}=\dfrac12\cdot \sqrt{18.56}\cdot\sqrt{17}\cdot\sin(81.71^\circ)\a pprox 8.7887\ m^2$

And now I hope that this reply doesn't come too late...

**mat** · October 29th 2008, 06:13 PM

thanks for replying but it was too late. The structure dimensions you drew were correct however the roof is slanting 22 degrees and the roof is a parelogram.he structure is a trapoiz prism like the first one you drew but tipped over.Thanks anyway.

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