# Thread: volume of a trapezoid

1. ## volume of a trapezoid

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.

2. 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.

3. i dont know where to put the numbers though

4. 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:

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

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

5. 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.

6. 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)

7. 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:

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

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

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

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

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

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

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

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

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

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

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

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

8. 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.