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 $\displaystyle V_T = 87.36\ m^3$
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.
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...