Issues with a 3D Trigonometry Problem

Question

I have the following shape:

Where the midpoint of $EF$ lies vertically above the intersection of the diagonals $AC$ and $BD$.

Now, I need to find the heights of the trapezium and triangles in order to calculate the total surface area of the sloping faces.

This was my working, but I keep getting an incorrect answer:

1. For the height of the triangle:

I did $tan(50)$ = $\frac{Height_{TRIANGLE}}{3.5}$

$Height_{TRIANGLE}$ = $3.5 * tan(50)$

2. For the height of the trapezium:

The hypotenuse of the triangle = the length of the slanted sides of the trapezium

$cos(50)$ = $\frac{3.5}{Hypotenuse_{TRIANGLE}}$

$Hypotenuse_{TRIANGLE}$ = $\frac{3.5}{cos(50)}$

And now to find the height of the trapezium.

If I made a triangle with the height of the trapezium as the opposite side.

$sin(50)$ = $\frac{Height_{TRAPEZIUM}}{Hypotenuse_{TRIANGLE}}$

$Height_{TRAPEZIUM}$ = $sin(50)$ * $Hypotenuse_{TRIANGLE}$

$Height_{TRAPEZIUM}$ = $sin(50)$ * $\frac{3.5}{cos(50)}$ = 3.5 * $\frac{sin(50)}{cos(50)}$ = $3.5 * tan(50)$

What am I missing here?

Answer 1

If you look at the figure horiztonally from the right side, then you will see a isosceles triangle with base $7$ and base angles $50^\circ$, so it height is

$ h = 3.5 \tan(50^\circ) $

And this is the elevation of $EF$ above $ABCD$.

From the above calculation, the height $h_1$ of $\triangle FDC $ is

$ h_1 = \dfrac{h}{\sin(50^\circ)} $

And similarly, the height of trapezium is

$h_2 = \dfrac{h}{\sin(50^\circ)} $

The segment $EF$ has a length given by

$ \ell_{EF} = 10 - 2 h_1 \cos(50^\circ) = 10 - 2 h \cot(50^\circ) = 10 - 7 \tan(50^\circ) \cot(50^\circ) = 3 $

Answer 2

Here is your diagram with a few extra lines added in blue and a few extra points labeled. Now there are some right triangles in the figure and you can start applying trigonometric functions to them. For example, $$ \tan(50^\circ) = \frac{FH}{HK}. $$

Accurate, precise labeling helps avoid errors. For example, what is "$Height_{TRIANGLE}$"? Some candidates are the length of segment $FG$ (the most likely candidate, as $\triangle CDF$ is one of the few triangles in the original figure and $FG$ is the altitude from the base $CD$) and the length of segment $FH$ (which is literally the "height" of the farthest point of $\triangle CDF$ above the plane $ABCD$, and which your formula implies is the intended length). You can avoid this confusion by simply writing $FG$ when you mean the length of the segment $FG$ and $FH$ when you mean the length of the segment $FH.$

The other advantage to precise, accurate labeling is that other people have a chance to actually understand what you're trying to say.

Answer 3

The height of both trapezium and triangle is the same and you only need to calculate the hypotenuse of triangle $PQR$.

You have the base and the angle.

$h = \dfrac{3.5}{\cos(50^\circ)}≈5.445m$

Answer 4

Assuming $ABCD$ is a rectangle whose center $P$ is right below midpoint $O$ of $EF$ which edge is parallel to base plane $ABCD.$

I suppose you started with a square pyramid 7 m base $ abcdO $ and extended either side faces keeping symmetry with 1.5 meters parallel translated/shifted colored planes shown. The red edges are parallel in this translation/shift.

It makes no difference to the central right triangle $OPQ$. Height $h$ is retained the same.

Accordingly from this right triangle $OPQ$ the height of the trapezium and two triangles is

$$ h=3.5 \tan 50^{\circ};$$