Question

Please try this, I forget absolutely everything about rhombuses. Thanks for all the help, shouldn't be too hard

Answer 1

So, check the picture below.  It has four sections.

the volume of a pyramid is, 1/3 ( base area)(height).

well, we know the base is a rhombus, and rhombuses diagonals meet at right-angles, therefore the center O has 4 right-angles where they meet.

since we know the angle at A, 50°, and since rhombuses are quadrilaterals with all equal sides, we know all the sides are 6 long, and the angle at C is 50° as well, and the remaining angles are 130°, as shown there.

now, if we tease out one of those triangles from the rhombus, as shown on the 1st section in the picture, we can say use the angle of 65° to get "x" and "y" lengths

$\bf cos(65^o)=\cfrac{x}{6}\implies 6cos(65^o)=x \\\\\\ sin(65^o)=\cfrac{y}{6}\implies 6sin(65^o)=y$

now, we'll use those to get the area of the rhombic base.

and then onto the second section

is a right-triangle with a perpendicular line through it, which creates 3 similar triangles, by using the "geometric mean" of "w" there.

anyhow, we used the medium-sized and large-sized triangles there, to draw proportions and get what the "w" distance is, and you can see there what "w" is.

and onto the 3rd section

now that we know how long "w" is, we can use the triangle containing the 40° angle, in order to get the pyramid's height of "h", by using the tangent function, as you see there.

and we also use the same triangle, to get the "z" distance, which is namely the slant-height of the pyramid, however, is also the "altitude" of the triangular faces.

and onto the 4 and last section in the picture

that's a triangular face by herself, it has a base of 6, and an altitude of "z".



now, all that said, let's plug those values to get the volume, and then the surface area, keeping in mind the surface area is just the area of the rhombus plus the area of all four triangles.


$\bf V=\cfrac{1}{3}\left[ \stackrel{\textit{area of the rhombic base}}{4\left( \frac{1}{2}xy \right)} \right](h) \\\\\\ SA=\left[ \stackrel{\textit{area of the rhombic base}}{4\left( \frac{1}{2}xy \right)} \right]~~+~~\left[ \stackrel{\textit{area of the four triangles}}{4\left[ \frac{1}{2}(6)(z) \right]} \right]$

x ≈ 2.54     y ≈ 5.44    w ≈ 2.298    h ≈ 1.93    z = 3

V ≈ 17.7265                         SA ≈ 63.58