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Need help finding Diagonals of a Rhombus

thatbluegsx90

New member
Dec 23, 2019
1
Not sure where exactly to post this but I think it fits in this catagory...I recently got interested in making a Terrarium and plan to make the glass enclosure myself.. for the life of me without having the physical thing in front of me I can't figure out one of the diagonal measurements of the top cover.. I apologize for the rough drawing but I'm looking for what A and B are(didn't mean to put the X in there)... I'm about 60% confident that B is 8 1/2 but I has been a while since I've done anything related to this
 

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HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
From your middle, side view, the back height of the terrarium is 16", the front is 8", and the base is 6". If you draw a line from the top of the front perpendicular to the back (so parallel to the base) you have a right triangle with height 16- 8= 8" and base 8". By the Pythagorean theorem, your distance "A" satisfies [tex]A^2= 8^2+ 8^2= 64+ 64= 120[/tex]. So [tex]A= \sqrt{120}= 2\sqrt{15}[/tex] or about 11" (10.954...). Your distance "B" is just the distance across which is 6+ 6= 12".
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,682
Not sure where exactly to post this but I think it fits in this catagory...I recently got interested in making a Terrarium and plan to make the glass enclosure myself.. for the life of me without having the physical thing in front of me I can't figure out one of the diagonal measurements of the top cover.. I apologize for the rough drawing but I'm looking for what A and B are(didn't mean to put the X in there)... I'm about 60% confident that B is 8 1/2 but I has been a while since I've done anything related to this
\begin{tikzpicture}
[scale=0.5]
\coordinate (A) at (-5,2) ;
\coordinate (B) at (1,3) ;
\coordinate (C) at (6,1) ;
\coordinate (D) at (0,0) ;
\coordinate [label=left: $B_1$] (E) at (-5,14) ;
\coordinate [label=above right: $A_1$] (F) at (1,19) ;
\coordinate [label=right: $B_2$] (G) at (6,13) ;
\coordinate [label=left: $A_2$] (H) at (0,8) ;
\draw [very thick] (G) --node
{$12$} (C) --node[below] {$6$} (D) --node[below] {$6$} (A) --node
{$12$} (E) -- (H) ;
\draw [very thick] (E) -- (F) -- (G) -- (H) --node
{$8$} (D) ;
\draw [dashed] (A) -- (B) -- (C) ;
\draw [dashed] (B) --node
{$16$} (F) ;

\end{tikzpicture}

In my picture, the distance between $A_1$ and $A_2$ is your distance $A$, and the distance between $B_1$ and $B_2$ is your distance $B$.

Both of the points $B_1$, $B_2$ are $12$ inches above diagonally opposite points of the base of the terrarium. So the distance between them is $6\sqrt2$, which is approximately $8.485$ (close to your estimate of 8 1/2, but a bit less).

The horizontal distance between $A_1$ and $A_2$ is again $6\sqrt2$, but there is also a vertical separation of $16 - 8 = 8$ between them. It follows from Pythagoras's theorem that $A = \sqrt{72 + 64} = \sqrt{136}$, which is approximately $11.66$.

If it helps you in cutting the glass, the sides of the rhombus are approximately $7.2$ inches. The angles at $A_1$ and $A_2$ are almost exactly $72^\circ$ (and the angles at $B_1$ and $B_2$ are almost exactly $108^\circ$).​