diagram of rectangle with sheared image

My calculation:

$$\tan(30^\circ)=\frac{2}{x_A'}$$ $$x_A'=\frac{2}{\tan(30^\circ)}$$ $$x_A'=\sqrt{3}$$

But the answer is:


I'm wondering what step I'm missing in regards to the $-1$.

  • $\begingroup$ If $x_{A}'$ is meant to be the $x$ co-ordinate of $C'$, then would you not need to subtract from the horizontal length (which is what your calculated $x_{A}'$ actually is) the distance that $D'$ is from the origin, which is $1$? $\endgroup$ Commented Mar 4, 2023 at 19:08
  • $\begingroup$ Note the $C'$ in my comment above should have been $A'$ instead. $\endgroup$ Commented Mar 4, 2023 at 19:18
  • $\begingroup$ Why was my question downvoted? If I made any mistakes in presenting my question let me know so I can avoid doing this again in the future. @JohnOmielan (and everyone who answered on this thread) -- thanks for the insights, it's been really helpful. $\endgroup$ Commented Mar 5, 2023 at 0:58
  • $\begingroup$ You're welcome for any insight my comment provided to you. Regarding the downvote of your question, since I didn't do that, I can't give any reason(s) why it it happened. $\endgroup$ Commented Mar 5, 2023 at 1:03

3 Answers 3


$$\tan(30^\circ)=\frac{2}{x_A'}$$ $$x_A'=\frac{2}{\tan(30^\circ)}=2 \sqrt{3}$$ Look at x- projection of $O$ and $A'$ $$ x _{O-A'}= x'A-1= 2 \sqrt{3}-1$$


The length of the top edge of the triangle is $x_A'+1$. $$\sqrt{3} = \tan(60^\circ) = \frac{x_A'+1}{2}$$ $$x_A'=2\sqrt{3}-1$$


The length of the hypotenuse $D'A'$ of right triangle $\triangle D'AA'$ can be found using the slope formula.

Given coordinates $(x_{D'},y_{D'})=(-1,0)$ and $(x_{A'},y_{A'})=(x_{A'},2)$ the slope of the hypotenuse $D'A'$ is given by

\begin{eqnarray} m&=&\frac{y_{A'}-y_{D'}}{x_{A'}-x_{D'}}\\ &=&\frac{2}{x_{A'}+1} \end{eqnarray}

But we are know also that $m=\tan{30^\circ}=\frac{\sqrt{3}}{3}$



Solving gives $x_{A'}=2\sqrt{3}-1$.


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