Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a 4-sided plane in a perspective view. Each side is equal in length to the side across from it. Given the length of two sides and the fore-shortened length of one side, how can we solve for the other two sides?

Assume the bottom side is not fore-shortened in this example.

Additional information:


Length: 2CM, Fore-shortened: 1 CM


Length: ?


Length: ?


Length: 2CM

share|cite|improve this question
No, you cannot solve for the lengths. There are some similar triangles in your second diagram, but you do not have not sufficient information overall. – Henry Apr 23 '12 at 7:36
I'm going to solve this when I get home. I'm at the school right now. – Garmen1778 Apr 23 '12 at 7:40
There are 2 variables here which determine the length of the sides, first one is the fore shortened length, and the other one is the angle of view, which apparently is missing from the data provided in the question. – Tomarinator Apr 23 '12 at 7:58
Very interesting problem. I am not sure if it is lack of information. But the answer for the left and right length must be way much longer than the top and bottom according to the fore-shortened length of the top. – Megan Apr 23 '12 at 9:26
I know for a fact that Photoshop Extended can solve this problem using its vanishing point tool. It will produce a 3d plane with the correct lengths with the only given being what you see above. However, I want to incorporate this feature into my own software and am wondering how this can be solved. – Abdulla Apr 23 '12 at 16:21

The far side must be twice as far from you as the front side. If it is not far away, you can make right triangles at your eye to the centers of the sides and the corners of the rectangle. If the half length of the side is $L$the near side is distance $d_1$ from your eye, you have $\frac L{d_1}=\tan \theta_1$ and the far side is $d_2$ from your eye then $\frac {L}{d_2}=\tan \theta_2$ and $\tan \theta_1=2\tan \theta_2$. Therefore $\frac {d_2}{d_1}=\frac{\tan \theta_1}{\tan \theta_2}=2$

share|cite|improve this answer

Triangle and Trigonometry. The perspective is a projection of the real 3D over a glass in front of your eye. What you see on the 2D glass is exactly what you see on your drawing above in 2D. The actual calculation uses angular size of the closer horizontal line, but for that, you need to know the distance between your eye to this line. Lets assume this Distance (D) is 5cm. A = CloserHorizontalLine. B = FarHorizontalLine. A Angular size = AAS = 2*arctan(size/2*distance). AAS on glass =2*arctan(2/10)=22.6°. BAS (on glass) = 2*arctan(1/10)=11.4°. Now, from the left side of the A to the left side of the B on glass, is 0.5cm, Half of difference A to B (HDAB). Now we have a rectangle triangle made by this HDAB a 90° line to the back of real rectangle (C) and the hypotenuse we don't care about, but it makes a 11.4°/2 with the C. Now it is easy to calculate C = HDAB/tan(BAS/2) = 0.5/tan(11.4/2) = 5. To make a formula of it, it will be C = ((A-B)/2)/arctan(tan(B/2D)), eliminating arctan and tan, it results in C=(A-B)*2 / B/2D = C=(A-B)*2D/2B = C=(A-B)*D/B. If A=2 B=1 D=5, C=5, If A=3 B= D=5, C=10. I may be wrong.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.