Repost: Perimeter of rectangle from diagonal and midpoint length

[robinc1402]

In rectangle ABCDABCD, MM is the midpoint of CD,AC=√73CD,AC=73 and AM=5AM=5 . What is the perimeter of ABCDABCD?

I'm very interested in this question and would like to know how to solve it in the future

 

[Deleted member 4993]

In rectangle ABCDABCD, MM is the midpoint of CD,AC=√73CD,AC=73 and AM=5AM=5 . What is the perimeter of ABCDABCD?

I'm very interested in this question and would like to know how to solve it in the future

We cannot read your problem as posted properly. Use plain text (e.g. without bold letters, writing out "square-root", etc.) and repost. Include your work.

 

[robinc1402]

Sorry for the double post; the formatting in my last post apparently couldn't be read by some. Anyway, I was wondering about the solution to this problem:

In rectangle ABCD, M is the midpoint of CD, AC = sqrt(73) and AM=5 . What is the perimeter of ABCD?

I had tried utilising the sine and cosine rules, and using 1/2ab sin(c), but I was not able to use them.

Is there any other way to do this kind of problem?

 

[tkhunny]

Why not the Pythagorean Theorem?

AD, DM, DC = 2DM, 5, \(\displaystyle \sqrt{73}\) lead immediately to the length of DM. This leads immediately to the length of AD and you're kind of done.

Let' see your workings.

 

[Dr.Peterson]

You appear to have pasted from some source that didn't copy properly, and failed to check what it looked like. I think you meant this:

In rectangle ABCD, M is the midpoint of CD,AC=√73 and AM=5. What is the perimeter of ABCD?

I'm very interested in this question and would like to know how to solve it in the future

I would look at the right triangles ADC and ADM, using the Pythagorean theorem to write two equations in x = AD and y = CD. Then you can solve the resulting system of equations.