Don;

Thanks for making the connection....!

Those professionals who have studied this problem all made a drastic error.
Problem #53 is not an isosceles triangle as they have assumed…and I was also lead to believe.
It required a lot of head scratching before I realized what the author of the Rhind had assembled.

To understand fully, you must read the complete section of the texts relating to this level/segment in geometry.
The section of the papyrus where these problems are located has the width of the scroll divided into six sections...each with a different problem to solve.
The clue is problem #49, the upper most "very simple" problem that does “not” require the illustration of a simple rectangle...but it's there. Yet the next problem down, problem #50, relates to the sophisticated 8/9 ratio of circle/square area; it is void of an illustration...very odd wouldn’t you think?

You must ask yourself why did the author not illustrate this important circle/square relationship?
For those who are reading this post...may I remind you that the measures/rulers used by these people of past were intricate instruments and a scribe would be very familiar with their use.


Knowing the approximate width of the papyrus then the scale of these illustrations can be calculated. I believe it uses a 5:1...fourth digit from the right on the ruler…I could be wrong since there is no data informing us of the two side lengths.

To confirm the intentional division of the papyrus into six segments the author drew the number five within the triangle of problem #53 (see the center of the right hand segment in black). The number is used in the calculations, but it also stipulates this problem set in the “fifth” section from the top.

Problem #51 and #53 both share the same shaped triangle and both have “identical” side lengths, but problem #53 has a shorter base.

Making the base of #51 = "7" then the base of #53 is 6...
A direction to the riddle is the problem in between...#52...it's base measures "4-1/2"...

Return to problem #53…it has two math lines that have stumped the best, one starts with the number "7" and the other with "4-1/2"…coincidence?

Of course...we all know that 7 x 4-1/2 = 31.5...right?
However...do we know...
The area of a circle being 31.5 then its radius is the square root of 10.0...3.16!
“BUT”
A volume of a cube being 31.5 then its side length measures 3.16...and we know from above that 3.16 is the square root of 10...!
In reality…
The square root of 10 is 3.16...and pi is 3.14
The average of these two is 3.15...or (7 x 4-1/2)/10.

Two birds...one stone.

Best.
Clive