B A Simpler Way to Find the Shaded Area?

Click For Summary
To determine the percentage of the shaded area in a square, the discussion revolves around finding a simpler geometric or trigonometric method, avoiding Cartesian coordinates. The calculated shaded area is confirmed to be 5%. The user faced challenges in proving the length of line segment EG, initially estimating it visually and later confirming it with a ruler. There is a desire for a more straightforward approach to solving the problem without complex calculations. The conversation highlights the need for efficient methods in geometric area calculations.
Saracen Rue
Messages
150
Reaction score
10
TL;DR
Is there an easier way to determine what percentage of this square is bound by the shaded area of 4 line segments
Consider the following scenario:
20230127_222828.png


Given that points ##M## and ##N## are the midpoints of their respective line segments, what would be the fastest way to determine what percentage of the squares total area is shaded purple?

I managed to determine that the purple shaded area is ##5\text{%}## as per my working below:
20230127_222843.png


The only real problem I had with this is that it took me a genuinely long time to figure out. After I drew it up by hand, I suspected that line segment ##EG## was roughly equal to ##\frac{a}{3}## from visual inspection alone, and I did use a ruler to confirm this. However, it took me quite a while to prove it was mathematically.

I feel as though there must be a simpler way to go about solving the question using just geometry/trigonometry (I do realise that it'd probably be easy enough to solve if you put this on a Cartesian Plane with ##A## at the origin, but I wanted to try avoiding that method if possible), and if so, can anyone point me in the right direction?
 
Mathematics news on Phys.org
Say square ABCD=1
quadrilateral EFGH = ##\triangle## ACM - ##\triangle## AEF - quadrilateral HCMG
where
##\triangle##ACM=1/4,
##\triangle## AEF = ##\triangle## AEN - ##\triangle## AFN = 1/12 - 1/20,
quadrilateral HCMG = ##\triangle##CHD-##\triangle## GDM = 1/4 - 1/12.
 
Last edited:
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

Replies
1
Views
3K
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
12
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
24K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
21
Views
5K
Replies
5
Views
5K