B A Simpler Way to Find the Shaded Area?

AI Thread 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 Summary
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:
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top