A Simpler Way to Find the Shaded Area?

Click For Summary
SUMMARY

The discussion focuses on calculating the percentage of a square's total area that is shaded purple, specifically identifying it as 5%. The user initially struggled with the geometric proof but confirmed that line segment EG is approximately equal to a/3 through visual inspection and measurement. The conversation suggests that a simpler method using geometry or trigonometry could exist, avoiding Cartesian coordinates, while providing specific area calculations for triangles and quadrilaterals involved in the problem.

PREREQUISITES
  • Understanding of basic geometry and trigonometry concepts
  • Familiarity with area calculations for triangles and quadrilaterals
  • Knowledge of midpoints and their properties in geometric figures
  • Ability to visualize geometric shapes and their relationships
NEXT STEPS
  • Research geometric proofs for area calculations in irregular shapes
  • Explore trigonometric methods for determining areas without Cartesian coordinates
  • Study the properties of midpoints in geometric figures
  • Learn about visual estimation techniques in geometry
USEFUL FOR

Mathematicians, geometry enthusiasts, educators, and students seeking efficient methods for area calculation in geometric problems.

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?
 
  • Like
Likes   Reactions: Drakkith
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:

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
24K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
15
Views
41K
  • · Replies 10 ·
Replies
10
Views
5K
Replies
10
Views
9K
  • · Replies 1 ·
Replies
1
Views
2K