https://brilliant.org/practice/joy-problem-solving-geometric-shortcuts/?p=4

Hi
I suggest a simpler solution for
https://brilliant.org/practice/joy-problem-solving-geometric-shortcuts/?p=4
instead of the two complicated solutions offered.

The area of triangles is base * height /2:
The area of triangle ACF is 1.
The area of triangle ECG is what remains if we subtract the questioned area (x) from the area of triangle ACF (which is 1): 1-x.
Triangles ECG and AEG have the same height, equal bases (1), thus the areas of ECG and AEG are the same (ECG is 1-x, thus AEG is also 1-x). AEG and AFG are symmetrical, thus have equal areas (AEG is 1-x, thus AFG is also 1-x).
The area of triangle ACF is 1, and triangle ACF comprises 3 triangles (AFG, AEG, ECG) with equal areas (1-x each), thus these equal areas (1-x) are ⅓, x=⅔.

#Geometry

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Comments

Thanks, that's a nice way of looking at it. I've added your solution in :)

Calvin Lin Staff - 4 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$