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Sum of shaded areas: The area of a rectangle ABCD is 1. Assume E is on the diagonal AC...
- Thread starter nanase
- Start date
If the ratio of AE:EC is x:1, then similar triangles tell us that the ratio of FE:EG is also x:1. Likewise, the ratio of AF:GC is also x:1.
The diagonal AC, the vertical side BC=FG, and the horizontal side DC=AB can be divided into [imath]\displaystyle\frac{x}{x+1}[/imath] and [imath]\displaystyle\frac{1}{x+1}[/imath] proportional parts.
Let the rectangle's vertical side BC=FG be [imath]l[/imath]. Since the rectangle's area is 1, then the horizontal side DC=AB must be [imath]\displaystyle\frac{1}{l}[/imath]
Break the vertical and horizontal sides down into their proportional parts:
AF must be [imath]\displaystyle\frac{x}{x+1}\cdot\frac{1}{l}[/imath]
GC must be [imath]\displaystyle\frac{1}{x+1}\cdot\frac{1}{l}[/imath]
FE must be [imath]\displaystyle\frac{x}{x+1}\cdot l[/imath]
EG must be [imath]\displaystyle\frac{1}{x+1}\cdot l[/imath]
Now you have the perpendicular legs of both shaded triangles. Add up their areas. The [imath]l[/imath]'s cancel, and you're left with an expression in [imath]x[/imath]
The diagonal AC, the vertical side BC=FG, and the horizontal side DC=AB can be divided into [imath]\displaystyle\frac{x}{x+1}[/imath] and [imath]\displaystyle\frac{1}{x+1}[/imath] proportional parts.
Let the rectangle's vertical side BC=FG be [imath]l[/imath]. Since the rectangle's area is 1, then the horizontal side DC=AB must be [imath]\displaystyle\frac{1}{l}[/imath]
Break the vertical and horizontal sides down into their proportional parts:
AF must be [imath]\displaystyle\frac{x}{x+1}\cdot\frac{1}{l}[/imath]
GC must be [imath]\displaystyle\frac{1}{x+1}\cdot\frac{1}{l}[/imath]
FE must be [imath]\displaystyle\frac{x}{x+1}\cdot l[/imath]
EG must be [imath]\displaystyle\frac{1}{x+1}\cdot l[/imath]
Now you have the perpendicular legs of both shaded triangles. Add up their areas. The [imath]l[/imath]'s cancel, and you're left with an expression in [imath]x[/imath]
Ah yes I missed on using similarity concept, Thank you for giving me the sides, now I can find the areas of each triangle.If the ratio of AE:EC is x:1, then similar triangles tell us that the ratio of FE:EG is also x:1. Likewise, the ratio of AF:GC is also x:1.
The diagonal AC, the vertical side BC=FG, and the horizontal side DC=AB can be divided into [imath]\displaystyle\frac{x}{x+1}[/imath] and [imath]\displaystyle\frac{1}{x+1}[/imath] proportional parts.
Let the rectangle's vertical side BC=FG be [imath]l[/imath]. Since the rectangle's area is 1, then the horizontal side DC=AB must be [imath]\displaystyle\frac{1}{l}[/imath]
Break the vertical and horizontal sides down into their proportional parts:
AF must be [imath]\displaystyle\frac{x}{x+1}\cdot\frac{1}{l}[/imath]
GC must be [imath]\displaystyle\frac{1}{x+1}\cdot\frac{1}{l}[/imath]
FE must be [imath]\displaystyle\frac{x}{x+1}\cdot l[/imath]
EG must be [imath]\displaystyle\frac{1}{x+1}\cdot l[/imath]
Now you have the perpendicular legs of both shaded triangles. Add up their areas. The [imath]l[/imath]'s cancel, and you're left with an expression in [imath]x[/imath]
But I am curious in how you obtain those sides so I dig on the similarity ratios and got this [imath]\frac{AE}{EC}=\frac{FE}{EG}=\frac{AF}{GC} [/imath] . It seems you use different ratios to obtain the sides.
Thank you so much, after much scribbling and working backward from your expression I got the idea how the comparison of sides is made to obtain the expression for the base and height of each triangle.
It seems you are using
[imath]\frac{AF}{AE}=\frac{DC}{AC} [/imath]
[imath]\frac{GC}{EC}=\frac{DC}{AC} [/imath]
[imath]\frac{FE}{AE}=\frac{CB}{AC} [/imath]
[imath]\frac{EG}{AC}=\frac{DA}{AC} [/imath]
I fail to see and make ratios from those, but now I see it can be done. Thank you!