Respuesta :
The given statement is not true.
Step-by-step explanation:
Here, the initial area of the sheet = 450 sq inches.
After folding the sheet in to thirds ONCE :
The area of the each rectangle so formed = [tex]\frac{1}{3} \times 450[/tex] = 150 sq. inches
So, the area of each rectangle = 150 sq inches
After folding the sheet in to thirds TWICE :
The area of the each rectangle so formed = [tex]\frac{1}{3} \times 150[/tex] = 50 sq. inches
So, the area of each rectangle = 50 sq inches
So, as we can see after two rounds of folding, we get in total 9 rectangles.
The area of each rectangle so formed = 50 sq inches.
50 sq inches = [tex](\frac{1}{9} )^{th}[/tex] of the total area.
Hence, the given statement is not true.
Since Han folds the paper into thirds, and again in thirds, then the statement believed by Han is wrong since his statement says that resultant folding's area is one third which means only one round of folding happened which isn't true.
Thus, No, it is not true that visible area of folded paper is 150 sq. inches(or one third of original area).
Given that:
- Original area of thin paper = 450 sq. inches.
- Han folds it into thirds two times.
First time folding the paper into thirds:
The visible area would be divided into three parts and only one part would be visible. Thus, one of third of the paper is visible only, or we can say that:
[tex]\text{Visible area} = \dfrac{450}{3} = 150 \: \rm inch^2[/tex]
Second time folding the folded paper into thirds:
By same above logic, we will have:
[tex]\text{Visible area} = \dfrac{150}{3} = 50 \: \rm inch^2[/tex]
Thus, the resultant folded paper's visible area would be 50 sq inch which is one-ninth of the original area.
Thus, No, it is not true that visible area of folded paper is 150 sq. inches(or one third of original area).
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