Series of Squares Under Triangle
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A line crosses the $x$ and $y$ axes at $(a,0)$ and $(0,1)$ respectively, where $a>0$. Square are placed successively inside the right angled triangle thus formed. What is the area enclosed by all squares when their number goes to infinity.
summation geometric-series
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A line crosses the $x$ and $y$ axes at $(a,0)$ and $(0,1)$ respectively, where $a>0$. Square are placed successively inside the right angled triangle thus formed. What is the area enclosed by all squares when their number goes to infinity.
summation geometric-series
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up vote
1
down vote
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up vote
1
down vote
favorite
A line crosses the $x$ and $y$ axes at $(a,0)$ and $(0,1)$ respectively, where $a>0$. Square are placed successively inside the right angled triangle thus formed. What is the area enclosed by all squares when their number goes to infinity.
summation geometric-series
A line crosses the $x$ and $y$ axes at $(a,0)$ and $(0,1)$ respectively, where $a>0$. Square are placed successively inside the right angled triangle thus formed. What is the area enclosed by all squares when their number goes to infinity.
summation geometric-series
summation geometric-series
edited 3 hours ago
mwt
546113
546113
asked 3 hours ago
Inquirer
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372210
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1 Answer
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Each square of side length $s$ has as "hat" a right triangle with horizontal leg $s$ and vertical leg $frac sa$. Within a single instance of "square + hat", the square occupies a share of $fracs^2s^2+frac12cdot scdot frac sa=frac11+frac12a=frac2a2a+1$. This proportion remains constant, no matter how many squares with hat we consider, hence also in the limit.
We conclude that the total area occupied by the squares is
$$ frac2a2a+1cdot frac a2=fraca^22a+1.$$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Each square of side length $s$ has as "hat" a right triangle with horizontal leg $s$ and vertical leg $frac sa$. Within a single instance of "square + hat", the square occupies a share of $fracs^2s^2+frac12cdot scdot frac sa=frac11+frac12a=frac2a2a+1$. This proportion remains constant, no matter how many squares with hat we consider, hence also in the limit.
We conclude that the total area occupied by the squares is
$$ frac2a2a+1cdot frac a2=fraca^22a+1.$$
add a comment |Â
up vote
5
down vote
accepted
Each square of side length $s$ has as "hat" a right triangle with horizontal leg $s$ and vertical leg $frac sa$. Within a single instance of "square + hat", the square occupies a share of $fracs^2s^2+frac12cdot scdot frac sa=frac11+frac12a=frac2a2a+1$. This proportion remains constant, no matter how many squares with hat we consider, hence also in the limit.
We conclude that the total area occupied by the squares is
$$ frac2a2a+1cdot frac a2=fraca^22a+1.$$
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Each square of side length $s$ has as "hat" a right triangle with horizontal leg $s$ and vertical leg $frac sa$. Within a single instance of "square + hat", the square occupies a share of $fracs^2s^2+frac12cdot scdot frac sa=frac11+frac12a=frac2a2a+1$. This proportion remains constant, no matter how many squares with hat we consider, hence also in the limit.
We conclude that the total area occupied by the squares is
$$ frac2a2a+1cdot frac a2=fraca^22a+1.$$
Each square of side length $s$ has as "hat" a right triangle with horizontal leg $s$ and vertical leg $frac sa$. Within a single instance of "square + hat", the square occupies a share of $fracs^2s^2+frac12cdot scdot frac sa=frac11+frac12a=frac2a2a+1$. This proportion remains constant, no matter how many squares with hat we consider, hence also in the limit.
We conclude that the total area occupied by the squares is
$$ frac2a2a+1cdot frac a2=fraca^22a+1.$$
answered 3 hours ago
Hagen von Eitzen
268k21260484
268k21260484
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