A line crosses the $x$ and $y$ axes at $(a,0)$ and $(0,1)$ respectively, where $a>0$. Squares are placed successively inside the right angled triangle thus formed as in the figure below. What is the area enclosed by all squares when their number goes to infinity?
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- Find the points of intersection of the line
$y= x+6$with the parabola$y=x^2.$ - Evaluate the sum
$\sum_{n=0}^{a}{2^n}.$ - What is the equation of the line with gradient
$5$that passes through the point$(2,5)?$
- Find the points of intersection of the line
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A line equation may help.
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What can you say about the
$x$and$y$coordinates of the top right corner of the largest square? -
… does that help to compute its side length?
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Is there a relationship between the large triangle and the one above the largest square?
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What about the successive triangles?
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What is then the sum of such numbers that are in such proportion, i.e. constant ratio?
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The equation of the line is
$y=1-\frac{x}{a}.$At the top right hand corner of the first square$y=x.$If we substitute this into the first equation we get$x = 1 - \frac{x}{a},$which we can solve to get the side length of the first square which is$\frac{a}{a+1}.$We notice that the triangle bounded by the lines
$x=\frac{a}{a+1},$$y = 1 - \frac{x}{a}$and$y=0$is similar to the larger triangle but with a height of$\frac{a}{a+1}.$The heights of successive nested triangles follow a geometric progression with ratio$\frac{a}{a+1}.$The squares contained within the triangles have side lengths equal to the height of the next triangle and the total area is a geometric series.
$$ \begin{aligned} \sum_{n=1}^{\infty}\bigg(\frac{a}{a+1}\bigg)^{2n} &= \frac{\big(\frac{a}{a+1}\big)^2}{1-\big(\frac{a}{a+1}\big)^2} \\[7pt]&= \frac{a^2}{2a+1} \end{aligned}$$