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# Practice Paper 3 Question 9

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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## Warm-up Questions

1. Find the points of intersection of the line $$y= x+6$$ with the parabola $$y=x^2.$$
2. Evaluate the sum $$\sum_{n=0}^{a}{2^n}.$$
3. What is the equation of the line with gradient $$5$$ that passes through the point $$(2,5)?$$

## Hints

• Hint 1
A line equation may help.\$
• Hint 2
What can you say about the $$x$$ and $$y$$ coordinates of the top right corner of the largest square?
• Hint 3
... does that help to compute its side length?
• Hint 4
Is there a relationship between the large triangle and the one above the largest square?
• Hint 5
What about the successive triangles?
• Hint 6
What is then the sum of such numbers that are in such proportion, i.e. constant ratio?

## Solution

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{align} \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{align}

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