# 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?

## Related topics

## Warm-up Questions

- 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 \((2,5)?\)

## Hints

- Hint 1A line equation may help.$
- Hint 2What 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 4Is there a relationship between the large triangle and the one above the largest square?
- Hint 5What about the successive triangles?
- Hint 6What 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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