This question is from the third sample paper for the CSAT (Computer Science Admissions Test) at Cambridge. My working is as follows:
First, I worked out the equation of the diagonal line touching the top right corner of each square as being: $y+\frac{1}{a}x=1 \to ay+x=a$
I then worked out the x-coordinate of the intersection point between the diagonal and the top right corner of the first square. I did this using simultaneous equations with the square's diagonal which is clearly $y=x$
$$ay+x=a$$ $$y=x$$ Sub in: $$ax+x=a$$ $$x(a+1)=a$$ $$x_1=\frac{a}{a+1}$$ Then clearly the area of the first square $A_1 = (\frac{a}{a+1})^2$
I then did the exact same thing to work out the top right corner of the second square. I used the equation $y-y_1=m(x-x_1)$ to work out the equation of the diagonal of the second square as $y=x-\frac{a}{a+1}$. Again I did simultaneous equations as follows:
$$ay+x=a$$ $$y=x-\frac{a}{a+1}$$ Sub in: $$a(x-\frac{a}{a+1})+x=a$$ $$ax-\frac{a^2}{a+1}+x=a$$ $$ax+x=a+\frac{a^2}{a+1}$$ $$x(a+1)=\frac{2a^2+a}{a+1}$$ $$x_2=\frac{a(2a+1)}{(a+1)^2}$$
Then, to work out the length of the second square, I did $x_2-x_1$ which yielded $(\frac{a}{a+1})^2$. I then saw a pattern between the two lengths so I repeated the working for the third fraction and the pattern held. The pattern I found is that the length of the square is always $(\frac{a}{a+1})^n$ and so the area of each square is $(\frac{a}{a+1})^{2n}$.
$\therefore$ the total area is $\sum_{n=1}^{\infty} (\frac{a}{a+1})^{2n}$
I have 2 questions:
Is my working correct?
Are there any better/more efficient ways of doing this? (I'm pre-calculus and haven't learnt summations etc.)
Look at the left-most trapezium. It consists of a square of sidelength $h$, say, surmounted by a right-angled triangle whose legs must measure $h$ and $h/a$. Therefore $h+h/a=1$, that is $h=a/(1+a)$. The area of the trapezium is $(h/2)(h+1)$, and so the proportion of its area contained in the square is $$\frac{h^2}{(h/2)(h+1)}=\frac{2h}{h+1}=\frac{2a}{1+2a}.$$ But that is the proportion of the whole triangle occupied by the squares. The squares' area must add up to $$\frac{2a}{1+2a}\frac{a}{2}=\frac{a^2}{1+2a}.$$