The question is: The figure shows two adjacent rectangles. Find a relationship between $x$ and $y$. (Hint: Use Pythagoras Theorem twice.) Diagram --> enter image description here

What I have attempted : I tried doing similar triangles

and doing

$$ \frac {x+y}{y} = \frac{y+3}{y-3}$$

$$ xy - 3x + y^2 - 3y = y^2 + 3y $$

$$ xy-3x = 6y $$

$$ x = \frac{6y}{y-3} $$

but that doesnt give me the answer I also tried doing this:

let the missing length of the base "a", and the missing length of the height "b". Then: $$x^2 = 9 + b^2$$, and $$y^2 = 16 + a^2$$ but I am not too sure how to continue

also the answer is $9y^2 + 16x^2 = x^2y^2 $ and I am not too sure how they got to that..

  • $\begingroup$ I don't see where $\frac {x+y}{y} = \frac{y+3}{y-3}$ comes from. $y$ is not in the same direction as $3$, so why can you add/subtract it? $\endgroup$ – Ross Millikan Mar 1 '16 at 5:02

As you say, we have $9+b^2=x^2, 16+a^2=y^2$. By similar triangles we have $\frac x3=\frac ya=\frac y{\sqrt{y^2-16}}$. Does that qualify as a relationship?

  • $\begingroup$ We have $(x^2-9)(y^2-16)=144.$ It is tempting to put $ x=y=5,$ but that is inconsistent with the diagram. We can't determine $x,y$ as we may draw the $xy$ diagonal at any angle between $ 0$ and $90^o ,$ and measure off $3$ on the horizontal, and then find the right point for the "$4$". $\endgroup$ – DanielWainfleet Mar 1 '16 at 5:32

One way to obtain the relationship you have given is to solve for a value of your variable $a$. $$y^2=16+a^2$$ $$a=\sqrt{y^2-16}$$

Now you can use similar triangles. $$\frac{x}{3}=\frac{y}{\sqrt{y^2-16}}$$ $$3y=x\sqrt{y^2-16}$$ $$9y^2=x^2(y^2-16)$$ $$16x^2+9y^2=x^2y^2$$


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