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Outer square with length 7

I have been given 2 squares here, and I supposed the ratio of (area of shaded region/ are of outer of square)

Obviously there are four right angled triangles here, once we have their size length we can calculate their are and get the final answer.

My book's answer says that all of the triangle 3-4-5 triangle, but I can't figure how to get the measurement of the base/altitude in triangle with altitude=3 and base=4 respectively. Can anyone tell me , How do I calculate the other side of triangles with given information.

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    $\begingroup$ PYthagoras is your friend. $\endgroup$ – Hagen von Eitzen Sep 16 '13 at 20:18
  • $\begingroup$ @HagenvonEitzen, I want to know how all of triangles are similar. mark's post explains it. $\endgroup$ – Dude Sep 16 '13 at 20:28
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Well, the obvious answer is that they are all $3,4,5$ triangles, but how might we see that?

Let's look at the angles which meet at the corner between the marked $3$ and $4$. Let the angle in the triangle with $3$ be $\alpha$, then the angle in the triangle with $4$ which is $90^{\circ}-\alpha$ because the unshaded area is a square and all its angles are right angles. You can trace the angles round and see that all the triangles are similar. And the fact that the outer figure is a square means they are all the same size (congruent).

Once that is established - and I agree that it needs to be checked, rather than assumed - you can use the $3,4,5$ fact with confidence.

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  • $\begingroup$ Thank You! This makes sense. You should see the other answer, I don't know what is that supposed to be. $\endgroup$ – Dude Sep 16 '13 at 20:30
  • $\begingroup$ @JoeDimaggio Everyone is trying to help. You noticed that the "obvious" required explanation. Occasionally that makes a real difference (lots of people thought that Alfred Kempe had solved the four colour problem until Heawood noticed the flaw in the "obvious" proof) $\endgroup$ – Mark Bennet Sep 16 '13 at 20:41
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$\sqrt{3^2+4^2}=?$ ${}{}{}{}{}{}{}{}$

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Consider the side of the bigger square where 3 and 4 are written. Now consider the two sides of the smaller square incident on this square. These two sides are of same length, hence the other side(not hypotenuse) of the right angled triangles should have alternate values. ($3^2+x^2=4^2+y^2$). And so on..

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  • $\begingroup$ yeah, and How will you show that x=4 and y=3 are the only solution? $\endgroup$ – Dude Sep 16 '13 at 20:32
  • $\begingroup$ congruency ofcourse. $\endgroup$ – user67773 Sep 17 '13 at 10:22
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I've seen this as a graphical representation of the Pythagorean Theorem. Given that both figures are squares, you should be able to prove all 4 triangles congruent by ASA. You know the side length of the outer square and calculating the area of the right triangles is simple: if one leg is the base, the length of the other leg is the altitude. This should be enough to solve your given problem.

As to proving the Pythagorean Theorem, consider a general case where the lengths have sides $a$ and $b$ and the hypoteneuse has length $c$. Then the area of the inner square is equal to the area of the outer square minus the area of the 4 triangles. So

$$c^2=(a+b)^2-(4\times\frac12ab)$$

$$c^2=a^2+2ab+b^2-2ab$$

$$c^2=a^2+b^2$$

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