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I have two right triangles One is a 6-8-10 and inside is a 3-4-5 and the space between the two triangles is a uniform amount.

I made a really awkward and weird pic of the diagram and I need to solve for X enter image description here

How would I go about solving this? I couldn't figure out any good approaches.

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Hint: draw in some lines and look for similar triangles inside the bigger triangle. –  Vitaly Lorman Mar 10 '11 at 1:07
    
@VitalyLorman , I'm not sure I follow –  qwertymk Mar 10 '11 at 1:15

4 Answers 4

up vote 14 down vote accepted

Draw lines connecting the "respective vertices". Add up the areas to solve for $x$.

enter image description here

Area of a right triangle with the non-hypotenuse sides being $a$ and $b$ is $\frac{1}{2}(a \times b)$ while the area of a trapezium with parallel sides being $a$ and $b$ and the height being $h$ is $\frac{1}{2}h(a+b)$

$$\frac{1}{2}x(5+10) + \frac{1}{2}x(3+6) + \frac{1}{2}x(4+8) + \frac{1}{2}(3 \times 4) = \frac{1}{2} (6 \times 8)$$

$$36x + 12 = 48 \Rightarrow x = 1$$

EDIT:

Let us try to look at a slightly general case. Take a triangle and scale it to another similar triangle with the scale factor being $t$.

enter image description here

Let the sides of the inner triangle be $a$, $b$ and $c$.

The perimeter and area of the inner triangle is $P$ and $A$ respectively.

The sides of the outer triangle are $ta$, $tb$ and $tc$ while the perimeter and area are $tP$ and $t^2A$.

As before join the "respective" vertices and summing the areas give us,

$$A + \frac{1}{2}x(ta + a) + \frac{1}{2}x(tb + b) + \frac{1}{2}x(tc + c) = t^2A$$

$$A + \frac{1}{2}x(t+1)P = t^2A \Rightarrow \frac{1}{2}x(t+1)P = (t^2-1)A$$

$$x = \frac{2A}{P}(t-1)$$

In the problem asked, $t=2$ with $P = 2A = 12$ and hence we get $x=1$.

Also, $t=1$ gives $x=0$ as expected.

This also gives a nice proof that the radius of the incircle of a triangle is $$r_{in} = \frac{2A}{P}$$

This is got by plugging in $t=0$ and realizing that $\left| x \right|$ is nothing but the radius of the incircle.

Hence, $$x = (t-1)r_{in}$$

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How are you splitting this? It seems that you made a bunch of trapezoids and added up their area but you seem to have added some areas more than once –  qwertymk Mar 10 '11 at 1:34
    
I thought of this after I posted my more complicated solution, and I edited without noticing that you had already done it, and with a nice picture even! I deleted my answer. +1. –  Jonas Meyer Mar 10 '11 at 1:50
    
@Sivaram: Very nice, +1. –  Eric Naslund Mar 10 '11 at 1:52
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@Jonas: I was typing out my solution and went away for a cup of coffee and posted it to find out your solution was similar to mine. You should not have deleted your solution since you actually posted before me. –  user17762 Mar 10 '11 at 2:00
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@Eric: True. Actually once I finished typing the details, I thought I should have actually "reverse engineered" to start from the incircle and proceed. –  user17762 Mar 10 '11 at 6:01

One way to solve this would be to set up a coordinate system with the bottom left corner of the outer triangle at $(0,0)$. The bottom left corner of the inner triangle would then be at $(t,t)$ for some positive real $t$. ($t$ is what you called $x$, but that's confusing in the coordinate context.) Then you just need to set $t$ so that the distance between the two hypotenuses are equal; see the Wikipedia article on parallel lines for the relevant formulas. (They're not too hard to derive on your own.)

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I get that but how would I go about figuring out what t is? Also where would the big triangle end on a y = x line? –  qwertymk Mar 10 '11 at 1:07
    
The bottom right corner of the inner triangle would be at $(t+4, t)$, not at $(t, t)$. –  Vitaly Lorman Mar 10 '11 at 1:21
    
I meant "bottom left". Sorry. –  Michael Lugo Mar 10 '11 at 1:43

Algebraic Solution:

Extend the lines making up the side length $3$ of the inner triangle until it hits the walls of the larger one, and consider the triangle formed. What is the length of this line particular line? Well it is $$x+3+(?)$$ where $(?)=\frac{5}{4}x$ by similar triangles. What about the base line? It has length $8-x$. And the hypotenuse? It has length $10-x\frac{5}{4}$ by similar triangles again. Then by the Pythagorean Theorem, $$\left(10-x\frac{5}{4}\right)^{2}=\left(8-x\right)^{2}+\left(3+\frac{9}{4}x\right)^{2}$$ and solving this quadratic yields $x=1$ as a solution.

Quick Solution:

Notice the side lengths of the larger triangle sum to equal its area, 24. (That is why we get 1) Draw straight tubes, a $x\times 6$ tube, a $x\times 8$ tube and a $x\times 10$ tube along the sides on the triangle, on the inside so that all of the area of the triangle except the missing middle piece is covered. (The middle piece is the smaller triangle) Then the overlapping parts, and the parts hanging outside the triangle can be moved around to make up the smaller inside triangle which is missing. Hence the areas are equal so $8x+6x+10x=\frac{6\cdot 8}{2}=24$ so $x=1$.

Remark: I also hope this is understandable without a picture.

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If we have right triangles with sides $a, b, c$ and $2a, 2b, 2c$, with $c, 2c$ the hypotenuses, then generalizing this algebraic approach gives $t = ab/(a+b+c) = 2ab/(2a+2b+2c)$ -- that is, the area of the larger triangle divided by its perimeter. I was able to show this algebraically; your "quick solution", I think, explains why this should be so. –  Michael Lugo Mar 10 '11 at 1:45
    
@Michael: That is interesting. I now see it follows much more easily from Sivarams solution –  Eric Naslund Mar 10 '11 at 2:10
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@Michael, @Eric: I have added some details on these lines in my updated post now @Michael: The area of the larger triangle is 4 times the area of the triangle inside and not twice. –  user17762 Mar 10 '11 at 5:54
    
Right. But the area of the larger triangle is $(2a)(2b)/2 = 2ab$. –  Michael Lugo Mar 10 '11 at 15:26

Find the lengths of the segments labeled A, B, C, and D. These should add up to 6. To find the lengths of C and D, use the fact that the triangles I have drawn in are similar to the inside (or outside) triangle.

enter image description here

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