For the life of me, I do not know how to help you with your work because you have not shown what your work is.
I understand where fractions come into play, but is your problem that you do not know how to handle algebraic fractions?
I also understand the relevance of the Pythagorean Theorem, but what is the Pythagorean method?
Please show what you have done so far so we can tell whether you are trying to solve the problem in a promising way.
The large triangle is of dimensions 6, 8, 10.
The top triangle is of dimensions $\displaystyle \begin{align*} x, 6 - y, \sqrt{ x^2 + (6- y)^2} \end{align*}$.
The bottom triangle is of dimensions $\displaystyle \begin{align*} 8-x,y,\sqrt{(8-x)^2+y^2} \end{align*}$.
Thus $\displaystyle \begin{align*} \sqrt{x^2 + (6-y)^2} + \sqrt{(8-x)^2+y^2} = 10 \end{align*}$.
We are also told $\displaystyle \begin{align*} xy = 12 \end{align*}$, so $\displaystyle \begin{align*} y = \frac{12}{x} \end{align*}$. Substituting in gives
$\displaystyle \begin{align*} \sqrt{x^2 + \left( 6 - \frac{12}{x} \right) ^2 }+ \sqrt{ (8-x)^2 + \left( \frac{12}{x} \right) ^2 } &= 10 \\ \sqrt{ x^2 + \frac{ (6x - 12)^2}{x^2} } + \sqrt{(8-x)^2 + \frac{144}{x^2}} &= 10 \\ \sqrt{ \frac{x^4 + (6x-12)^2}{x^2} } + \sqrt{ \frac{x^2 \left( 8 - x \right) ^2 + 144}{x^2} } &= 10 \\ \frac{\sqrt{ x^4 + (6x-12)^2}}{x} + \frac{\sqrt{x^2 \left( 8-x \right) ^2 + 144}}{x} &= 10 \\ \sqrt{ x^4 + (6x-12)^2 } + \sqrt{ x^2 \left( 8-x \right) ^2 + 144 } &= 10x \\ \sqrt{ x^4 + 36x^2 - 144x + 144 } + \sqrt{ x^4 - 16x^3 + 64x^2 + 144} &= 10x \end{align*}$
Now try to solve this equation.
Perhaps an easier way to go is to look at areas.
$Area\ of\ big\ triangle = \dfrac{1}{2} * 6 * 8 = 24.$
$Area\ of\ rectangle = 12 = xy.$
$area\ of\ small\ triangle\ above\ rectangle = \dfrac{1}{2} * x * (6 - y) = \dfrac{6x - xy}{2} = \dfrac{6x - 12}{2} = 3x - 6.$
$area\ of\ small\ triangle\ beside\ rectangle = \dfrac{1}{2} * (8 - x) * y = \dfrac{8y - xy}{2} = \dfrac{8y - 12}{2} = 4y - 6.$
Now what is the relationship among these four areas?
Does that allow you to determine y in terms of x?
That leads to a quadratic equation, which you may find easier to address than the equation in the previous post. But both equations give the same result.
I think it is easier to take a look at the areas. The whole triangle has an area of (1/2)(6)(8) = 24 cm^2. Thus the area of the top and bottom triangles, plus the rectangle is 24:
You can multiply that all out and note that whenever you have an xy you can replace that with 12. (xy = 12). You will get a simple equation in terms of x and y. Then use y = 12/x.
If you have any troubles with this on the way, just let us know.
-Dan
Hello, xwy!
Yet another approach . . .
29. A rectangle of area 12 cm^2 is inscribed in right triangle ABC.
What are its dimensions?
Code:- A o : | * : | * E : D o - - - - - o 6 | | * : | |y * : | x | 8-x * - o - - - - - o - - - - - - - o B F C : - - - - - - 8 - - - - - - :
We have right triangle $ABC\!:\;AB = 6,\;BC = 8,$
inscribed rectangle $BFED\!:\;BF = DE = x,\;DB = EF = y.$
Then $FC = 8-x.$
We have similar triangles: .
Hence: .$\dfrac{y}{8-x} \:=\:\dfrac{6}{8} \quad\Rightarrow\quad y \:=\:\dfrac{24-3x}{4}$ .[1]
The area of $BFED$ is 12 cm^{2}: .$xy \,=\,12$
Substitute [1]: .$x\left(\dfrac{24-3x}{4}\right) \:=\:12 $
Simplify: .$x^2 - 8x + 12 \:=\:0 \quad\Rightarrow\quad (x-2)(x-6) \:=\:0$
Hence: .$x \:=\:2,\,6 \quad\Rightarrow\quad y \,=\,6,\,2$
The dimensions are: .$2\times6.$
Soroban and Idea are quite right that you can use similar triangles as well as area or the Pythagorean Theorem, but Soroban made a slight arithmetic mistake in working out the similar triangle idea. (He skipped some steps: even a skilled mathematician like Soroban can make mistakes when skipping steps.)
$x\left(\dfrac{24 - 3x}{4}\right) = 12 \implies \dfrac{24x - 3x^2}{4} = 12 \implies 24x - 3x^2 = 48 \implies$
$3x^2 - 24x = - 48 \implies 3x^2 - 24x + 48 = 0 \implies x^2 - 8x + 16 = 0 \implies (x - 4)^2 = 0 \implies x = 4.$
$\therefore xy = 12 \implies 4y = 12 \implies y = 3.$
So you have been given three different ways to attack this problem.
In the spirit of a challenge problem, I have been thinking about this way of solving this problem without putting too much strain on my shaky arithmetic.
When I get to $\sqrt{x^2 + (6 - y)^2} + \sqrt{(8 - x)^2 + y^2} = 10$, I can see this is going to end up as a polynomial of degree 8 Yuck.
Just expanding the equation above gives me a decent probability of making an arithmetic error, and applying Newton-Raphson just about guarantees that I shall do so at least once. Of course, the rational root theorem may save the day, but the list of possible roots may be scarily large. But wait, if the rational root theorem works, perhaps x and y are integers. In that case, because they are lengths, both must be positive integers.
$x, y \in \mathbb Z^+\ and\ xy = 12 \implies (x,\ y) = (1,\ 12),\ (2,\ 6),\ (3,\ 4),\ (4,\ 3),\ (6,\ 2),\ or\ (12,\ 1).$
$(x,\ y) = (1,\ 12) \implies \sqrt{x^2 + (6 - y)^2} + \sqrt{(8 - x)^2 + y^2} > \sqrt{12^2} = 12 > 10.$ Not (1, 12).
$(x,\ y) = (12,\ 1) \implies \sqrt{x^2 + (6 - y)^2} + \sqrt{(8 - x)^2 + y^2} > \sqrt{12^2} = 12 > 10.$ Not (12, 1).
$(x,\ y) = (2,\ 6) \implies \sqrt{x^2 + (6 - y)^2} + \sqrt{(8 - x)^2 + y^2} = \sqrt{2^2 + 4^2} + \sqrt{6^2 + 6^2} = \sqrt{20} + \sqrt{72} > 4 + 8 > 10.$ Not (2, 6).
$(x,\ y) = (6,\ 2) \implies \sqrt{x^2 + (6 - y)^2} + \sqrt{(8 - x)^2 + y^2} = \sqrt{6^2 + 4^2} + \sqrt{2^2 + 2^2} = \sqrt{52} + \sqrt{8} = $
$\sqrt{52 + 2\sqrt{52 * 8} + 8} = \sqrt{60 + 2\sqrt{4 * 13 * 4 * 2}} = \sqrt{60 + 8\sqrt{26}} > \sqrt{60 + 8\sqrt{25}} = \sqrt{100} = 10.$ (Not 6, 2).
$(x,\ y) = (3,\ 4) \implies \sqrt{x^2 + (6 - y)^2} + \sqrt{(8 - x)^2 + y^2} = \sqrt{3^2 + 2^2} + \sqrt{5^2 + 4^2} = \sqrt{13} + \sqrt{41} = $
$\sqrt{13 + 2\sqrt{13 * 41} + 41} = \sqrt{54 + 2\sqrt{533}} > \sqrt{54 + 2\sqrt{529}} = \sqrt{54 + 2 *23} = \sqrt{100} = 10.$ (Not 3, 4).
$(x,\ y) = (4,\ 3) \implies \sqrt{x^2 + (6 - y)^2} + \sqrt{(8 - x)^2 + y^2} = \sqrt{4^2 + 3^2} + \sqrt{4^2 + 3^2} = \sqrt{25} + \sqrt{25} = 10.$
x = 4 and y = 3 are the only integer answers.
The problem with this method of course is that it leaves open the possibility that there are positive, real answers that are not integers. The quadratic methods show that x = 4 and y = 3 are the only answers.