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$(1) \quad \dfrac{256}{17}$

$(2) \quad \dfrac{39}{4}$

$(3) \quad \dfrac{483}{8}$

$(4) \quad \dfrac{52}{4}$

$(5) \quad \dfrac{492}{18}$

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    $\begingroup$ I don't think there is enough information to know. The triangle could be drawn at any angle, giving different bounding box rectangles. $\endgroup$ – Jaap Scherphuis Oct 22 at 9:57
  • $\begingroup$ The triangle is not unique. However, the area of rectangle is bounded between $12$ and $16$. So only the first or fourth answer is possible. $\endgroup$ – achille hui Oct 22 at 11:25
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    $\begingroup$ Interesting problem, but if you want your question to remain open, please see how to ask a good question and edit your work accordingly. It's not just 'include your work' on there. $\endgroup$ – Toby Mak Oct 22 at 11:28
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A unique triangle cannot be constructed so no area can be associated with it.

Shown here are two cases of rectangles (red and blue borders). The circles have radii $(5,4)$ for side lengths and a side of tangent length $3$ units between them.

EDIT1:

Area is not invariant. To calculate when hypotenuse makes an angle $u$ to vertical,

$$ (3 \cos u + 4 \sin u)(3 \sin u + +4 \cos u)=(12+12.5 \sin2u) $$

It has maximum value 24.5 when hypotenuse makes $45^{\circ}$ to vertical, and minimum 12 at time of vertical/horizontal positions of the shorter sides.

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  • $\begingroup$ True. The next question would be: is the area an invariant? $\endgroup$ – cvanaret Oct 22 at 11:12
  • $\begingroup$ Please ask that in another question. $\endgroup$ – Narasimham Oct 22 at 11:14
  • $\begingroup$ It's not my post, I'm just reacting. $\endgroup$ – cvanaret Oct 22 at 11:16
  • $\begingroup$ The rigid yellow triangle triangle has constant area 6 sq. units..but let me formulate $\endgroup$ – Narasimham Oct 22 at 11:18
  • $\begingroup$ No. when hypotenuse makes $ 45^0$ to vertical,the area is maximum at 24.5; and minimum 12 at time of vertical/horizontal positions of the shorter sides. Generally the area is $ (12 +12.5 \sin 2u) $. $\endgroup$ – Narasimham Oct 22 at 11:27
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Let $AE=a$, $BE=b$, $AF=c$, $DF=d$.

Now

\begin{align} a^2+c^2 &= 3^2 \tag{1} \\ d^2+(a+b)^2 &= 4^2 \tag{2} \\ b^2+(c+d)^2 &= 5^2 \tag{3} \\ a^2+c^2+d^2+(a+b)^2 &= b^2+(c+d)^2 \tag{$3^2+4^2=5^2$} \\ a(a+b) &= cd \tag{4} \\ d^2+\left( \frac{cd}{a} \right)^2 &= 4^2 \\ d^2(a^2+c^2) &= 4^2 a^2 \\ 3^2d^2 &= 4^2a^2 \\ d &= \frac{4a}{3} \\ c &= \sqrt{3^2-a^2} \\ a+b &= \frac{4c}{3} \\ (a+b)(c+d) &= \frac{4}{3} \sqrt{3^2-a^2} \left( \frac{4a}{3}+\sqrt{3^2-a^2} \right) \\ \end{align}

Unless $a+b=c+d$, that is a square, you cannot have a proper choice.

For the square case,

$$(a+b)(c+d)=\frac{256}{17}$$

I'm leaving the missing steps for above result as an exercise.

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  • $\begingroup$ If you are going to post an answer, it would be better to post an incomplete solution since the OP hasn't shown any work yet. $\endgroup$ – Toby Mak Oct 22 at 11:18
  • $\begingroup$ This is a problematic problem, it's normal for most people unable to solve it. That's an exceptional case. Also, I leave the special case as an exercise. $\endgroup$ – Ng Chung Tak Oct 22 at 11:21
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    $\begingroup$ It seems likely that the missing information is that the triangle is inscribed in a square, since this gives one of the choices. $\endgroup$ – David K Oct 22 at 11:46
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    $\begingroup$ It took me a while to realize you meant "if" when you wrote "unless". $\endgroup$ – David K Oct 22 at 11:47

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