<s>[INST] Area of similar triangle Suppose that we are given a triangle whose area is known. put a circle C of radius r inside that triangle. How can we find the area of a triangle similar to the first one and whose inscribed circle is C?
 [/INST] The area $a$ of the smaller triangle whose inscribed circle has radius $r$ cannot be defined in an univocal manner only knowing the area $A$ of the larger triangle. For a given value of $A$, there are infinite possible values of $a$, since the area of the smaller triangle depends on the "shape" of the larger triangle. 
Let us make an example. Imagine that the larger triangle is an equilateral one with side $\displaystyle 2 \sqrt{\frac{A}{\sqrt {3}}    } $. Its height is  therefore $\displaystyle  \sqrt{A \sqrt {3}} $, and its area is $\displaystyle \sqrt{\frac{A}{\sqrt {3}} } \cdot  \sqrt{A \sqrt {3}} =A $. In this case, the smaller triangle is also equilateral. It is not difficult to show that an equilateral triangle whose inscribed circle has radius $r$ has side equal to $2 \sqrt{3} \,r$ and height equal to $3r$, so that its area is $a=r^2 \cdot 3\sqrt{3} \approx r^2 \cdot 5.196...$. 
Now let us consider another initial larger triangle, with unchanged area and different shape. For instance, we could choose an isosceles right triangle where the length of each leg is $\sqrt{2A}$, so that the area is still equal to $A$. In this case, the smaller triangle is also an isosceles right triangle. We can find its leg length $s$ by noting that the area is $\displaystyle \frac{s^2}{2}$ and the perimeter is $\displaystyle (2+\sqrt{2})s$. Applying the standard formula for the calculation of the radius of the inscribed circle (area/semiperimeter) we get $\displaystyle \frac{s^2}{   s(2+ \sqrt{2})}=r$, and then $s=r(2+\sqrt{2})$. The area of the smaller triangle is therefore given by $\displaystyle a=\frac{r^2}{2}(2+\sqrt{2})^2\approx r^2 \cdot 5.828...$. 
In conclusion, to determine the area $a$ of the smaller triangle, some other information about the larger triangle - in addition to the area $A$ - is necessary. In particular, the shape of the larger triangle has to be unambiguously defined. The simplest information that could allow to identify the shape of the larger triangle is its semiperimeter $P$. Knowing this information, we could directly calculate the radius of the circle inscribed in the larger triangle as $\displaystyle R=\frac{A}{P}$, and obtain the scale factor $\displaystyle \frac{r}{R}= \frac{Pr}{A}$. Since the ratio of the areas of two similar triangles is equal to the square of the scale factor, we can conclude that $\displaystyle a=A (\frac{Pr}{A})^2=\frac{(Pr)^2}{A}$.
As a confirmation, we can apply this formula to the triangles of the two mentioned examples. In the first case, that of the equilateral larger triangle, we have $P=\displaystyle 3 \sqrt{\frac{A}{\sqrt {3}}    } $, and the formula yields the correct value of $\displaystyle (3 \sqrt{\frac{A}{\sqrt {3}}    }r )^2/A= r^2 \cdot 3\sqrt{3}$. In the second case, that of the isosceles right triangle, we have $P=\frac{1}{2}(2+\sqrt{2})\sqrt{2A}$, and again the formula yields the correct value of $\displaystyle a= \frac
{[\frac{1}{2}(2+\sqrt{2})\sqrt{2A}]^2r^2}{A}=    \frac{r^2}{2}(2+\sqrt{2})^2$.
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