The problem states that a triangle is circumscribed about a circle. We are given two side lengths of the triangle, with the segments of the sides from the vertices to the point of tangency of the inscribed circle being 7 and 9. We need to find the perimeter of the triangle.

1. Problem Description

The problem states that a triangle is circumscribed about a circle. We are given two side lengths of the triangle, with the segments of the sides from the vertices to the point of tangency of the inscribed circle being 7 and

9. We need to find the perimeter of the triangle.

2. Solution Steps

Let the triangle be $ABC$. Let the circle be tangent to $AB$ at $D$, to $BC$ at $E$, and to $AC$ at $F$.

We are given that $AD = 7$ and $BE = 9$.

Since tangents from a point to a circle are equal in length, we have $AD = AF = 7$ and $BE = BD = 9$.

Let $CF = CE = x$.

Then the sides of the triangle are $AB = AD + DB = 7 + 9 = 16$, $BC = BE + EC = 9 + x$, and $AC = AF + FC = 7 + x$.

The perimeter of the triangle is $AB + BC + AC = 16 + (9 + x) + (7 + x) = 16 + 9 + 7 + 2x = 32 + 2x$.

However, without knowing the length of the missing side, we cannot determine the exact value of $x$. The sides are $16$, $9+x$, $7+x$.

Looking at the diagram again, it is more likely that the side segment 7 is adjacent to the side segment

9. So let $AD = 7$ and $CE = 9$.

Since tangents from a point to a circle are equal in length, we have $AD = AF = 7$ and $CE = BE = 9$.

Let $BD = BF = x$.

Then the sides of the triangle are $AB = AD + DB = 7 + x$, $BC = BE + EC = 9 + 9 = 18$, and $AC = AF + FC = 7 + 9 = 16$.

The perimeter of the triangle is $AB + BC + AC = (7 + x) + 18 + 16 = 41 + x$. This seems incorrect.

It looks like the image is indicating that the segment of length 7 and 9 are adjacent to the same vertex.

So we have $AD=AF=7$ and $BD=BE=9$. If we let $CF=CE=x$,

The sides are $7+9=16$, $9+x$, $7+x$.

Then the perimeter $P = 16 + 9+x + 7+x = 32 + 2x$.

From the triangle inequality, we have:

$16 + (9+x) > 7+x \implies 25 > 7$

$16 + (7+x) > 9+x \implies 23 > 9$

$(9+x) + (7+x) > 16 \implies 16+2x > 16 \implies 2x>0 \implies x>0$.

However, looking at the figure, it seems that $CD=7$.

Let $CF=CE=7$.

Then $AD=AF=7$ and $BE=BD=9$.

The sides are $16$, $9+7=16$, $7+7=14$.

Then the perimeter $P = 16 + 16 + 14 = 46$.

3. Final Answer

46