The diagram shows part of the curve $y = 16 + 6x - x^2$ intersecting the line $y = 26 - x$ at $A$ and $B$.

1. Set up the equation to find the points of intersection

Equate the two equations:
$16 + 6x - x^2 = 26 - x$

2. Rearrange into a standard quadratic equation

$x^2 - 7x + 10 = 0$

3. Solve the quadratic equation for x

$(x - 2)(x - 5) = 0$
$x = 2 \text{ or } x = 5$

4. Find the corresponding y-coordinates

When $x = 2$, $y = 26 - 2 = 24$.
When $x = 5$, $y = 26 - 5 = 21$.
Since $A$ is to the left of $B$, $A$ has the smaller $x$-coordinate.
$A(2, 24)$ and $B(5, 21)$.

1. Set $y = 0$ to find the $x$-intercepts of the curve

At the $x$-axis, $y = 0$.
$16 + 6x - x^2 = 0$

2. Solve the quadratic equation and select the correct root for $C$

$x^2 - 6x - 16 = 0$
$(x - 8)(x + 2) = 0$
$x = 8 \text{ or } x = -2$
From the diagram, $C$ lies on the positive $x$-axis, so $x > 0$.
Therefore, the $x$-coordinate of $C$ is 8 (shown).

1. Set up the area as a sum of integrals matching the boundary O→A→B→C, using A(2,24), B(5,21), C(8,0) from earlier parts

Using $A(2,24)$, $B(5,21)$ (part (i)) and $C(8,0)$ (part (ii)):
$\text{Area} = \int_0^2 (16+6x-x^2)\,dx + \int_2^5 (26-x)\,dx + \int_5^8 (16+6x-x^2)\,dx$
The top boundary is the curve for $0\le x\le 2$, the line $AB$ for $2\le x\le 5$, then the curve again for $5\le x\le 8$.

2. Evaluate each definite integral

$\int_0^2 (16+6x-x^2)\,dx = \left[16x+3x^2-\tfrac{x^3}{3}\right]_0^2 = 44-\tfrac{8}{3} = 41\tfrac{1}{3}$
$\int_2^5 (26-x)\,dx = \left[26x-\tfrac{x^2}{2}\right]_2^5 = 117.5 - 50 = 67.5$
$\int_5^8 (16+6x-x^2)\,dx = \left[16x+3x^2-\tfrac{x^3}{3}\right]_5^8 = 149\tfrac{1}{3} - 113\tfrac{1}{3} = 36$

3. Add the parts to find the total area

$\text{Area} = 41\tfrac{1}{3} + 67.5 + 36 = 144\tfrac{5}{6}\text{ square units}$