The area of the triangle formed by the lines $y = 4$, $x/5 + y/4 = 1$ and $x/5 - y/8 = 1$ is

The two slanted lines meet where $\frac{x}{5} + \frac{y}{4} = 1$ and $\frac{x}{5} - \frac{y}{8} = 1$.

Subtracting the second from the first:

$\frac{y}{4} + \frac{y}{8} = 0$

$\frac{3y}{8} = 0$

$y = 0$

Substituting $y = 0$ in Line 2: $\frac{x}{5} = 1 \implies x = 5$

Vertex $A = (5, 0)$

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Substituting $y = 4$ in Line 2: $\frac{x}{5} + \frac{4}{4} = 1$

$\frac{x}{5} = 0$

$x = 0$

Vertex $B = (0, 4)$

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Substituting $y = 4$ in Line 3: $\frac{x}{5} - \frac{4}{8} = 1$

$\frac{x}{5} = \frac{3}{2}$

$x = 7.5$

Vertex $C = (7.5, 4)$

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Both $B$ and $C$ lie on the line $y = 4$, so the base $= 7.5 - 0 = 7.5$

The height is the perpendicular distance from $A(5, 0)$ to the line $y = 4$, which is $4 - 0 = 4$.

Area $= \dfrac{1}{2} \times 7.5 \times 4 = 15$