The ratio of the radius and height of a cone is 3:4. Its volume is $37\frac{5}{7}$ cm³. The slant height of the cone is

The ratio of radius to height is $3:4$, and the volume is $37\frac{5}{7}$ cm³.

Let the radius be $r = 3x$ and height be $h = 4x$ for some constant $x$.

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Converting the mixed fraction to an improper fraction:

$37\frac{5}{7} = \frac{37 \times 7 + 5}{7}$

$= \frac{259 + 5}{7}$

$= \frac{264}{7}$ cm³

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The volume of a cone is given by $V = \frac{1}{3}\pi r^2 h$.

Substituting the values with $\pi = \frac{22}{7}$:

$\frac{264}{7} = \frac{1}{3} \times \frac{22}{7} \times (3x)^2 \times 4x$

$\frac{264}{7} = \frac{1}{3} \times \frac{22}{7} \times 9x^2 \times 4x$

$\frac{264}{7} = \frac{1}{3} \times \frac{22}{7} \times 36x^3$

$\frac{264}{7} = \frac{22}{7} \times 12x^3$

$\frac{264}{7} = \frac{264}{7} \times x^3$

$x^3 = 1$

$x = 1$ cm

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The actual dimensions are:

Radius $= 3x = 3 \times 1 = 3$ cm

Height $= 4x = 4 \times 1 = 4$ cm

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The slant height forms the hypotenuse of a right triangle with the radius and height.

Using the Pythagorean theorem:

$l = \sqrt{r^2 + h^2}$

$l = \sqrt{3^2 + 4^2}$

$l = \sqrt{9 + 16}$

$l = \sqrt{25}$

$l = 5$ cm

Therefore, the slant height of the cone is $5$ cm.