Here is a circle and a right-angled triangle. The diameter of the circle is 27 cm. Show that the circumference of the circle is greater than the perimeter of the triangle.

The diameter of the circle is given as $D = 27 \text{ cm}$.
The formula for the circumference $C$ of a circle is $C = \pi D$.

$$C = \pi \times 27 \text{ cm}$$

Using $\pi \approx 3.14159$:

$$C \approx 3.14159 \times 27 \text{ cm}$$ $$C \approx 84.823 \text{ cm}$$

The lengths of the two shorter sides (legs) of the right-angled triangle are $a = 21 \text{ cm}$ and $b = 28 \text{ cm}$.
Using the Pythagorean theorem, $a^2 + b^2 = c^2$, where $c$ is the hypotenuse:

$$c^2 = (21 \text{ cm})^2 + (28 \text{ cm})^2$$ $$c^2 = 441 \text{ cm}^2 + 784 \text{ cm}^2$$ $$c^2 = 1225 \text{ cm}^2$$ $$c = \sqrt{1225 \text{ cm}^2}$$ $$c = 35 \text{ cm}$$

The perimeter $P_T$ of the triangle is the sum of its three sides:

$$P_T = 21 \text{ cm} + 28 \text{ cm} + 35 \text{ cm}$$ $$P_T = 84 \text{ cm}$$

Circumference of the circle $C \approx 84.823 \text{ cm}$.
Perimeter of the triangle $P_T = 84 \text{ cm}$.
Since $84.823 \text{ cm} > 84 \text{ cm}$, the circumference of the circle is greater than the perimeter of the triangle.

We have shown that the circumference of the circle ($84.823 \text{ cm}$) is greater than the perimeter of the triangle ($84 \text{ cm}$).