The given figure is composed of a right triangle and a quarter circle. Let us calculate its area and perimeter step by step.

Given:

• The base $AB = 6 \, \text{cm}$, and the quarter circle is based on this diameter.
• $AC$ is the hypotenuse of the triangle, while $BC$ is the height of the triangle.

Step 1: Area

(a) Area of the right triangle ($\triangle ABC$):

Using the formula for the area of a triangle: $\text{Area of } \triangle ABC = \frac{1}{2} \times \text{base} \times \text{height}.$

The base $AB = 6 \, \text{cm}$ and the height $BC = 6 \, \text{cm}$ (since the quarter circle's radius equals the height): $\text{Area of } \triangle ABC = \frac{1}{2} \times 6 \times 6 = 18 \, \text{cm}^2.$

(b) Area of the quarter circle:

The formula for the area of a quarter circle is: $\text{Area of a quarter circle} = \frac{1}{4} \pi r^2,$ where $r$ is the radius. The radius of the quarter circle is: $r = \frac{\text{diameter}}{2} = \frac{6}{2} = 3 \, \text{cm}.$

Substituting $r = 3$: $\text{Area of quarter circle} = \frac{1}{4} \pi (3)^2 = \frac{9\pi}{4} \, \text{cm}^2.$

Total Area:

$\text{Total Area} = \text{Area of triangle} + \text{Area of quarter circle}.$ $\text{Total Area} = 18 + \frac{9\pi}{4}.$

So, the total area is: $\boxed{\text{Total Area} = 18 + \frac{9\pi}{4} \, \text{cm}^2}.$

Step 2: Perimeter

The perimeter consists of:

1. The hypotenuse ($AC$),
2. The curved edge of the quarter circle,
3. The straight side ($BC$).

(a) Length of $AC$ (Hypotenuse):

Using the Pythagorean theorem: $AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \, \text{cm}.$

(b) Length of the quarter-circle arc:

The formula for the arc length of a quarter circle is: $\text{Arc length} = \frac{1}{4} \times 2\pi r = \frac{\pi r}{2}.$ With $r = 3 \, \text{cm}$: $\text{Arc length} = \frac{\pi (3)}{2} = \frac{3\pi}{2} \, \text{cm}.$

(c) Total Perimeter:

$\text{Perimeter} = AC + BC + \text{Arc length}.$ Substitute the values: $\text{Perimeter} = 6\sqrt{2} + 6 + \frac{3\pi}{2}.$

So, the perimeter is: $\boxed{\text{Perimeter} = 6\sqrt{2} + 6 + \frac{3\pi}{2} \, \text{cm}}.$

Final Answer:

1. Area: $\boxed{18 + \frac{9\pi}{4} \, \text{cm}^2}$.
2. Perimeter: $\boxed{6\sqrt{2} + 6 + \frac{3\pi}{2} \, \text{cm}}$.

Would you like further clarifications or examples? 😊

Related Questions:

1. How do you calculate the area of a semicircle with a given radius?
2. What is the formula for the hypotenuse in a right triangle?
3. How is the arc length of a quarter circle derived?
4. If the base and height of a triangle are doubled, what happens to its area?
5. How does changing the radius of a circle affect the perimeter of arcs?

Tip:

Always ensure measurements of all components are clearly understood when solving composite shapes.