Math worksheet for calculating areas of regular polygons and composite figures, including triangles, rectangles, hexagons, and combined shapes.

Let's solve each problem step by step and find the area of each regular polygon or composite figure.

---

Problem 1: Rectangle with a diagonal

- Given: A rectangle with:
- Base = 10.9 m
- Height = 7.5 m (this is the perpendicular height from base to top)
- This is a rectangle, so area is:

$$
\text{Area} = \text{base} \times \text{height} = 10.9 \times 7.5 = 81.75 \, \text{m}^2
$$

✔ Answer: 81.75 m²

---

Problem 2: Regular Pentagon

- Side length = 13.2 yd
- Apothem (distance from center to midpoint of side) = 3.5 yd

For any regular polygon:
$$
\text{Area} = \frac{1}{2} \times \text{Perimeter} \times \text{Apothem}
$$

First, perimeter = number of sides × side length
Pentagon has 5 sides:
$$
P = 5 \times 13.2 = 66 \, \text{yd}
$$

Now,
$$
\text{Area} = \frac{1}{2} \times 66 \times 3.5 = 33 \times 3.5 = 115.5 \, \text{yd}^2
$$

✔ Answer: 115.5 yd²

---

Problem 3: Regular Hexagon

- Side length = 12 cm
- Apothem = $ 5\sqrt{3} $ cm

Regular hexagon can be split into 6 equilateral triangles.

Use formula:
$$
\text{Area} = \frac{1}{2} \times \text{Perimeter} \times \text{Apothem}
$$

Perimeter = $ 6 \times 12 = 72 $ cm

$$
\text{Area} = \frac{1}{2} \times 72 \times 5\sqrt{3} = 36 \times 5\sqrt{3} = 180\sqrt{3} \, \text{cm}^2
$$

Approximate value: $ 180 \times 1.732 \approx 311.76 \, \text{cm}^2 $

But exact answer is better:

✔ Answer: $ 180\sqrt{3} \, \text{cm}^2 $

---

Problem 4: Triangle

- Base = 22 in
- Height = 16 in

$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 22 \times 16 = 11 \times 16 = 176 \, \text{in}^2
$$

✔ Answer: 176 in²

---

Problem 5: Composite Figure (Rectangle + Semicircle)

- Rectangle: 32 ft wide, 18 ft tall
- Semicircle on one end: diameter = 32 ft → radius = 16 ft

Area of rectangle:
$$
32 \times 18 = 576 \, \text{ft}^2
$$

Area of semicircle:
$$
\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (16)^2 = \frac{1}{2} \pi (256) = 128\pi \, \text{ft}^2
$$

Total area:
$$
576 + 128\pi \, \text{ft}^2
$$

Approximate: $ 128 \times 3.1416 \approx 402.12 $
So total ≈ $ 576 + 402.12 = 978.12 \, \text{ft}^2 $

But exact form is better:

✔ Answer: $ 576 + 128\pi \, \text{ft}^2 $
(Approximately: 978.12 ft²)

---

Problem 6: Composite Figure (Triangle + Semicircle)

- The figure looks like an isosceles triangle on top of a semicircle
- Triangle: base = 9 mm, height = 12 mm
- Semicircle: diameter = 9 mm → radius = 4.5 mm

Area of triangle:
$$
\frac{1}{2} \times 9 \times 12 = 54 \, \text{mm}^2
$$

Area of semicircle:
$$
\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (4.5)^2 = \frac{1}{2} \pi (20.25) = 10.125\pi \, \text{mm}^2
$$

Total area:
$$
54 + 10.125\pi \, \text{mm}^2
$$

Approximate: $ 10.125 \times 3.1416 \approx 31.81 $
So total ≈ $ 54 + 31.81 = 85.81 \, \text{mm}^2 $

✔ Answer: $ 54 + 10.125\pi \, \text{mm}^2 $
(Approximately: 85.81 mm²)

---

Problem 7: Composite Figure (Rectangle + Right Triangle)

- Rectangle: 10 km × 6 km
- Attached right triangle: leg = 4.8 km, other leg = 6 km (same as rectangle height)

Wait — look at the diagram: the triangle is attached to the right side, and the dashed line shows that it’s a right triangle with:
- One leg = 4.8 km (horizontal)
- Other leg = 6 km (vertical)

So the triangle is right-angled, legs = 6 km and 4.8 km

Area of rectangle:
$$
10 \times 6 = 60 \, \text{km}^2
$$

Area of triangle:
$$
\frac{1}{2} \times 6 \times 4.8 = 3 \times 4.8 = 14.4 \, \text{km}^2
$$

Total area:
$$
60 + 14.4 = 74.4 \, \text{km}^2
$$

✔ Answer: 74.4 km²

---

Problem 8: Composite Figure (Square + Triangle)

- Square: side = 16 cm
- Triangle: base = 16 cm, height = 8.5 cm? Wait — check diagram.

Wait: The triangle is below the square, and it has:
- Two equal sides marked with ticks → isosceles triangle
- Base = 16 cm (same as square)
- Height = 11.5 cm? But label says "8.5 cm" and "11.5 cm"

Looking carefully:
- The triangle has two equal sides marked with "X"
- The base is 16 cm
- The height is drawn from apex to base, labeled 8.5 cm
- The side of the triangle is labeled 11.5 cm, but that may be redundant.

So, triangle:
- Base = 16 cm
- Height = 8.5 cm

Area of triangle:
$$
\frac{1}{2} \times 16 \times 8.5 = 8 \times 8.5 = 68 \, \text{cm}^2
$$

Area of square:
$$
16 \times 16 = 256 \, \text{cm}^2
$$

Total area:
$$
256 + 68 = 324 \, \text{cm}^2
$$

✔ Answer: 324 cm²

---

✔ Final Answers Summary:

| Problem | Area |
|--------|------|
| 1 | 81.75 m² |
| 2 | 115.5 yd² |
| 3 | $ 180\sqrt{3} $ cm² |
| 4 | 176 in² |
| 5 | $ 576 + 128\pi $ ft² |
| 6 | $ 54 + 10.125\pi $ mm² |
| 7 | 74.4 km² |
| 8 | 324 cm² |

Let me know if you'd like these answers rounded or formatted for a worksheet!