Solved GEOMETRY Name Area (apothem) worksheet Date HR Find | Chegg.com - Free Printable

Let's solve each problem step by step using the formula for the area of a regular polygon:

$$
\text{Area} = \frac{1}{2} \times P \times a
$$

Where:
- $ P $ is the perimeter of the polygon,
- $ a $ is the apothem (the distance from the center to the midpoint of a side),
- The polygon is regular, meaning all sides and angles are equal.

We'll go through each figure one by one.

---

1) Regular Octagon (8 sides), side length = 3 in

- Number of sides: $ n = 8 $
- Side length: $ s = 3 $ in
- Perimeter: $ P = n \times s = 8 \times 3 = 24 $ in
- Apothem $ a $: Not given directly, but we can calculate it using geometry or use known formulas.

But wait — in this case, no apothem is shown. However, since the side length is given, and it’s a regular octagon, we can compute the apothem using trigonometry.

For a regular polygon:
$$
a = \frac{s}{2 \tan\left(\frac{180^\circ}{n}\right)}
$$

For $ n = 8 $, $ s = 3 $:

$$
a = \frac{3}{2 \tan(22.5^\circ)} \approx \frac{3}{2 \times 0.4142} \approx \frac{3}{0.8284} \approx 3.62 \text{ in}
$$

So:
- $ P = 24 $ in
- $ a \approx 3.62 $ in
- Area $ = \frac{1}{2} \times 24 \times 3.62 = 12 \times 3.62 = 43.44 $ in²

But wait — let's check if the image provides more information.

Looking at the second figure (2), it shows a line labeled "3 in" from the center to the side — that's the apothem.

But in Figure 1, no apothem is drawn. So likely, we are supposed to assume that the apothem is not given, but maybe we're missing something?

Wait — actually, in Figure 2, there is a line labeled "3 in" from center to side → that’s the apothem.

But in Figure 1, only side length is given as 3 in.

So for Figure 1, we must calculate apothem.

Let’s proceed carefully.

---

Problem 1: Regular Octagon, side = 3 in

- $ n = 8 $
- $ s = 3 $
- $ P = 8 \times 3 = 24 $ in

Use formula for apothem:
$$
a = \frac{s}{2 \tan\left(\frac{\pi}{n}\right)} = \frac{3}{2 \tan(22.5^\circ)}
$$

$ \tan(22.5^\circ) = \sqrt{2} - 1 \approx 0.4142 $

So:
$$
a = \frac{3}{2 \times 0.4142} = \frac{3}{0.8284} \approx 3.62 \text{ in}
$$

Now area:
$$
A = \frac{1}{2} \times P \times a = \frac{1}{2} \times 24 \times 3.62 = 12 \times 3.62 = 43.44 \text{ in}^2
$$

So:
- $ P = 24 $
- $ a \approx 3.62 $
- Area $ \approx 43.44 $ in²

But perhaps they expect exact value?

Alternatively, maybe the figures are meant to be interpreted differently.

Let’s look at Figure 2: It has a regular octagon with a vertical line from center to side labeled 3 in — that’s the apothem.

So for Figure 2, apothem $ a = 3 $ in.

But what is the side length? Not given.

Wait — maybe each figure is independent, and some have side lengths, others have apothems.

Let’s analyze each one carefully.

---

## Let's go figure by figure:

---

1) Regular Octagon, side = 3 in

- Side length: $ s = 3 $ in
- Number of sides: $ n = 8 $
- Perimeter: $ P = 8 \times 3 = 24 $ in
- Apothem: Need to find

Formula:
$$
a = \frac{s}{2 \tan\left(\frac{180^\circ}{n}\right)} = \frac{3}{2 \tan(22.5^\circ)}
$$

Use $ \tan(22.5^\circ) = \sqrt{2} - 1 \approx 0.4142 $

So:
$$
a = \frac{3}{2 \times 0.4142} = \frac{3}{0.8284} \approx 3.62 \text{ in}
$$

Then:
$$
\text{Area} = \frac{1}{2} \times 24 \times 3.62 = 43.44 \text{ in}^2
$$

So:
- $ P = 24 $
- $ a \approx 3.62 $
- Area $ \approx 43.44 $

But let’s keep more decimals or see if exact form is expected.

Alternatively, use calculator:

$ \tan(22.5^\circ) \approx 0.41421356 $

So $ a = 3 / (2 \times 0.41421356) = 3 / 0.82842712 \approx 3.623 $

Area $ = 0.5 \times 24 \times 3.623 = 12 \times 3.623 = 43.476 $ in²

We’ll round to nearest hundredth: 43.48 in²

---

2) Regular Octagon, apothem = 3 in

- Apothem $ a = 3 $ in
- But side length not given
- We need to find perimeter, so we need side length

Use the same formula:
$$
a = \frac{s}{2 \tan(22.5^\circ)} \Rightarrow s = 2a \tan(22.5^\circ)
$$

$ \tan(22.5^\circ) \approx 0.4142 $

So:
$$
s = 2 \times 3 \times 0.4142 = 6 \times 0.4142 = 2.4852 \text{ in}
$$

Perimeter $ P = 8 \times s = 8 \times 2.4852 = 19.8816 $ in

Area $ = \frac{1}{2} \times P \times a = \frac{1}{2} \times 19.8816 \times 3 = 9.9408 \times 3 = 29.8224 $ in² ≈ 29.82 in²

So:
- $ P \approx 19.88 $
- $ a = 3 $
- Area $ \approx 29.82 $

---

3) Regular Octagon, triangle formed with apothem = 3 in

This figure shows a triangle from center to two adjacent vertices, with a line from center to side labeled 3 in — again, that’s the apothem, so $ a = 3 $ in

Same as Figure 2? Possibly.

But now, it shows a triangle from center to two vertices, and apothem = 3 in.

But still, unless side length is implied, we need to use the same method.

Wait — is the side length given?

No. But perhaps the triangle helps us?

Actually, the triangle connects center to two adjacent vertices — that forms an isosceles triangle with vertex angle $ 360^\circ / 8 = 45^\circ $

And the apothem is the height to the base (side).

So we can use this triangle to find side length.

Let’s do that.

In the central triangle:
- Vertex angle = 45°
- Two equal sides = radius $ r $
- Height from apex to base = apothem $ a = 3 $ in
- Base = side length $ s $

Split into two right triangles:
- Each has angle = 22.5°
- Opposite side = $ s/2 $
- Adjacent side = $ a = 3 $
- So $ \tan(22.5^\circ) = \frac{s/2}{3} \Rightarrow s/2 = 3 \tan(22.5^\circ) $

So:
$$
s = 2 \times 3 \times \tan(22.5^\circ) = 6 \times 0.4142 = 2.4852 \text{ in}
$$

Same as before!

So:
- $ s = 2.4852 $
- $ P = 8 \times s = 19.8816 $ in
- $ a = 3 $
- Area $ = \frac{1}{2} \times 19.8816 \times 3 = 29.8224 $ in²

So same as Figure 2.

Thus:
- $ P \approx 19.88 $
- $ a = 3 $
- Area $ \approx 29.82 $

But wait — why are Figures 2 and 3 both giving same values? Probably because both show apothem = 3 in.

Possibly typo or different interpretations.

But in Figure 2, the apothem is drawn vertically; in Figure 3, it's drawn inside a triangle — same thing.

So likely, both have $ a = 3 $ in, and we must compute side from that.

So both 2 and 3 are identical in data.

---

4) Regular Pentagon, apothem = 6 cm

- Regular pentagon: $ n = 5 $
- Apothem $ a = 6 $ cm
- Need side length

Use:
$$
a = \frac{s}{2 \tan(36^\circ)} \quad \text{(since } 180^\circ / 5 = 36^\circ \text{)}
$$

$ \tan(36^\circ) \approx 0.7265 $

So:
$$
s = 2a \tan(36^\circ) = 2 \times 6 \times 0.7265 = 12 \times 0.7265 = 8.718 \text{ cm}
$$

Perimeter $ P = 5 \times s = 5 \times 8.718 = 43.59 $ cm

Area $ = \frac{1}{2} \times P \times a = \frac{1}{2} \times 43.59 \times 6 = 21.795 \times 6 = 130.77 $ cm²

So:
- $ P \approx 43.59 $
- $ a = 6 $
- Area $ \approx 130.77 $ cm²

---

5) Regular Pentagon, triangle with apothem = 6 cm

Shows a triangle from center to two vertices, with apothem = 6 cm.

Same as above.

Vertex angle = $ 360^\circ / 5 = 72^\circ $

Split into two right triangles: angle = 36°

Apothem = adjacent = 6 cm

Then:
$$
\tan(36^\circ) = \frac{s/2}{6} \Rightarrow s/2 = 6 \tan(36^\circ) \approx 6 \times 0.7265 = 4.359
\Rightarrow s = 8.718 \text{ cm}
$$

Same as before.

So:
- $ P = 5 \times 8.718 = 43.59 $ cm
- $ a = 6 $
- Area $ = \frac{1}{2} \times 43.59 \times 6 = 130.77 $ cm²

Again, same as Figure 4.

---

6) Regular Pentagon, side = 6 cm

Now, side length is given: $ s = 6 $ cm

Number of sides: $ n = 5 $

Perimeter: $ P = 5 \times 6 = 30 $ cm

Now find apothem:
$$
a = \frac{s}{2 \tan(36^\circ)} = \frac{6}{2 \times 0.7265} = \frac{6}{1.453} \approx 4.13 \text{ cm}
$$

Then area:
$$
A = \frac{1}{2} \times 30 \times 4.13 = 15 \times 4.13 = 61.95 \text{ cm}^2
$$

So:
- $ P = 30 $
- $ a \approx 4.13 $
- Area $ \approx 61.95 $ cm²

---

## Final Answers:

1) Octagon, side = 3 in

- $ P = 24 $ in
- $ a \approx 3.62 $ in
- Area $ \approx 43.48 $ in²

2) Octagon, apothem = 3 in

- $ P \approx 19.88 $ in
- $ a = 3 $ in
- Area $ \approx 29.82 $ in²

3) Octagon, apothem = 3 in (same as 2)

- $ P \approx 19.88 $ in
- $ a = 3 $ in
- Area $ \approx 29.82 $ in²

4) Pentagon, apothem = 6 cm

- $ P \approx 43.59 $ cm
- $ a = 6 $ cm
- Area $ \approx 130.77 $ cm²

5) Pentagon, apothem = 6 cm (same as 4)

- $ P \approx 43.59 $ cm
- $ a = 6 $ cm
- Area $ \approx 130.77 $ cm²

6) Pentagon, side = 6 cm

- $ P = 30 $ cm
- $ a \approx 4.13 $ cm
- Area $ \approx 61.95 $ cm²

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## Summary Table:

| # | Shape | Given | P | a | Area |
|---|-------------|--------------|-----------|-----------|--------------|
| 1 | Octagon | side = 3 in | 24 in | ~3.62 in | ~43.48 in² |
| 2 | Octagon | a = 3 in | ~19.88 in | 3 in | ~29.82 in² |
| 3 | Octagon | a = 3 in | ~19.88 in | 3 in | ~29.82 in² |
| 4 | Pentagon | a = 6 cm | ~43.59 cm | 6 cm | ~130.77 cm² |
| 5 | Pentagon | a = 6 cm | ~43.59 cm | 6 cm | ~130.77 cm² |
| 6 | Pentagon | side = 6 cm | 30 cm | ~4.13 cm | ~61.95 cm² |

---

Notes:

- For problems where only side is given, we use $ a = \frac{s}{2 \tan(180^\circ/n)} $
- When apothem is given, we use $ s = 2a \tan(180^\circ/n) $
- Always use $ \text{Area} = \frac{1}{2} P a $
- Values rounded to two decimal places

Let me know if you want exact expressions (e.g., using radicals).