Area of Composite Figures Worksheets - Math Monks - Free Printable

Let’s solve each problem one by one. We’ll calculate the area for each figure using the right formulas and given measurements.

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Problem 1: Rectangle with a triangle inside?

Wait — looking at the diagram, it shows a rectangle with a diagonal line from top-left to bottom-right corner, and a perpendicular height labeled 7.5 m from the center point to the base (10.9 m). But actually, this looks like they’re showing the apothem of a regular polygon? Wait — no, it's a rectangle? That doesn’t make sense.

Actually, re-examining: The shape is a rectangle, but there’s a dot in the middle with a perpendicular dropped to the base (7.5 m), and the base is 10.9 m. But that would be half the rectangle? No — perhaps it’s meant to be a regular polygon? Wait, problem #1 is just a rectangle? But then why show an apothem?

Hold on — let me check the title: “Areas of Regular Polygons and Composite Figures”. So maybe #1 is not a rectangle? Actually, looking again — it’s drawn as a rectangle, but with a central point and a perpendicular to the side — that suggests it might be intended as a regular polygon, but rectangles aren’t regular unless square. Hmm.

Wait — perhaps it’s a square? But sides are different: 10.9 m base, and height isn't given directly. But they give 7.5 m as the distance from center to side — that’s the apothem.

Ah! I think I see — even though it’s drawn like a rectangle, if it’s labeled as a regular polygon problem, and they give apothem and side length, we should treat it as a regular polygon where:

Area = (1/2) × perimeter × apothem

But for a rectangle? That formula only works for regular polygons. A rectangle is not regular unless it’s a square.

Wait — look at the drawing: It has four sides, opposite sides equal, angles 90° — so it’s a rectangle. But they give “7.5 m” as the perpendicular from center to base — which for a rectangle would be half the height. And base is 10.9 m.

So if the apothem (distance from center to side) is 7.5 m, and it’s perpendicular to the base of 10.9 m, then for a rectangle, the full height would be 2 × 7.5 = 15 m? Because the center is midway.

Then area of rectangle = base × height = 10.9 × 15 = ?

Let me compute:
10.9 × 15 = 10.9 × 10 + 10.9 × 5 = 109 + 54.5 = 163.5 m²

But wait — is that correct? Let me verify.

In a rectangle, the distance from center to a side is half the dimension perpendicular to that side. So if the apothem to the base (which is horizontal) is 7.5 m, that means the vertical half-height is 7.5 m, so full height is 15 m. Base is 10.9 m. So yes, area = 10.9 × 15 = 163.5 m².

Alternatively, if they meant it as a regular polygon, but it’s not regular — so probably it’s just a rectangle, and the 7.5 m is half the height.

I think that’s the intention. So Area = 163.5 m².

But let’s hold on — maybe I’m overcomplicating. Let’s look at other problems.

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Problem 2: Pentagon

It’s a regular pentagon? They give side length 13.2 yd, and apothem 3.5 yd.

For any regular polygon: Area = (1/2) × perimeter × apothem

Perimeter = 5 × side = 5 × 13.2 = 66 yd

Apothem = 3.5 yd

Area = (1/2) × 66 × 3.5 = 33 × 3.5

33 × 3 = 99, 33 × 0.5 = 16.5 → total 115.5 yd²

Okay.

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Problem 3: Hexagon

Regular hexagon, side 12 cm, apothem given as 5√3 cm.

Again, Area = (1/2) × perimeter × apothem

Perimeter = 6 × 12 = 72 cm

Apothem = 5√3 cm

Area = (1/2) × 72 × 5√3 = 36 × 5√3 = 180√3 cm²

We can leave it like that, or approximate? But since it’s exact value given, probably keep as 180√3.

But let’s confirm: In a regular hexagon, apothem = (s√3)/2, so for s=12, apothem = (12√3)/2 = 6√3, but here it’s given as 5√3 — that’s inconsistent.

Wait — problem says apothem is 5√3 cm, but for side 12 cm, actual apothem should be 6√3 ≈ 10.392, but 5√3 ≈ 8.66 — so either it’s not regular, or typo? But the problem says "regular polygons", so likely we should use the given values regardless.

So proceed with given: Area = (1/2)*perimeter*apothem = 0.5 * 72 * 5√3 = 180√3 cm²

Perhaps they want numerical value? But let’s see other problems.

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Problem 4: Triangle

Base 22 in, height 16 in (since the line from apex to base is perpendicular, labeled 16 in).

Area of triangle = (1/2) × base × height = 0.5 × 22 × 16

22 × 16 = 352, half is 176 in²

Easy.

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Problem 5: Composite figure — rectangle + semicircle

Rectangle: width 32 ft, height 18 ft? Wait — the arrow says 32 ft for the top, and left side is 18 ft. Then on the right, there’s a semicircle attached.

The semicircle is on the right end, so its diameter must be equal to the height of the rectangle, which is 18 ft. So radius = 9 ft.

Area = area of rectangle + area of semicircle

Rectangle: 32 × 18 = let’s compute: 30×18=540, 2×18=36, total 576 ft²

Semicircle: (1/2) π r² = 0.5 × π × 81 = 40.5π ft²

So total area = 576 + 40.5π ft²

If they want numerical, π≈3.14, 40.5×3.14=127.17, total ≈703.17, but probably leave in terms of π? Or check context.

Looking at other problems, some have radicals, some integers. Probably acceptable to leave as 576 + 40.5π, but let’s see.

Actually, in worksheet, often they expect numerical approximation. But since no instruction, I’ll compute both ways later. For now, note it.

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Problem 6: Composite — triangle + semicircle

Triangle on top, semicircle at bottom.

Given: triangle has two equal sides marked (so isosceles), base of triangle is same as diameter of semicircle, which is 9 mm? Wait — the dashed line is 9 mm, and it’s the base of the triangle and diameter of semicircle.

Also, the height of the whole figure is 12 mm, but that includes the triangle and the semicircle.

The semicircle has diameter 9 mm, so radius 4.5 mm, so height of semicircle is 4.5 mm.

Therefore, height of triangle = total height - height of semicircle = 12 - 4.5 = 7.5 mm

Now, area = area of triangle + area of semicircle

Triangle: base 9 mm, height 7.5 mm → area = 0.5 × 9 × 7.5 = 4.5 × 7.5

4.5 × 7 = 31.5, 4.5 × 0.5 = 2.25, total 33.75 mm²

Semicircle: (1/2) π r² = 0.5 × π × (4.5)^2 = 0.5 × π × 20.25 = 10.125π mm²

Total area = 33.75 + 10.125π mm²

Again, may need numerical.

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Problem 7: Rectangle + triangle

Rectangle: 10 km by 6 km? Left side 6 km, bottom 10 km.

On the right, there’s a triangle attached. The triangle has two sides marked equal, and one side is 4.8 km — probably the slant side.

The triangle is attached to the right side of the rectangle, which is 6 km high. So the base of the triangle is 6 km? But it’s drawn with the 4.8 km as the equal sides.

Looking: the triangle has vertices at top-right, bottom-right of rectangle, and a point to the right. The two equal sides are from that point to top and bottom of rectangle, each 4.8 km. So it’s an isosceles triangle with two sides 4.8 km, and base equal to the height of rectangle, which is 6 km.

Is that possible? Check triangle inequality: 4.8 + 4.8 > 6? 9.6 > 6 yes. 4.8 + 6 > 4.8 yes.

So base of triangle = 6 km, equal sides 4.8 km.

To find area of triangle, we need height. Since it’s isosceles, drop perpendicular from apex to base, bisects base.

So half-base = 3 km.

Then height h = √(4.8² - 3²) = √(23.04 - 9) = √14.04

Calculate: 14.04, sqrt? 3.747 approximately? But let’s keep exact.

4.8 = 48/10 = 24/5, 3=3

h = √[(24/5)^2 - 3^2] = √[576/25 - 9] = √[576/25 - 225/25] = √(351/25) = √351 / 5

√351 = √(9×39) = 3√39

So h = 3√39 / 5 km

Then area of triangle = (1/2) × base × height = 0.5 × 6 × (3√39 / 5) = 3 × (3√39 / 5) = 9√39 / 5 km²

This is messy. Perhaps I misinterpreted.

Another way: maybe the 4.8 km is the height? But it’s labeled on the side.

Look back: in the diagram, the 4.8 km is along the slanted side, and there are tick marks on the two slanted sides, indicating they are equal, and the base is the side of the rectangle, 6 km.

But calculating height gives irrational number, while other problems have nice numbers. Perhaps the 4.8 km is not the side, but something else.

Wait — in problem 7, the rectangle is 10 km long, 6 km high. On the right, a triangle is attached. The triangle has a side labeled 4.8 km, and it’s the hypotenuse? Or leg?

Perhaps the triangle is right-angled? But no right angle marked.

Another thought: maybe the 4.8 km is the length of the extension, but that doesn't help.

Perhaps the triangle is such that its base is not 6 km, but the rectangle's width is 10 km, and the triangle is attached to the end, so the base of the triangle is the same as the rectangle's height, 6 km, and the 4.8 km is the equal sides.

But then area is complicated.

Perhaps for area, we can use Heron's formula.

Sides of triangle: a=4.8, b=4.8, c=6

Semi-perimeter s = (4.8+4.8+6)/2 = 15.6/2 = 7.8

Area = √[s(s-a)(s-b)(s-c)] = √[7.8(7.8-4.8)(7.8-4.8)(7.8-6)] = √[7.8 × 3 × 3 × 1.8]

Compute inside: 7.8 × 3 = 23.4, times 3 = 70.2, times 1.8 = 126.36

√126.36 = ? 11.24 or something? 11.24^2 = 126.3376, close.

But 126.36 = 12636/100, sqrt = √12636 / 10

12636 ÷ 4 = 3159, so √12636 = 2√3159, not nice.

Perhaps the 4.8 km is the height of the triangle? But it's labeled on the side.

Let's look at the diagram description: "6 km" on left, "10 km" on bottom, "4.8 km" on the slanted side of the triangle on the right.

And the triangle has two sides with tick marks, so isosceles with legs 4.8 km, base 6 km.

But then area is not nice. Perhaps in the context, they expect us to use the given and calculate.

Maybe the 4.8 km is the distance from the rectangle to the apex, but that doesn't make sense.

Another idea: perhaps the triangle is right-angled at the rectangle's corner. But no indication.

Let's move to problem 8 and come back.

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Problem 8: Square + triangle

Square: 16 cm by 16 cm.

Below it, a triangle attached. The triangle has base same as square's side, 16 cm? But in diagram, the triangle is below, and it's divided into two parts by a vertical line, with labels 8.5 cm and 11.5 cm on the sides, and the base is not labeled, but the two sides are 8.5 cm and 11.5 cm, and it's isosceles? No, the tick marks are on the two sides of the triangle, but 8.5 and 11.5 are different, so not isosceles.

Look: the triangle has vertices at bottom-left, bottom-right of square, and a point below. The two sides from that point to the square's corners are labeled 8.5 cm and 11.5 cm, and there are tick marks on them, but they are different lengths, so probably the tick marks indicate something else, or perhaps it's a mistake.

In the diagram, it says "8.5 cm" and "11.5 cm" on the two sides of the triangle, and there are double tick marks on both, which usually means they are equal, but 8.5 ≠ 11.5, so contradiction.

Perhaps the 8.5 cm and 11.5 cm are not the sides, but segments.

Read carefully: "8.5 cm" is written near the left part of the triangle's base? No, in the text: "8.5 cm" and "11.5 cm" are on the sides of the triangle.

Perhaps the triangle is not isosceles, and the tick marks are for something else, but typically tick marks mean equal length.

Another possibility: the 8.5 cm and 11.5 cm are the lengths of the two segments when the triangle is divided, but the diagram shows a vertical line from the apex to the base, dividing the base into two parts, but the base is not labeled.

In problem 8, the square is 16 cm x 16 cm. Below it, a triangle with apex down. There is a vertical line from the apex to the base of the triangle (which is the bottom of the square), and this vertical line is labeled? No, the labels are "8.5 cm" on the left side of the triangle, "11.5 cm" on the right side, and there are double tick marks on both sides, which is confusing because 8.5 ≠ 11.5.

Perhaps the tick marks are on the base segments. Let me assume that the vertical line divides the base into two parts: left part 8.5 cm, right part 11.5 cm, but 8.5 + 11.5 = 20 cm, but the square's side is 16 cm, so the base of the triangle should be 16 cm, but 8.5+11.5=20>16, impossible.

Unless the base is not the entire bottom, but that doesn't make sense.

Perhaps the 8.5 cm and 11.5 cm are the lengths of the two sides of the triangle, and the base is 16 cm, and the tick marks are erroneous or for decoration.

That makes more sense. So triangle with sides 8.5 cm, 11.5 cm, and base 16 cm.

Then we can use Heron's formula.

Semi-perimeter s = (8.5 + 11.5 + 16)/2 = 36/2 = 18 cm

Area = √[s(s-a)(s-b)(s-c)] = √[18(18-8.5)(18-11.5)(18-16)] = √[18 × 9.5 × 6.5 × 2]

Compute step by step:

18 × 2 = 36

9.5 × 6.5 = (10-0.5)(6+0.5) = 10*6 +10*0.5 -0.5*6 -0.5*0.5 = 60 +5 -3 -0.25 = 61.75

Then 36 × 61.75

First, 36 × 60 = 2160, 36 × 1.75 = 36×1 + 36×0.75 = 36 + 27 = 63, so total 2160+63=2223

So area = √2223

Simplify: 2223 ÷ 3 = 741, 741 ÷ 3 = 247, 247 ÷ 13 = 19, so 2223 = 3×3×13×19 = 9×247

√2223 = 3√247, not nice.

This is getting messy. Perhaps I have misinterpreted several problems.

Let me go back to problem 1.

In problem 1, it's a rectangle with base 10.9 m, and from the center, a perpendicular to the base is 7.5 m. In a rectangle, the center is at intersection of diagonals, so the distance from center to a side is half the adjacent side.

So if distance to the base (length 10.9 m) is 7.5 m, that means the height of the rectangle is 2 × 7.5 = 15 m, as I had.

So area = 10.9 × 15 = 163.5 m²

Similarly, for problem 2, pentagon, area = (1/2) * perimeter * apothem = 0.5 * 5*13.2 * 3.5 = 0.5*66*3.5 = 33*3.5 = 115.5 yd²

For problem 3, hexagon, area = 0.5 * 6*12 * 5√3 = 0.5*72*5√3 = 36*5√3 = 180√3 cm²

For problem 4, triangle, 0.5*22*16 = 176 in²

For problem 5, rectangle 32 ft by 18 ft, area 576 ft², semicircle with diameter 18 ft, radius 9 ft, area (1/2)π(81) = 40.5π ft², total 576 + 40.5π ft²

For problem 6, triangle base 9 mm, height of triangle is total height minus radius of semicircle. Total height 12 mm, semicircle radius 4.5 mm (since diameter 9 mm), so triangle height = 12 - 4.5 = 7.5 mm, area triangle 0.5*9*7.5 = 33.75 mm², semicircle 0.5*π*(4.5)^2 = 0.5*π*20.25 = 10.125π mm², total 33.75 + 10.125π mm²

For problem 7, rectangle 10 km by 6 km, area 60 km². Triangle on right: isosceles with two sides 4.8 km, base 6 km. As calculated, area = √[7.8*3*3*1.8] = √[7.8*9*1.8] = √[126.36] = 11.24 km² approximately, but let's calculate exactly.

From earlier, s=7.8, s-a=3, s-b=3, s-c=1.8, so area = √(7.8 * 3 * 3 * 1.8) = √(7.8 * 9 * 1.8) = √(126.36)

126.36 = 12636/100, sqrt = √12636 / 10

12636 = 4 * 3159, 3159 ÷ 3 = 1053, 1053 ÷ 3 = 351, 351 ÷ 3 = 117, 117 ÷ 3 = 39, 39÷3=13, so 12636 = 4 * 3^5 * 13 = 4 * 243 * 13 = wait, 3^5=243, 243*13=3159, yes, so 12636 = 4 * 3159 = 4 * 3^5 * 13

So √12636 = 2 * 3^{2} * √(3*13) = 2*9*√39 = 18√39

So area = 18√39 / 10 = 9√39 / 5 km²

Then total area for problem 7 = rectangle + triangle = 60 + 9√39/5 km²

This is ugly, but perhaps correct.

For problem 8, square 16x16 = 256 cm². Triangle with sides 8.5 cm, 11.5 cm, and base 16 cm. s=18, area = √[18*9.5*6.5*2] = as before, √2223 cm²

2223 = 9*247, 247=13*19, so √2223 = 3√247 cm²

Total area = 256 + 3√247 cm²

This is not satisfactory. Perhaps in problem 8, the 8.5 cm and 11.5 cm are not the sides, but the segments of the base.

Let me reinterpret problem 8.

In the diagram, there is a square 16 cm x 16 cm. Below it, a triangle with apex down. There is a vertical line from the apex to the base (which is the bottom of the square), and this vertical line is the height of the triangle. The base of the triangle is divided into two parts by this vertical line: left part 8.5 cm, right part 11.5 cm. But 8.5 + 11.5 = 20 cm, but the square's side is 16 cm, so the base of the triangle should be 16 cm, but 20 > 16, impossible.

Unless the base is not the entire bottom, but that doesn't make sense.

Perhaps the 8.5 cm and 11.5 cm are the lengths of the two sides, and the base is 16 cm, and we have to live with it.

But let's check the sum: 8.5 + 11.5 = 20 > 16, ok, and |8.5-11.5|=3 < 16, so triangle inequality holds.

Area by Heron's formula is √[18(18-8.5)(18-11.5)(18-16)] = √[18*9.5*6.5*2]

Calculate numerically: 18*2=36, 9.5*6.5=61.75, 36*61.75=2223, sqrt(2223) = 47.15 approximately, since 47^2=2209, 48^2=2304, 47.1^2=2218.41, 47.2^2=2227.84, so ~47.15 cm²

Then total area = 256 + 47.15 = 303.15 cm², but not exact.

Perhaps the vertical line is the height, and 8.5 cm and 11.5 cm are the distances from the foot to the corners, but then the base would be 8.5 + 11.5 = 20 cm, but the square is 16 cm, so inconsistency.

Unless the triangle's base is 20 cm, but the square is 16 cm, so it extends beyond, but in the diagram, it's attached to the square, so base should be 16 cm.

I think there might be a mistake in my interpretation or in the problem.

Another idea for problem 8: perhaps the 8.5 cm and 11.5 cm are the lengths of the two segments of the height or something, but unlikely.

Let's look at the label: "8.5 cm" is written on the left side of the triangle, "11.5 cm" on the right side, and there are double tick marks on both, which typically means they are equal, but 8.5 ≠ 11.5, so perhaps it's a typo, and it's 8.5 and 8.5 or 11.5 and 11.5.

Or perhaps the 8.5 and 11.5 are not the sides, but the base segments.

Assume that the vertical line from apex to base divides the base into two parts: left 8.5 cm, right 11.5 cm, but then base = 8.5 + 11.5 = 20 cm, but the square is 16 cm, so the triangle's base is 20 cm, which is larger than the square's side, so it must be that the triangle is wider, but in the diagram, it's attached to the square, so probably not.

Perhaps the square is on top, and the triangle is below, and the base of the triangle is the same as the square's side, 16 cm, and the 8.5 cm and 11.5 cm are the lengths of the two sides, and we have to use that.

For the sake of progressing, I'll assume that for problem 7 and 8, we use the given values and calculate as is.

But let's try to see if there's a better way.

For problem 7, perhaps the 4.8 km is the height of the triangle, not the side.

In the diagram, if the triangle is attached to the right side of the rectangle, and the 4.8 km is the horizontal distance from the rectangle to the apex, then it would be the base of the triangle if it's right-angled, but not specified.

Assume that the triangle is right-angled at the bottom-right corner of the rectangle. Then the two legs are: vertical leg is the height of the rectangle, 6 km, horizontal leg is 4.8 km, then area of triangle = 0.5 * 6 * 4.8 = 14.4 km², and rectangle 10*6=60 km², total 74.4 km².

And the hypotenuse would be √(6^2 + 4.8^2) = √(36 + 23.04) = √59.04 = 7.684, not 4.8, but in the diagram, 4.8 is labeled on the slanted side, so if it's the hypotenuse, then it should be longer, but 4.8 is given, while legs would be shorter.

If 4.8 is the hypotenuse, and one leg is 6 km, then other leg = √(4.8^2 - 6^2) = √(23.04 - 36) = √(-12.96) impossible.

So not right-angled.

Perhaps the 4.8 km is the length of the median or something.

I think for the purpose of this, I'll use the isosceles triangle with sides 4.8, 4.8, base 6, and area 9√39/5 km² for the triangle, but that's approximately 9*6.245/5 = 56.205/5 = 11.241 km², so total 60 + 11.241 = 71.241 km², but let's keep exact.

Similarly for problem 8, area of triangle = √2223 cm² ≈ 47.15 cm², total 256 + 47.15 = 303.15 cm².

But perhaps in the worksheet, they expect numerical answers or simplified.

Let's list what we have:

1. 163.5 m²

2. 115.5 yd²

3. 180√3 cm² or approximately 311.76 cm², but since given as 5√3, probably leave as 180√3

4. 176 in²

5. 576 + 40.5π ft² or approximately 576 + 127.23 = 703.23 ft²

6. 33.75 + 10.125π mm² or approximately 33.75 + 31.81 = 65.56 mm²

7. 60 + 9√39/5 km² or approximately 60 + 11.24 = 71.24 km²

8. 256 + √2223 cm² or approximately 256 + 47.15 = 303.15 cm²

But this is not clean. Perhaps for problem 3, the apothem is given as 5√3, but for a regular hexagon with side 12, apothem should be (12*√3)/2 = 6√3, so perhaps it's a typo, and it's 6√3, then area = 0.5*72*6√3 = 216√3 cm².

Similarly, for problem 1, if it's a regular polygon, but it's not.

Another thought for problem 1: perhaps it's a rhombus or something, but drawn as rectangle.

I recall that in some contexts, for a rectangle, if they give "apothem" , it might be misnomer, but in this case, with the perpendicular from center to side, for a rectangle, it's half the other side.

So I think my initial calculation is correct.

Perhaps for problem 7, the 4.8 km is the height of the triangle, and the base is 6 km, then area = 0.5*6*4.8 = 14.4 km², total 60 + 14.4 = 74.4 km².

And the slanted sides would be √(3^2 + 4.8^2) = √(9 + 23.04) = √32.04 = 5.66, not 4.8, but in the diagram, 4.8 is labeled on the side, so if it's the height, it should be perpendicular, but it's labeled on the slanted side.

In the diagram, the 4.8 km is along the slanted side, so it's the side length.

I think I have to go with the calculations.

Perhaps for problem 8, the 8.5 cm and 11.5 cm are the lengths from the apex to the corners, and the base is 16 cm, and the vertical line is the height, but then we can find the height.

Let me denote the triangle with base BC = 16 cm, apex A. Let D be the foot of perpendicular from A to BC. Let BD = x, DC = 16-x.

Then AB = 8.5 cm, AC = 11.5 cm, or vice versa.

Assume AB = 8.5, AC = 11.5, BC = 16.

Then by Pythagoras, AD^2 + BD^2 = AB^2, AD^2 + DC^2 = AC^2

So AD^2 + x^2 = 8.5^2 = 72.25

AD^2 + (16-x)^2 = 11.5^2 = 132.25

Subtract the first from the second: [AD^2 + (16-x)^2] - [AD^2 + x^2] = 132.25 - 72.25

So (256 - 32x + x^2) - x^2 = 60

256 - 32x = 60

32x = 256 - 60 = 196

x = 196/32 = 49/8 = 6.125 cm

Then AD^2 = 72.25 - (6.125)^2 = 72.25 - 37.515625 = 34.734375

AD = √34.734375 = 5.894 cm approximately

Then area of triangle = 0.5 * 16 * 5.894 = 8 * 5.894 = 47.152 cm², same as before with Heron's formula.

So consistent.

So for problem 8, area of triangle is approximately 47.15 cm², total 256 + 47.15 = 303.15 cm².

But perhaps they want exact form.

Since the problem asks for area, and in worksheets, often they expect numerical values or simplified radicals, but here it's messy.

Perhaps in problem 7, the 4.8 km is the height, and the base is 6 km, and the 4.8 is mislabeled, but in the diagram, it's on the side.

I think for the sake of completing, I'll provide the answers as per calculations, and for those with π or radicals, leave them, or approximate if needed.

But let's see the user's request: "solve the problem accurately", and "verify your answer".

Perhaps for problem 3, since it's a regular hexagon, and apothem is given as 5√3, but for side 12, it should be 6√3, so maybe it's a different polygon, but it's labeled as hexagon.

Another idea: in problem 3, the 5√3 cm is not the apothem, but something else, but the diagram shows it as the perpendicular from center to side, so it is the apothem.

Perhaps the side is not 12 cm for the hexagon, but the diagram shows 12 cm on the top side.

I think I have to proceed.

Let me write the answers as:

1. 163.5 m²

2. 115.5 yd²

3. 180√3 cm²

4. 176 in²

5. 576 + 40.5π ft²

6. 33.75 + 10.125π mm²

7. 60 + \frac{9\sqrt{39}}{5} km² or approximately 71.24 km²

8. 256 + \sqrt{2223} cm² or approximately 303.15 cm²

But for the final answer, perhaps they want numerical values.

Maybe for problem 5 and 6, use π=3.14.

Let me calculate numerical values where possible.

For problem 3: 180 * 1.732 = 311.76 cm²

For problem 5: 40.5 * 3.14 = 127.17, so 576 + 127.17 = 703.17 ft²

For problem 6: 10.125 * 3.14 = 31.7925, so 33.75 + 31.7925 = 65.5425 mm²

For problem 7: √39 ≈ 6.245, so 9*6.245/5 = 56.205/5 = 11.241, so 60 + 11.241 = 71.241 km²

For problem 8: √2223 ≈ 47.15, so 256 + 47.15 = 303.15 cm²

But still not integer.

Perhaps in problem 1, the 7.5 m is the full height, not half. Let's check.

In problem 1, if the 7.5 m is the height of the rectangle, then area = 10.9 * 7.5 = 81.75 m², but then why show the perpendicular from center? If it's from center to base, and it's 7.5 m, for a rectangle, that would imply height is 15 m, as I had.

Unless the "center" is not the geometric center, but that doesn't make sense.

I think my initial interpretation is correct.

Perhaps for problem 2, the 3.5 yd is not the apothem, but the diagram shows it as perpendicular from center to side, so it is.

I recall that for a regular pentagon, area can be calculated as (1/4) * √(5(5+2√5)) * s^2, but here they give apothem, so better to use (1/2)*P*a.

So I think 115.5 yd² is correct.

Let's box the answers as per calculation.

For the final answer, since the user asked for each area, I'll list them.

But the response should have Final Answer with all areas.

Perhaps write:

After careful calculation, the areas are:

1. 163.5 m²

2. 115.5 yd²

3. 180\sqrt{3} cm²

4. 176 in²

5. 576 + 40.5\pi ft²

6. 33.75 + 10.125\pi mm²

7. 60 + \frac{9\sqrt{39}}{5} km²

8. 256 + \sqrt{2223} cm²

To make it neat, perhaps simplify or use decimals.

For problem 7, \frac{9\sqrt{39}}{5} = \frac{9}{5} \sqrt{39} , and \sqrt{39} is already simplified.

For problem 8, \sqrt{2223} = \sqrt{9*247} = 3\sqrt{247}, and 247=13*19, so 3\sqrt{247} cm².

So area = 256 + 3\sqrt{247} cm²

Similarly for others.

But for the worksheet, perhaps they expect numerical values.

Maybe in problem 6, the 12 mm is the height of the triangle only, not including the semicircle.

Let's read problem 6: "12 mm" with arrow from top to bottom, and "9 mm" for the diameter.

If 12 mm is the height of the triangle, then area of triangle = 0.5*9*12 = 54 mm², semicircle 0.5*π*(4.5)^2 = 10.125π mm², total 54 + 10.125π mm².

But in the diagram, the 12 mm arrow goes from the top of the triangle to the bottom of the semicircle, so it includes both.

In many such problems, the total height is given, so my initial calculation is correct.

I think I have to submit with the calculations.

So for the Final Answer, I'll put the areas as calculated.

Perhaps for problem 1, it's a parallelogram or something, but it's drawn as rectangle.

Another idea: in problem 1, the shape is a rectangle, and the 7.5 m is the distance from the center to the side, which is half the height, so height 15 m, area 10.9*15=163.5 m².

I think it's correct.

So I'll go with that.

Final Answer:

1. 163.5 m²
2. 115.5 yd²
3. 180\sqrt{3} cm²
4. 176 in²
5. 576 + 40.5\pi ft²
6. 33.75 + 10.125\pi mm²
7. 60 + \frac{9\sqrt{39}}{5} km²
8. 256 + 3\sqrt{247} cm²

To make it cleaner, perhaps write numerical approximations, but since the problem has exact values like 5√3, probably keep exact.

For problem 5 and 6, leave with π.

So I think this is as good as it gets.

Final Answer

1. \( 163.5 \text{m}^2 \)
2. \( 115.5 \text{yd}^2 \)
3. \( 180\sqrt{3} \text{cm}^2 \)
4. \( 176 \text{in}^2 \)
5. \( 576 + 40.5\pi \text{ft}^2 \)
6. \( 33.75 + 10.125\pi \text{mm}^2 \)
7. \( 60 + \frac{9\sqrt{39}}{5} \text{km}^2 \)
8. \( 256 + 3\sqrt{247} \text{cm}^2 \)