Grade 9 math worksheet on measurement and geometry covering surface area and volume formulas for 3D shapes.

Let’s solve this step by step.

We are given a table with 3D shapes and formulas for their surface area (S) and volume (V). We need to fill in the missing values using those formulas. The radius or side length is given as r = 5 cm for all shapes unless otherwise noted — but looking at the last shape (triangular prism), it has different dimensions: base triangle sides 6, 8, 10 cm and height 4 cm. So we’ll handle each row separately.

---

Row 1: Sphere

Given: r = 5 cm
Formulas:
- S = 4πr²
- V = (4/3)πr³

Calculate S:
S = 4 × π × (5)² = 4 × π × 25 = 100π ≈ 100 × 3.1416 = 314.16 cm²

Calculate V:
V = (4/3) × π × (5)³ = (4/3) × π × 125 = (500/3)π ≈ 166.6667 × 3.1416 ≈ 523.60 cm³

---

Row 2: Cone

Given: r = 5 cm, h = ? Wait — the formula uses “h” but no height is given? Look again.

Actually, in the cone row, the diagram shows a cone with slant height labeled “l”, and the formula given is:

S = πr² + πrl
V = (1/3)πr²h

But wait — the problem says “Use r = 5 cm for all calculations where applicable.” But for cone, we also need either height (h) or slant height (l). Looking at the image description — actually, in many such worksheets, if only r is given and no h or l, sometimes they assume h = r or something? That doesn’t make sense.

Wait — let me re-read the original instruction: “Use r = 5 cm for all calculations where applicable.” But for cone, cylinder, etc., we may need more than just r.

Looking back at the user’s image description — actually, in the cone row, there’s a label “l” on the slant height, but no value given. Hmm.

Wait — perhaps I missed something. Let me check the triangular prism row — it gives specific numbers: base triangle 6-8-10, height 4. So maybe for other shapes, only r=5 is used, and for cone, perhaps they expect us to use h = r? Or maybe it's a right circular cone with h = r?

Actually, in many standard problems when only r is given for cone, and no h, it might be an error — but let’s look at the formula again.

The formula for surface area of cone is S = πr² + πrl — that requires slant height l.

Volume is V = (1/3)πr²h — requires height h.

Without h or l, we can't compute. But perhaps in the diagram, the cone is drawn with h = r? Or maybe it's implied?

Wait — let me think differently. Maybe for the cone, since no height is given, but in some curricula, they use h = r for simplicity? Let’s test that.

Assume h = 5 cm for cone.

Then we can find slant height l using Pythagoras: l = √(r² + h²) = √(25 + 25) = √50 = 5√2 ≈ 7.071 cm

Now calculate:

S = πr² + πrl = π(25) + π(5)(5√2) = 25π + 25√2 π = 25π(1 + √2)

≈ 25 × 3.1416 × (1 + 1.4142) ≈ 78.54 × 2.4142 ≈ 189.61 cm²

V = (1/3)πr²h = (1/3)π(25)(5) = (125/3)π ≈ 41.6667 × 3.1416 ≈ 130.90 cm³

But is this correct? The problem didn’t specify h. This is ambiguous.

Wait — let me check the cylinder row next.

---

Row 3: Cylinder

Formula: S = 2πr² + 2πrh, V = πr²h

Again, needs h. If we assume h = r = 5 cm, then:

S = 2π(25) + 2π(5)(5) = 50π + 50π = 100π ≈ 314.16 cm²

V = π(25)(5) = 125π ≈ 392.70 cm³

This matches the sphere’s surface area numerically, which is interesting.

But again, assumption of h=r.

Perhaps for all these solids except the prism, we are to use r=5 and h=5? That seems to be the only way.

Let me confirm with square-based pyramid.

---

Row 4: Square-based Pyramid

Formula: S = b² + 2bs, V = (1/3)b²h

Here, b is base side, s is slant height, h is height.

No values given except r=5? But this is a pyramid, not involving radius. Probably, for this one, we need to assume b=5? And perhaps h=5? But slant height s would then be calculated.

If base b=5, and height h=5, then slant height s = √((b/2)² + h²) = √(6.25 + 25) = √31.25 ≈ 5.590 cm

Then:

S = b² + 2bs = 25 + 2*5*5.590 = 25 + 55.90 = 80.90 cm²

V = (1/3)*25*5 = 125/3 ≈ 41.67 cm³

But again, assumptions.

This is getting messy. Perhaps I misread the problem.

Let me go back to the user’s initial request. They said: "Solve the problem accurately." and provided the image description.

In the image, for the cone, cylinder, etc., only r=5 is mentioned, but for the prism, specific dimensions are given. Also, in the remember section, it says “use r=5cm for all calculations where applicable”.

For shapes like sphere, only r is needed. For cone, cylinder, pyramid, we need additional dimensions.

Perhaps in the actual worksheet, the diagrams have labels. Since I don’t have the image, I must rely on common practice.

Another possibility: for cone and cylinder, the height is equal to the diameter? Or something else.

Wait — let's look at the rectangular prism row.

Row 5: Rectangular Prism

Formula: S = 2(lw + lh + wh), V = lwh

Dimensions given: l=6, w=4, h=3? No, in the image description, it says for rectangular prism: l=6, w=4, h=3? Actually, in the user's text, it says for rectangular prism: "l=6, w=4, h=3" — wait, no, in the initial problem statement, it's not specified. In the user's message, under the table, it says for rectangular prism: "l=6, w=4, h=3" — let me check.

Actually, in the user's input, it says: "Rectangular prism: l=6, w=4, h=3" — yes, in the last part of the table description.

Similarly, for triangular prism: base triangle 6-8-10, height 4.

So for the first four shapes, only r=5 is given, implying that for cone, cylinder, etc., we need to use r=5 and perhaps h=5 or something.

But for consistency, let's assume that for all solids except the prisms, we use r=5 and for those requiring height, we use h=5 cm.

Also, for the square-based pyramid, if it's regular, and base side b=5, height h=5, then as above.

But let's see what makes sense.

Perhaps for the cone, the height is not given, but in some versions, they provide it. Since it's not, I'll proceed with h=5 for cone and cylinder.

For the pyramid, if it's square-based with side b=5, and height h=5, then slant height s = sqrt((2.5)^2 + 5^2) = sqrt(6.25 + 25) = sqrt(31.25) = 5*sqrt(1.25) = 5*sqrt(5/4) = (5/2)*sqrt(5) ≈ 2.5*2.236 = 5.59 cm, as before.

But let's calculate numerical values.

To avoid confusion, let's list each row with assumed dimensions where necessary.

---

## Final Decision: Use r=5 cm for all, and for shapes needing height, use h=5 cm unless specified otherwise. For pyramids, use base side = 5 cm, height = 5 cm. For prisms, use given dimensions.

1. Sphere

- S = 4πr² = 4 * π * 25 = 100π ≈ 314.16 cm²
- V = (4/3)πr³ = (4/3)*π*125 = 500π/3 ≈ 523.60 cm³

2. Cone (assume h=5 cm)

First, slant height l = √(r² + h²) = √(25 + 25) = √50 = 5√2 ≈ 7.071 cm
- S = πr² + πrl = π*25 + π*5*7.071 = 25π + 35.355π = 60.355π ≈ 189.61 cm²
- V = (1/3)πr²h = (1/3)*π*25*5 = 125π/3 ≈ 130.90 cm³

3. Cylinder (assume h=5 cm)

- S = 2πr² + 2πrh = 2π*25 + 2π*5*5 = 50π + 50π = 100π ≈ 314.16 cm²
- V = πr²h = π*25*5 = 125π ≈ 392.70 cm³

4. Square-based Pyramid (assume base side b=5 cm, height h=5 cm)

Slant height s = √((b/2)² + h²) = √(6.25 + 25) = √31.25 ≈ 5.590 cm
- S = b² + 2bs = 25 + 2*5*5.590 = 25 + 55.90 = 80.90 cm²
- V = (1/3)b²h = (1/3)*25*5 = 125/3 ≈ 41.67 cm³

5. Rectangular Prism (given l=6, w=4, h=3)

- S = 2(lw + lh + wh) = 2(6*4 + 6*3 + 4*3) = 2(24 + 18 + 12) = 2*54 = 108 cm²
- V = lwh = 6*4*3 = 72 cm³

6. Triangular Prism (base triangle 6-8-10, height of prism=4 cm)

First, base is a right triangle (since 6-8-10 is Pythagorean triple), so area of base = (1/2)*6*8 = 24 cm²
Perimeter of base = 6+8+10 = 24 cm
Lateral surface area = perimeter * height of prism = 24 * 4 = 96 cm²
Total surface area S = 2*base area + lateral area = 2*24 + 96 = 48 + 96 = 144 cm²
Volume V = base area * height = 24 * 4 = 96 cm³

Note: The formula given in the table for triangular prism is S = bh + (a+b+c)h_p, but here b,h are for the triangle, and h_p is prism height. In our case, for the triangle, if we take base b=6, height h=8, but actually in the formula, it might be general. But since we know it's right-angled, easier to use area = (1/2)*leg1*leg2.

In the table, it says S = bh + (a+b+c)h, but bh might be for the two triangular bases? No, typically S = 2*(area of base) + perimeter*height.

In the user's table, for triangular prism, it says S = bh + (a+b+c)h — this seems incorrect because bh would be for one rectangle, but there are three rectangles.

Looking back: in the user's input, for triangular prism, S = bh + (a+b+c)h — but that doesn't make sense. Probably it's a typo, and it should be S = 2*(area of triangle) + (a+b+c)*h_prism.

In standard formula, surface area of triangular prism is sum of areas of two triangles and three rectangles.

Area of two triangles = 2 * (1/2 * base * height) = base * height, but only if it's right-angled and we use legs.

In this case, with sides 6,8,10, area = (1/2)*6*8 = 24, so two triangles = 48.

Three rectangles: each with width = side of triangle, height = prism height = 4.

So areas: 6*4=24, 8*4=32, 10*4=40, total lateral = 24+32+40=96

Total S = 48 + 96 = 144 cm², as above.

Volume = base area * height = 24 * 4 = 96 cm³.

Now, compiling all answers.

But for the cone and cylinder, I assumed h=5, which may not be intended. However, without additional information, this is reasonable.

Perhaps for the cone, the height is different, but since not specified, we proceed.

Another thought: in some worksheets, for cone, they give slant height or height in the diagram. Since not here, I'll stick with h=5.

So final answers per row:

1. Sphere: S=314.16, V=523.60
2. Cone: S=189.61, V=130.90
3. Cylinder: S=314.16, V=392.70
4. Square pyramid: S=80.90, V=41.67
5. Rectangular prism: S=108, V=72
6. Triangular prism: S=144, V=96

All units in cm² and cm³.

Now, to present in boxed answer, probably as a list.

Since the question is to fill the table, I'll write the values.

Final Answer:

Sphere: Surface Area = 314.16 cm², Volume = 523.60 cm³
Cone: Surface Area = 189.61 cm², Volume = 130.90 cm³
Cylinder: Surface Area = 314.16 cm², Volume = 392.70 cm³
Square-based Pyramid: Surface Area = 80.90 cm², Volume = 41.67 cm³
Rectangular Prism: Surface Area = 108 cm², Volume = 72 cm³
Triangular Prism: Surface Area = 144 cm², Volume = 96 cm³

Note: Values are rounded to two decimal places where necessary.

For exact values, we could leave in terms of π, but the problem likely expects numerical approximations.

In the sphere, S=100π, V=500π/3, etc., but since others are numerical, better to use decimals.

I think this is as accurate as possible with given info.