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Geometry and Trigonometry - Volume and capacity of additional shapes-extended

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The volume of any pyramid or cone is exactly one-third of the volume of a prism or cylinder with the same base area and vertical height. This relationship is defined by the formula V=13×Abase×hV = \frac{1}{3} \times A_{base} \times h.

Diagram of a pyramid showing vertical height h and base area A.
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For spheres and hemispheres, the volume depends solely on the radius rr. A hemisphere is exactly half of a sphere. Capacity refers to the volume of liquid a container can hold, often requiring conversion: 1000 cm3=1 liter1000 \text{ cm}^3 = 1 \text{ liter}.

Diagram of a hemisphere with radius r.
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In right cones, the vertical height hh, radius rr, and slant height ll form a right-angled triangle. Applying the Pythagorean theorem allows us to find hh if only rr and ll are given: h=l2−r2h = \sqrt{l^2 - r^2}.

Right-angled triangle within a cone showing h, r, and l.
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Complex or 'composite' shapes are solved by decomposing the object into simpler solids (e.g., a cylinder topped with a hemisphere) and summing their individual volumes.

📐Formulae

Vpyramid=13AbasehV_{pyramid} = \frac{1}{3} A_{base} h

Vcone=13πr2hV_{cone} = \frac{1}{3} \pi r^2 h

Vsphere=43πr3V_{sphere} = \frac{4}{3} \pi r^3

Vhemisphere=23πr3V_{hemisphere} = \frac{2}{3} \pi r^3

l2=r2+h2 (for cones where l is slant height)l^2 = r^2 + h^2 \text{ (for cones where } l \text{ is slant height)}

1 L=1000 cm31 \text{ L} = 1000 \text{ cm}^3

💡Examples

Problem 1:

A cone has a radius of 5 cm5 \text{ cm} and a slant height of 13 cm13 \text{ cm}. Calculate its volume in terms of π\pi.

Solution:

  1. Find the vertical height hh using Pythagoras: h2+52=132h^2 + 5^2 = 13^2 h2+25=169h^2 + 25 = 169 h2=144  ⟹  h=12 cmh^2 = 144 \implies h = 12 \text{ cm}

  2. Use the volume formula: V=13π(5)2(12)V = \frac{1}{3} \pi (5)^2 (12) V=13π(25)(12)V = \frac{1}{3} \pi (25) (12) V=100π cm3V = 100\pi \text{ cm}^3

Explanation:

First, the vertical height is calculated using the relationship between the radius, height, and slant height. Then, the volume formula for a cone is applied.

Problem 2:

A hemispherical bowl has a diameter of 20 cm20 \text{ cm}. Find the capacity of the bowl in liters (rounded to 2 decimal places).

Solution:

  1. Radius r=202=10 cmr = \frac{20}{2} = 10 \text{ cm}.

  2. Calculate volume of the hemisphere: V=23π(10)3V = \frac{2}{3} \pi (10)^3 V=23π(1000)V = \frac{2}{3} \pi (1000) V≈2094.40 cm3V \approx 2094.40 \text{ cm}^3

  3. Convert to liters: Capacity=2094.401000≈2.09 L\text{Capacity} = \frac{2094.40}{1000} \approx 2.09 \text{ L}

Explanation:

The radius is half the diameter. The volume of a hemisphere is calculated, and then the result is divided by 10001000 to convert from cubic centimeters to liters.

Problem 3:

Calculate the volume of a square-based pyramid with a base side length of 6 m6 \text{ m} and a vertical height of 10 m10 \text{ m}.

Solution:

  1. Calculate the base area AbaseA_{base}: Abase=6×6=36 m2A_{base} = 6 \times 6 = 36 \text{ m}^2

  2. Calculate the volume: V=13×36×10V = \frac{1}{3} \times 36 \times 10 V=12×10V = 12 \times 10 V=120 m3V = 120 \text{ m}^3

Explanation:

The area of the square base is found first, then multiplied by the height and divided by 33 according to the pyramid volume formula.

Problem 4:

A composite solid consists of a cylinder of radius 3 cm3 \text{ cm} and height 10 cm10 \text{ cm} with a cone of the same radius and a vertical height of 4 cm4 \text{ cm} attached to the top. Calculate the total volume of the solid.

A cylinder topped with a cone.

Solution:

Vcylinder=πr2h1=π×32×10=90π cm3V_{cylinder} = \pi r^2 h_1 = \pi \times 3^2 \times 10 = 90\pi \text{ cm}^3 Vcone=13πr2h2=13π×32×4=12π cm3V_{cone} = \frac{1}{3} \pi r^2 h_2 = \frac{1}{3} \pi \times 3^2 \times 4 = 12\pi \text{ cm}^3 Vtotal=90π+12π=102π≈320.44 cm3V_{total} = 90\pi + 12\pi = 102\pi \approx 320.44 \text{ cm}^3

Explanation:

First, calculate the volume of the cylindrical base. Then, calculate the volume of the conical top. The total volume is the sum of these two parts.

Problem 5:

A glass sphere has a radius of 6 cm6 \text{ cm}. It is melted down to cast several small solid cones, each with a radius of 2 cm2 \text{ cm} and a height of 3 cm3 \text{ cm}. How many such cones can be made?

A large sphere being converted into small cones.

Solution:

Vsphere=43π(6)3=43π×216=288π cm3V_{sphere} = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi \times 216 = 288\pi \text{ cm}^3 Vcone=13π(2)2(3)=4π cm3V_{cone} = \frac{1}{3} \pi (2)^2 (3) = 4\pi \text{ cm}^3 Number of cones=VsphereVcone=288π4π=72\text{Number of cones} = \frac{V_{sphere}}{V_{cone}} = \frac{288\pi}{4\pi} = 72

Explanation:

Find the total volume of glass available from the sphere. Then find the volume required for one cone. Divide the total volume by the volume of one cone to find the quantity.