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Geometry and Trigonometry - Area and volume of 3D shapes

Grade 10IB

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

🔑Concepts

Prisms and Uniform Cross-sections: A prism is a 3D solid with a constant cross-section throughout its length. Imagine a 2D shape, like a triangle or hexagon, being stretched along a straight line. To find its volume, you simply multiply the area of this base shape by the height or length of the prism.

Cylinders as Circular Prisms: A cylinder consists of two parallel circular bases connected by a curved surface. If you were to 'unroll' the curved side of a cylinder, it would form a flat rectangle. The length of this rectangle corresponds to the circumference of the circle (2πr2\pi r), and its width is the height (hh) of the cylinder.

Pyramids and Cones: Unlike prisms, pyramids and cones taper to a single point called the apex. Visually, a pyramid has a polygon base (like a square or triangle) and flat triangular sides, while a cone has a circular base and a smooth curved side. Their volume is always 13\frac{1}{3} of a prism or cylinder with the same base and height.

The Relationship in Cones (Slant Height): In a right cone, the vertical height (hh), the radius of the base (rr), and the slant height (ll) form a right-angled triangle. This allows us to use the Pythagorean theorem, l=r2+h2l = \sqrt{r^2 + h^2}, to find missing dimensions needed for surface area or volume calculations.

Spheres and Hemispheres: A sphere is a perfectly round 3D object where every point on the surface is the same distance (rr) from the center. A hemisphere is exactly half of a sphere. When calculating the surface area of a solid hemisphere, you must remember to add the area of the circular flat face (πr2\pi r^2) to the curved half-surface (2πr22\pi r^2).

Surface Area vs. Volume: Volume measures the capacity or the amount of 3D space an object occupies, measured in cubic units like cm3cm^3. Surface area is the total area of all the outer faces of the object, measured in square units like cm2cm^2. For composite shapes, volume is additive, but surface area requires careful subtraction of the faces that are touching or hidden.

Units and Conversions: Always ensure dimensions are in the same units before calculating. Remember that for area, 1 m2=10,000 cm21\text{ m}^2 = 10,000\text{ cm}^2 (100×100100 \times 100), and for volume, 1 m3=1,000,000 cm31\text{ m}^3 = 1,000,000\text{ cm}^3 (100×100×100100 \times 100 \times 100).

📐Formulae

Volume of a Prism: V=Abase×hV = A_{base} \times h

Volume of a Cylinder: V=πr2hV = \pi r^2 h

Total Surface Area of a Cylinder: A=2πr2+2πrhA = 2\pi r^2 + 2\pi rh

Volume of a Sphere: V=43πr3V = \frac{4}{3} \pi r^3

Surface Area of a Sphere: A=4πr2A = 4\pi r^2

Volume of a Cone: V=13πr2hV = \frac{1}{3} \pi r^2 h

Curved Surface Area of a Cone: Acurved=πrlA_{curved} = \pi r l (where ll is slant height)

Volume of a Pyramid: V=13Abase×hV = \frac{1}{3} A_{base} \times h

💡Examples

Problem 1:

A solid metal cone has a base radius of 5 cm5\text{ cm} and a vertical height of 12 cm12\text{ cm}. Calculate (a) the slant height of the cone and (b) the total surface area of the cone, leaving your answer in terms of π\pi.

Solution:

Step 1: Find the slant height (ll) using Pythagoras' theorem. l=r2+h2l = \sqrt{r^2 + h^2} l=52+122=25+144=169=13 cml = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}

Step 2: Use the total surface area formula, which is the sum of the base area and the curved surface area. A=πr2+πrlA = \pi r^2 + \pi r l A=π(52)+π(5)(13)A = \pi (5^2) + \pi (5)(13) A=25π+65π=90π cm2A = 25\pi + 65\pi = 90\pi\text{ cm}^2

Explanation:

We first identified the relationship between the height, radius, and slant height to find ll. Then, we applied the surface area formula, keeping the units consistent and the value in terms of π\pi as requested.

Problem 2:

A hemisphere has a volume of 486π cm3486\pi\text{ cm}^3. Find its radius.

Solution:

Step 1: Set up the volume formula for a hemisphere. Since a sphere's volume is 43πr3\frac{4}{3}\pi r^3, a hemisphere is half of that. V=23πr3V = \frac{2}{3} \pi r^3

Step 2: Substitute the known volume into the equation. 486π=23πr3486\pi = \frac{2}{3} \pi r^3

Step 3: Divide both sides by π\pi and solve for r3r^3. 486=23r3486 = \frac{2}{3} r^3 r3=486×32=729r^3 = 486 \times \frac{3}{2} = 729

Step 4: Take the cube root of both sides. r=7293=9 cmr = \sqrt[3]{729} = 9\text{ cm}

Explanation:

To find the radius from the volume, we set the given value equal to the hemisphere volume formula. We simplified the algebraic equation by canceling π\pi and then isolated rr by taking the cube root.