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Mensuration - Surface Area and Volume of 3D Solids

Grade 12A Level

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

🔑Concepts

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The Volume and Surface Area of a Cylinder depend on the radius rr and the perpendicular height hh. The Total Surface Area includes both the curved face (a rectangle when unrolled) and the two circular ends.

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A Right Circular Cone is defined by its radius rr, vertical height hh, and slant height ll. These three lengths form a right-angled triangle, such that l2=r2+h2l^2 = r^2 + h^2.

Cone with radius r, height h, and slant height l
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For mathematically similar solids, the ratio of their volumes is the cube of the ratio of their corresponding lengths: V1V2=(L1L2)3\frac{V_1}{V_2} = \left(\frac{L_1}{L_2}\right)^3

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A composite solid is formed by joining two or more basic 3D shapes. To find the total volume, sum the volumes of the individual parts. To find the surface area, ensure you subtract the 'hidden' faces where the shapes meet.

📐Formulae

Cuboid: V=l×w×hV = l \times w \times h; TSA=2(lw+wh+lh)TSA = 2(lw + wh + lh)

Cylinder: V=πr2hV = \pi r^2 h; CSA=2πrhCSA = 2\pi rh; TSA=2πr2+2πrhTSA = 2\pi r^2 + 2\pi rh

Cone: V=13πr2hV = \frac{1}{3}\pi r^2 h; CSA=πrlCSA = \pi rl (where ll is slant height r2+h2\sqrt{r^2 + h^2})

Sphere: V=43πr3V = \frac{4}{3}\pi r^3; SurfaceArea=4πr2Surface Area = 4\pi r^2

Pyramid: V=13×Base Area×Perpendicular HeightV = \frac{1}{3} \times \text{Base Area} \times \text{Perpendicular Height}

Similar Solids: V1V2=(l1l2)3\frac{V_1}{V_2} = (\frac{l_1}{l_2})^3 and A1A2=(l1l2)2\frac{A_1}{A_2} = (\frac{l_1}{l_2})^2

💡Examples

Problem 1:

A solid toy is made of a hemisphere of radius 3 cm topped by a cone of the same radius and a height of 4 cm. Calculate the total volume of the toy.

Solution:

Vtotal=Vhemisphere+Vcone=(23πr3)+(13πr2h)=(23π×33)+(13π×32×4)=18π+12π=30π≈94.25 cm3V_{total} = V_{hemisphere} + V_{cone} = (\frac{2}{3} \pi r^3) + (\frac{1}{3} \pi r^2 h) = (\frac{2}{3} \pi \times 3^3) + (\frac{1}{3} \pi \times 3^2 \times 4) = 18\pi + 12\pi = 30\pi \approx 94.25 \text{ cm}^3

Explanation:

To find the volume of a composite solid, calculate the volume of each component part separately and sum them. Note that the volume of a hemisphere is half that of a sphere.

Problem 2:

Two mathematically similar cylinders have heights of 5 cm and 10 cm. If the smaller cylinder has a surface area of 40 cm², find the surface area of the larger cylinder.

Solution:

k=h2h1=105=2k = \frac{h_2}{h_1} = \frac{10}{5} = 2. Therefore, Area Ratio =k2=22=4= k^2 = 2^2 = 4. A2=A1×4=40×4=160 cm2A_2 = A_1 \times 4 = 40 \times 4 = 160 \text{ cm}^2

Explanation:

When objects are similar, the ratio of their areas is the square of the ratio of their corresponding linear dimensions (the scale factor kk).

Problem 3:

A cone has a radius of 5 cm and a perpendicular height of 12 cm. Find its curved surface area.

Solution:

l=r2+h2=52+122=25+144=13 cml = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13 \text{ cm}. CSA=πrl=π×5×13=65π≈204.2 cm2CSA = \pi r l = \pi \times 5 \times 13 = 65\pi \approx 204.2 \text{ cm}^2

Explanation:

To find the Curved Surface Area of a cone, you must first find the slant height (ll) using Pythagoras' theorem with the radius (rr) and the vertical height (hh).

Problem 4:

A spherical ball of radius 66 cm is melted down and recast into a cylinder with a radius of 44 cm. Calculate the height hh of the cylinder.

Recasting a sphere into a cylinder

Solution:

Vsphere=43π(6)3=43π(216)=288π cm3V_{sphere} = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = 288\pi \text{ cm}^3 Since the volume remains constant: Vcylinder=πr2h=π(4)2h=16πhV_{cylinder} = \pi r^2 h = \pi (4)^2 h = 16\pi h Set the volumes equal: 16πh=288π16\pi h = 288\pi h=28816=18 cmh = \frac{288}{16} = 18 \text{ cm}

Explanation:

Calculate the volume of the sphere first using V=43πr3V = \frac{4}{3}\pi r^3. Since the material is recast, the volume of the cylinder must be equal to the volume of the sphere. Solve for hh in the cylinder volume formula.

Problem 5:

A hollow pipe is 2020 cm long. The external radius is 55 cm and the internal radius is 33 cm. Calculate the volume of the material used to make the pipe.

Cross-section of a hollow pipe showing inner and outer radii

Solution:

Vexternal=πR2h=π(5)2(20)=500πV_{external} = \pi R^2 h = \pi (5)^2 (20) = 500\pi Vinternal=πr2h=π(3)2(20)=180πV_{internal} = \pi r^2 h = \pi (3)^2 (20) = 180\pi Vmaterial=Vexternal−VinternalV_{material} = V_{external} - V_{internal} Vmaterial=500π−180π=320π≈1005.31 cm3V_{material} = 500\pi - 180\pi = 320\pi \approx 1005.31 \text{ cm}^3

Explanation:

To find the volume of a hollow cylinder, subtract the volume of the inner empty cylinder from the volume of the outer cylinder.