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Geometry and Measurement - Surface Area and Volume of Prisms, Pyramids, Cones, Spheres, and Compound Solids

Grade 8IB

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

🔑Concepts

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A prism is a 3D solid with a constant cross-section. To find the volume, calculate the area of this cross-section and multiply it by the length or height. For a rectangular prism, the surface area is the sum of the areas of its six rectangular faces.

Rectangular prism showing length l, width w, and height h.
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A pyramid's volume is exactly one-third of the volume of a prism with the same base area and height. The slant height (ll) is used to calculate the surface area of the triangular faces, while the vertical height (hh) is used for the volume.

Square-based pyramid showing the vertical height h from the apex to the center of the base.
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For a cone, the slant height ll is the distance from the apex to any point on the circumference of the base. It forms a right-angled triangle with the vertical height hh and radius rr, satisfying l2=r2+h2l^2 = r^2 + h^2.

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Compound solids are formed by joining two or more basic 3D shapes. To find the total volume, add the volumes of the individual components. To find the total surface area, add the external surface areas, ensuring any overlapping faces are subtracted.

📐Formulae

Vprism=Base Area×hV_{\text{prism}} = \text{Base Area} \times h

TSAcylinder=2πr2+2πrhTSA_{\text{cylinder}} = 2\pi r^2 + 2\pi rh

Vcylinder=πr2hV_{\text{cylinder}} = \pi r^2 h

Vpyramid=13×Base Area×hV_{\text{pyramid}} = \frac{1}{3} \times \text{Base Area} \times h

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

TSAcone=πr2+πrl (where l=slant height)TSA_{\text{cone}} = \pi r^2 + \pi rl \text{ (where } l = \text{slant height)}

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

SAsphere=4πr2SA_{\text{sphere}} = 4\pi r^2

💡Examples

Problem 1:

Calculate the volume of a cone with a radius of 33 cm and a vertical height of 77 cm. Use π≈3.142\pi \approx 3.142.

Solution:

V=13πr2hV = \frac{1}{3} \pi r^2 h V=13×3.142×32×7V = \frac{1}{3} \times 3.142 \times 3^2 \times 7 V=13×3.142×9×7V = \frac{1}{3} \times 3.142 \times 9 \times 7 V=3.142×3×7=65.982 cm3V = 3.142 \times 3 \times 7 = 65.982 \text{ cm}^3

Explanation:

Substitute the given radius r=3r = 3 and height h=7h = 7 into the volume formula for a cone.

Problem 2:

A sphere has a surface area of 616 cm2616 \text{ cm}^2. Find its radius. (Take π=227\pi = \frac{22}{7})

Solution:

SA=4πr2SA = 4\pi r^2 616=4×227×r2616 = 4 \times \frac{22}{7} \times r^2 616=887×r2616 = \frac{88}{7} \times r^2 r2=616×788r^2 = \frac{616 \times 7}{88} r2=7×7=49r^2 = 7 \times 7 = 49 r=49=7 cmr = \sqrt{49} = 7 \text{ cm}

Explanation:

We use the surface area formula for a sphere and solve for the unknown variable rr by rearranging the equation.

Problem 3:

A compound solid consists of a cylinder of radius 22 cm and height 55 cm, topped with a hemisphere of the same radius. Calculate the total volume of the solid.

Solution:

Vcylinder=πr2h=π×22×5=20πV_{\text{cylinder}} = \pi r^2 h = \pi \times 2^2 \times 5 = 20\pi Vhemisphere=12×43πr3=23π×23=163πV_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi \times 2^3 = \frac{16}{3}\pi Vtotal=20π+163π=60π+16π3=76π3≈79.59 cm3V_{\text{total}} = 20\pi + \frac{16}{3}\pi = \frac{60\pi + 16\pi}{3} = \frac{76\pi}{3} \approx 79.59 \text{ cm}^3

Explanation:

The total volume is the sum of the volume of the cylinder and the volume of the hemisphere. A hemisphere is half of a sphere.

Problem 4:

A square-based pyramid has a base side length of 1010 cm and a slant height of 1313 cm. Calculate the total surface area of the pyramid.

Square pyramid with base 10cm and slant height 13cm indicated.

Solution:

Area of base=10×10=100 cm2\text{Area of base} = 10 \times 10 = 100 \text{ cm}^2 Area of one triangular face=12×base×slant height=12×10×13=65 cm2\text{Area of one triangular face} = \frac{1}{2} \times \text{base} \times \text{slant height} = \frac{1}{2} \times 10 \times 13 = 65 \text{ cm}^2 Total Surface Area=Base Area+4×(Area of triangle)\text{Total Surface Area} = \text{Base Area} + 4 \times (\text{Area of triangle}) TSA=100+4×65=100+260=360 cm2TSA = 100 + 4 \times 65 = 100 + 260 = 360 \text{ cm}^2

Explanation:

The total surface area consists of the square base and four identical isosceles triangles. We use the slant height for the triangle area, not the vertical height.

Problem 5:

A cylindrical tank has a radius of 33 m and a height of 1010 m. It is half-filled with water. Calculate the volume of the water in terms of π\pi.

Cylinder with radius 3m and height 10m, showing a water line halfway up.

Solution:

Volume of Cylinder=πr2h\text{Volume of Cylinder} = \pi r^2 h V=π×32×10=π×9×10=90π m3V = \pi \times 3^2 \times 10 = \pi \times 9 \times 10 = 90\pi \text{ m}^3 Volume of water=12×90π=45π m3\text{Volume of water} = \frac{1}{2} \times 90\pi = 45\pi \text{ m}^3

Explanation:

First, find the total capacity of the cylinder using the volume formula. Since it is half-filled, divide the total volume by 2.

Surface Area and Volume of Prisms, Pyramids, Cones, Spheres, and Compound Solids Grade 8 Notes &…