krit.club logo

Mensuration - Surface area and volume of 3D solids

Grade 11A Level

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

🔑Concepts

•

A cylinder is a solid with two parallel congruent circular bases and a curved surface. The volume is calculated by multiplying the base area by the height, while the total surface area includes the curved surface and two circular lids.

•

A cone is a solid with a circular base and a single vertex. The perpendicular height hh and the radius rr form a right-angled triangle with the slant height ll. The relationship is given by l2=r2+h2l^2 = r^2 + h^2.

Diagram of a cone showing perpendicular height h, radius r, and slant height l
•

A sphere is a perfectly symmetrical 3D object where all points on the surface are at an equal distance (radius rr) from the center. Its volume is 43πr3\frac{4}{3}\pi r^3 and its surface area is 4πr24\pi r^2.

•

A pyramid is a polyhedron formed by connecting a polygonal base and a point, called the apex. The volume is always one-third of the product of the base area and the perpendicular height.

•

For similar solids, if the ratio of corresponding lengths is kk, then the ratio of their surface areas is k2k^2 and the ratio of their volumes is k3k^3.

📐Formulae

Cylinder Volume: V=πr2hV = \pi r^2 h

Cylinder Total Surface Area: TSA=2πrh+2πr2TSA = 2\pi rh + 2\pi r^2

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

Cone Curved Surface Area: CSA=πrlCSA = \pi rl (where ll is slant height)

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

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

Pyramid Volume: V=13×base area×perpendicular heightV = \frac{1}{3} \times \text{base area} \times \text{perpendicular height}

Slant height of a cone: l=r2+h2l = \sqrt{r^2 + h^2}

💡Examples

Problem 1:

A solid metal cone has a radius of 5 cm and a perpendicular height of 12 cm. Calculate its total surface area. (Take π=3.142\pi = 3.142)

Solution:

l=52+122=25+144=13l = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13 cm. CSA=π×5×13=65πCSA = \pi \times 5 \times 13 = 65\pi. Base Area = π×52=25π\pi \times 5^2 = 25\pi. TSA=65π+25π=90π≈282.78TSA = 65\pi + 25\pi = 90\pi \approx 282.78 cm².

Explanation:

First, find the slant height (ll) using Pythagoras' theorem. Then, calculate the Curved Surface Area and the base area separately before adding them for the Total Surface Area.

Problem 2:

Two similar spheres have radii in the ratio 2:3. If the volume of the smaller sphere is 16π16\pi cm³, find the volume of the larger sphere.

Solution:

Linear scale factor k=32k = \frac{3}{2}. Volume scale factor =k3=(32)3=278= k^3 = (\frac{3}{2})^3 = \frac{27}{8}. Volume of larger sphere =16π×278=2π×27=54π= 16\pi \times \frac{27}{8} = 2\pi \times 27 = 54\pi cm³.

Explanation:

Use the property that the ratio of volumes of similar solids is the cube of the ratio of their corresponding lengths.

Problem 3:

A hemisphere has a radius of 6 cm. Calculate its volume in terms of π\pi.

Solution:

V=12×(43πr3)=23π(6)3=23π(216)=2π(72)=144πV = \frac{1}{2} \times (\frac{4}{3} \pi r^3) = \frac{2}{3} \pi (6)^3 = \frac{2}{3} \pi (216) = 2 \pi (72) = 144\pi cm³.

Explanation:

A hemisphere is half of a sphere. Use the sphere volume formula and divide by 2.

Problem 4:

A square-based pyramid has a base side length of 10 cm10\text{ cm} and a perpendicular height of 12 cm12\text{ cm}. Calculate the volume of the pyramid.

Square-based pyramid with height 12 and base side 10

Solution:

Base Area=side×side\text{Base Area} = \text{side} \times \text{side} Base Area=10×10=100 cm2\text{Base Area} = 10 \times 10 = 100\text{ cm}^2

Volume=13×Base Area×h\text{Volume} = \frac{1}{3} \times \text{Base Area} \times h V=13×100×12V = \frac{1}{3} \times 100 \times 12 V=100×4V = 100 \times 4 V=400 cm3V = 400\text{ cm}^3

Explanation:

The volume of any pyramid is one-third of the base area times the vertical height. Here, the base is a square, so its area is 10210^2. Multiplying by the height (1212) and dividing by 33 gives the final volume.

Problem 5:

A cylindrical water tank has a radius of 33 m and a height of 77 m. Calculate the volume of the tank and its total surface area (including the top lid). Take π=227\pi = \frac{22}{7}.

Cylinder with radius 3m and height 7m

Solution:

  1. Volume (VV): V=πr2hV = \pi r^2 h V=227×32×7V = \frac{22}{7} \times 3^2 \times 7 V=22×9V = 22 \times 9 V=198 m3V = 198\text{ m}^3

  2. Total Surface Area (TSATSA): TSA=2πrh+2πr2TSA = 2\pi r h + 2\pi r^2 TSA=(2×227×3×7)+(2×227×32)TSA = (2 \times \frac{22}{7} \times 3 \times 7) + (2 \times \frac{22}{7} \times 3^2) TSA=(44×3)+(447×9)TSA = (44 \times 3) + (\frac{44}{7} \times 9) TSA=132+3967TSA = 132 + \frac{396}{7} TSA≈132+56.57TSA \approx 132 + 56.57 TSA≈188.57 m2TSA \approx 188.57\text{ m}^2

Explanation:

To find the volume, we use the formula for a cylinder which is the base area (circle) multiplied by the height. For the total surface area, we calculate the area of the curved surface (2πrh2\pi rh) and add the areas of the two circular faces (top and bottom, 2πr22\pi r^2).