krit.club logo

Mensuration - Surface Area and Volume of 3D Solids

Grade 9IGCSE

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

🔑Concepts

•

A Prism is a 3D solid with a constant cross-section. Its volume is calculated by multiplying the area of this cross-section by the length or height of the prism. Common examples include cuboids and triangular prisms.

Diagram of a triangular prism showing the cross-sectional area and the length.
•

For a Right-Circular Cone, the relationship between the vertical height hh, the radius rr, and the slant height ll is defined by the Pythagorean theorem: l2=r2+h2l^2 = r^2 + h^2. This is essential for finding the Curved Surface Area (CSA) when only vertical height is given.

Diagram of a cone showing radius, vertical height, and slant height forming a right-angled triangle.
•

A Pyramid has a base that can be any polygon, and its lateral faces are triangles meeting at a common vertex called the apex. The volume is exactly one-third of the volume of a prism with the same base and height: V=13×Base Area×hV = \frac{1}{3} \times \text{Base Area} \times h.

•

A Sphere is perfectly symmetrical. All points on its surface are at an equal distance rr from the center. Its surface area is 4πr24\pi r^2, which is exactly four times the area of its great circle.

Sphere with radius r marked from the center.

📐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=2πr(r+h)TSA = 2\pi r(r + h)

Prism: V=Area of cross-section×LV = \text{Area of cross-section} \times L

Sphere: V=43πr3V = \frac{4}{3} \pi r^3, SA=4πr2SA = 4\pi r^2

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

Pyramid: V=13×Base Area×hV = \frac{1}{3} \times \text{Base Area} \times h

Hemisphere: V=23πr3V = \frac{2}{3} \pi r^3, CSA=2πr2CSA = 2\pi r^2, TSA=3πr2TSA = 3\pi r^2

💡Examples

Problem 1:

A cylinder has a radius of 33 cm and a height of 77 cm. Calculate its Total Surface Area. (Take π=3.142\pi = 3.142)

Solution:

TSA=2πr2+2πrh=2(3.142)(32)+2(3.142)(3)(7)=56.556+131.964=188.52TSA = 2\pi r^2 + 2\pi rh = 2(3.142)(3^2) + 2(3.142)(3)(7) = 56.556 + 131.964 = 188.52 cm²

Explanation:

To find the Total Surface Area of a cylinder, you must add the area of the two circular bases (2πr22\pi r^2) to the area of the curved side (2πrh2\pi rh).

Problem 2:

A right-circular cone has a radius of 55 cm and a vertical height of 1212 cm. Find its Volume.

Solution:

V=13πr2h=13×π×52×12=100π≈314.16V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times 5^2 \times 12 = 100\pi \approx 314.16 cm³

Explanation:

The volume of a cone is exactly one-third the volume of a cylinder with the same radius and height. Plug the radius (55) and height (1212) into the formula.

Problem 3:

A metal sphere with a radius of 66 cm is melted down and recast into a solid cylinder with a radius of 44 cm. Find the height of the cylinder.

Solution:

Volume of Sphere = 43π(6)3=288π\frac{4}{3} \pi (6)^3 = 288\pi. Volume of Cylinder = π(4)2h=16πh\pi (4)^2 h = 16\pi h. Set volumes equal: 288π=16πh⇒h=288/16=18288\pi = 16\pi h \Rightarrow h = 288 / 16 = 18 cm.

Explanation:

In recasting problems, the volume remains constant. Calculate the volume of the original shape (sphere) and set it equal to the formula for the volume of the new shape (cylinder) to solve for the missing dimension.

Problem 4:

A square-based pyramid has a base side length of 1010 cm and a vertical height of 1212 cm. Calculate the Total Surface Area of the pyramid.

Square-based pyramid with height 12cm and base side 10cm.

Solution:

  1. Find the slant height (ll) of the triangular faces using the right-angled triangle formed by the vertical height and half the base length: l=h2+(s2)2=122+52=144+25=169=13 cml = \sqrt{h^2 + (\frac{s}{2})^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm}
  2. Calculate the area of the square base: Base Area=10×10=100 cm2\text{Base Area} = 10 \times 10 = 100 \text{ cm}^2
  3. Calculate the area of the four triangular faces: Lateral Area=4×(12×base×l)=4×(12×10×13)=260 cm2\text{Lateral Area} = 4 \times (\frac{1}{2} \times \text{base} \times l) = 4 \times (\frac{1}{2} \times 10 \times 13) = 260 \text{ cm}^2
  4. Total Surface Area (TSA): TSA=100+260=360 cm2TSA = 100 + 260 = 360 \text{ cm}^2

Explanation:

To find the surface area, we must sum the area of the square base and the area of the four identical triangular side faces. The height of these triangles is the 'slant height' of the pyramid, found using Pythagoras.

Problem 5:

A composite solid consists of a hemisphere of radius 33 cm joined to the top of a cylinder of radius 33 cm and height 1010 cm. Calculate the total volume of the solid in terms of π\pi.

A cylinder with a hemisphere on top.

Solution:

  1. Calculate the volume of the cylinder: Vcylinder=πr2h=π(3)2(10)=90π cm3V_{\text{cylinder}} = \pi r^2 h = \pi (3)^2 (10) = 90\pi \text{ cm}^3
  2. Calculate the volume of the hemisphere: Vhemisphere=23πr3=23π(3)3=23π(27)=18π cm3V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (3)^3 = \frac{2}{3} \pi (27) = 18\pi \text{ cm}^3
  3. Total Volume: Vtotal=90π+18π=108π cm3V_{\text{total}} = 90\pi + 18\pi = 108\pi \text{ cm}^3

Explanation:

For composite solids, calculate the volume of each individual simple solid and then sum them together. Ensure the radius is consistent for both parts.