10.3 Perimeter, Area, Surface Area & 3D Volume Calculations

Perimeter, Area, Surface Area & 3D Volume Calculations

Measurement questions on the NTS GAT General test your ability to calculate boundary lengths (perimeter and circumference), flat region coverage (area), and spatial capacity (volume, surface area, and space diagonals). Mastering 2D plane formulas, 3D solid mensuration, dimensional scaling principles, and inscribed figure relationships ensures rapid, error-free problem solving.

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1. 2D Perimeter and Area Master Reference

Perimeter ($P$) represents the continuous linear boundary length around a two-dimensional shape, while Area ($A$) measures the number of square units contained within that boundary.

Composite 2D Region Calculations

To find the area of a complex composite figure, decompose the overall shape into simple non-overlapping standard shapes (rectangles, triangles, semicircles) and sum their areas. For shaded border regions lying between nested shapes, subtract the inner unshaded area from the total outer area.

Worked NTS GAT Example 1

Problem: A rectangular field has a length of $20\text{ meters}$ and a width of $14\text{ meters}$. A semicircular flower bed is cut out from one of its $14\text{-meter}$ width ends. Find the area of the remaining field. (Use $\pi = \frac{22}{7}$).

Solution:

1. Area of Rectangle ($A_{\text{rect}}$): $A_{\text{rect}} = l \times w = 20 \times 14 = 280\text{ m}^2$
2. Area of Semicircular Cutout ($A_{\text{semi}}$):
   • Semicircle diameter $d = 14\text{ m} \implies \text{radius } r = 7\text{ m}$. $A_{\text{semi}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7^2 = \frac{1}{2} \times \frac{22}{7} \times 49 = \frac{1}{2} \times 154 = 77\text{ m}^2$
3. Remaining Area ($A_{\text{remaining}}$): $A_{\text{remaining}} = A_{\text{rect}} - A_{\text{semi}} = 280 - 77 = 203\text{ m}^2$
4. Answer: The area of the remaining field is $203\text{ square meters}$.

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2. 3D Solid Mensuration: Volume, Surface Area & Diagonals

Three-dimensional solids occupy spatial capacity (Volume, measured in cubic units $u^3$) and possess exterior boundary faces (Surface Area, measured in square units $u^2$).

Core 3D Solid Formulas

1. Rectangular Prism (Box)

• Volume ($V$): $V = l \cdot w \cdot h$
• Total Surface Area ($SA$): $SA = 2(lw + lh + wh)$
• Space Diagonal ($d$): $d = \sqrt{l^2 + w^2 + h^2}$

2. Cube (Edge length $s$)

• Volume ($V$): $V = s^3$
• Total Surface Area ($SA$): $SA = 6s^2$
• Space Diagonal ($d$): $d = s\sqrt{3}$

3. Right Circular Cylinder

• Volume ($V$): $V = \pi r^2 h$
• Curved (Lateral) Surface Area ($LSA$): $LSA = 2\pi rh$
• Total Surface Area ($TSA$): $TSA = 2\pi r^2 + 2\pi rh = 2\pi r(r + h)$

4. Right Circular Cone

• Slant Height ($l$): $l = \sqrt{r^2 + h^2}$
• Volume ($V$): $V = \frac{1}{3}\pi r^2 h$
• Total Surface Area ($TSA$): $TSA = \pi r^2 + \pi rl = \pi r(r + l)$

Scale Factor Transformation Rules

When all linear dimensions of a 3D solid are scaled by a constant factor $k$:

• Linear metrics (edges, perimeter) change by factor $k$.
• Surface Area metrics change by factor $k^2$.
• Volume metrics change by factor $k^3$.

Worked NTS GAT Example 2

Problem: A solid cylindrical tank has a radius of $7\text{ meters}$ and a height of $10\text{ meters}$. If the radius of the tank is doubled while its height is halved, how does the new volume compare to the original volume?

Solution:

1. Original Volume ($V_1$): $V_1 = \pi r^2 h = \pi (7^2)(10) = 490\pi\text{ m}^3$
2. New Dimensions: $r' = 2(7) = 14\text{ m}$, $h' = \frac{10}{2} = 5\text{ m}$.
3. New Volume ($V_2$): $V_2 = \pi (r')^2 (h') = \pi (14^2)(5) = \pi (196)(5) = 980\pi\text{ m}^3$
4. Volume Comparison: $\frac{V_2}{V_1} = \frac{980\pi}{490\pi} = 2$.
5. Proportionality Rule: Since $V \propto r^2 h$, doubling $r$ multiplies volume by $2^2 = 4$, and halving $h$ multiplies volume by $\frac{1}{2}$. Combined factor = $4 \times \frac{1}{2} = 2$.
6. Answer: The volume of the new tank is twice ($2\times$) the original volume.

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3. Spheres and Hemispheres

A sphere is a 3D surface where every boundary point is equidistant (radius $r$) from the center point.

• Volume of Sphere ($V$): $V = \frac{4}{3}\pi r^3$
• Surface Area of Sphere ($SA$): $SA = 4\pi r^2$
• Volume of Solid Hemisphere: $V = \frac{2}{3}\pi r^3$
• Total Surface Area of Solid Hemisphere ($TSA$): Includes flat circular base: $TSA = 2\pi r^2 + \pi r^2 = 3\pi r^2$

Worked NTS GAT Example 3

Problem: A solid metal sphere of radius $6\text{ cm}$ is melted down and recast into small solid spheres each of radius $2\text{ cm}$. How many small spheres can be formed?

Solution:

1. Volume of original sphere: $V_{\text{large}} = \frac{4}{3}\pi (6^3) = \frac{4}{3}\pi (216)$.
2. Volume of one small sphere: $V_{\text{small}} = \frac{4}{3}\pi (2^3) = \frac{4}{3}\pi (8)$.
3. Number of small spheres $N = \frac{V_{\text{large}}}{V_{\text{small}}} = \frac{216}{8} = 27$.
4. Direct Scaling Shortcut: $N = \left(\frac{R}{r}\right)^3 = \left(\frac{6}{2}\right)^3 = 3^3 = 27$.
5. Answer: Exactly 27 small spheres can be formed.

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4. Inscribed and Circumscribed Geometric Figures

Questions involving nested figures require establishing algebraic links between inner and outer geometric dimensions.

Key Inscribed Configurations

1. Square Inscribed in a Circle:

   • Square diagonal equals circle diameter: $d = s\sqrt{2} = 2r \implies s = r\sqrt{2}$.
   • Area of inscribed square $A_{\text{square}} = s^2 = (r\sqrt{2})^2 = 2r^2$.
   • Area of circle $A_{\text{circle}} = \pi r^2$.
   • Area ratio: $\frac{A_{\text{square}}}{A_{\text{circle}}} = \frac{2}{\pi}$.

2. Circle Inscribed in a Square:

   • Circle diameter equals square side: $d = 2r = s \implies r = \frac{s}{2}$.
   • Area of square $A_{\text{square}} = s^2$.
   • Area of inscribed circle $A_{\text{circle}} = \pi r^2 = \frac{\pi s^2}{4}$.
   • Area ratio: $\frac{A_{\text{circle}}}{A_{\text{square}}} = \frac{\pi}{4}$.

Worked NTS GAT Example 4

Problem: A square is inscribed inside a circle of radius $7\text{ cm}$. Calculate the area of the region inside the circle but outside the square. (Use $\pi = \frac{22}{7}$).

Solution:

1. Area of Outer Circle ($A_{\text{circle}}$): $A_{\text{circle}} = \pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 154\text{ cm}^2$
2. Area of Inscribed Square ($A_{\text{square}}$):
   • Circle radius $r = 7\text{ cm} \implies$ Circle diameter = Square diagonal $d = 14\text{ cm}$.
   • Area of square $A_{\text{square}} = \frac{1}{2} d^2 = \frac{1}{2} (14^2) = \frac{1}{2}(196) = 98\text{ cm}^2$.
3. Area of Outer Region ($A_{\text{shaded}}$): $A_{\text{shaded}} = A_{\text{circle}} - A_{\text{square}} = 154 - 98 = 56\text{ cm}^2$
4. Answer: The area of the region outside the square is $56\ square centimeters}$.