10.2 Area, Perimeter, and Volume

Measurement on the USTET

After angle and triangle relationships, the next high-frequency Mathematics cluster is measurement: how long around, how much area inside, and how much space inside a solid. These items are formula-driven. If your formula table is automatic and you convert units before multiplying, you can finish most of them in well under a minute — valuable when Mathematics shares time with Mental Ability, English, and Science in a single USTET sitting.

Perimeter is the total length of the boundary. Circumference is the perimeter of a circle. Area is the measure of a planar region (square units). Surface area is the total area of all faces of a solid. Volume is the space a solid occupies (cubic units).

Perimeter and Circumference

Use π ≈ 3.14, 22/7 (when r or d is a multiple of 7), or leave answers in terms of π when options do. Match the approximation implied by the choices.

Worked Example 1. A rectangular lot in Sampaloc is 24 m by 15 m. Find its perimeter.

P = 2(24) + 2(15) = 48 + 30 = 54 m.

Worked Example 2. A circular track has diameter 28 m. Using π = 22/7, find the circumference.

C = πd = (22/7)(28) = 88 m. (Using radius 14: C = 2πr = 2 × 22/7 × 14 = 88 m — same result.)

Area Formulas You Must Own

Worked Example 3. A triangular garden has base 18 m and height 7 m. Find its area.

A = ½(18)(7) = 9 × 7 = 63 m². The trap option 126 m² comes from forgetting the ½.

Worked Example 4. A trapezoidal plaza has parallel sides 12 m and 20 m and height 9 m. Find its area.

A = ½(12 + 20)(9) = ½(32)(9) = 16 × 9 = 144 m². Skipping the ½ yields 288 m² — almost always listed as a distractor.

Worked Example 5. A circle has diameter 10 cm. Using π = 3.14, find its area.

First r = 5 cm. A = 3.14 × 25 = 78.5 cm². Plugging d = 10 directly into πr² gives 314 cm² — four times too large, because area scales with r² and (2r)² = 4r².

Worked Example 6. A rhombus has diagonals 10 cm and 24 cm. Find its area.

A = ½(10)(24) = 120 cm². The diagonals of a rhombus are perpendicular bisectors of each other, so you can also think of four right triangles with legs 5 and 12 (a 5-12-13 family), but the diagonal formula is faster on test day.

Composite and Shaded Regions

USTET-style items love composite figures: a rectangle with a semicircle attached, a square with a circular hole, a path around a garden. Strategy:

1. Sketch and label every dimension you are given (and any you can deduce).
2. Decide whether the shaded region is a sum of parts or a difference (outer minus inner).
3. Compute each region with the correct formula, then add or subtract.
4. Keep units consistent before you multiply.

Worked Example 7 (shaded region). A rectangular courtyard measures 20 m by 14 m and contains a circular fountain of radius 3 m. What walking area remains? Use π = 3.14.

Rectangle: 20 × 14 = 280 m². Fountain: 3.14 × 9 = 28.26 m². Remaining: 280 − 28.26 = 251.74 m².

Worked Example 8 (sum of parts). A running track is a rectangle 100 m by 40 m with a semicircle of diameter 40 m attached to each short end. Find the total area enclosed. Use π = 3.14.

Rectangle: 100 × 40 = 4,000 m². Two semicircles make one full circle of radius 20 m: A = 3.14 × 400 = 1,256 m². Total: 4,000 + 1,256 = 5,256 m².

Surface Area of Common Solids

Surface area appears less often than volume on entrance exams, but the formulas are short:

Worked Example 9. A closed cylindrical can has radius 7 cm and height 10 cm. Using π = 22/7, find its total surface area.

TSA = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 10) = 2 × 22 × 17 = 748 cm².

Volume Formulas

Think in terms of base area B times height when the solid has uniform cross-section, and one-third of that for pointed solids.

Worked Example 10. A rectangular water tank is 2 m long, 1.5 m wide, and 1.2 m high. Find its volume in cubic meters and in liters (1 m³ = 1,000 L).

V = 2 × 1.5 × 1.2 = 3.6 m³ = 3.6 × 1,000 = 3,600 L.

Worked Example 11. A cylinder has radius 7 cm and height 15 cm. Using π = 22/7, find its volume.

V = πr²h = (22/7)(49)(15) = 22 × 7 × 15 = 2,310 cm³.

Worked Example 12. A cone and a cylinder have the same base radius and the same height. How do their volumes compare?

Cone volume is one-third of cylinder volume: V_cone = (1/3)πr²h while V_cylinder = πr²h. If the cylinder holds 300 cm³, the matching cone holds 100 cm³.

Worked Example 13. Find the volume of a sphere with diameter 6 cm. Leave the answer in terms of π.

r = 3. V = (4/3)π(27) = 36π cm³. Using diameter 6 inside r³ without halving produces (4/3)π(216) = 288π — eight times too large, because volume scales with r³ and 2³ = 8.

Units and Scaling

• Convert before multiplying: 80 cm × 1.2 m is illegal until both are meters or both are centimeters.
• Linear scale factor k multiplies lengths by k, areas by k², and volumes by k³. If each edge of a cube doubles, surface area ×4 and volume ×8.
• Answer units must match the quantity: perimeter in m, area in m², volume in m³.

Common Traps

• Forgetting ½ in triangle and trapezoid areas.
• Using diameter as radius in πr² or (4/3)πr³.
• Using slant height as perpendicular height in parallelogram, triangle, pyramid, or cone volume.
• Confusing lateral surface area with total surface area (forgetting the bases).
• Applying V = Bh to a pyramid or cone without the factor 1/3.

Section Takeaway

Keep one clean formula table for perimeter, area, surface area, and volume. Composite problems are sum or difference of known shapes. Prisms and cylinders: V = Bh. Pyramids and cones: V = (1/3)Bh. Sphere: V = (4/3)πr³. Always confirm whether you were given r or d before you substitute.