Look around the room you're in right now. The rectangular floor, the square tiles, the cylindrical glass of water, the circular clock on the wall — every one of them has a size you can measure: how much floor to sweep, how much water the glass holds, how much ribbon to wrap a gift. That measuring of shapes is exactly what geometry and mensuration are about, and it is one of the most concrete, formula-friendly topics on the Civil Service Exam. If you can remember a compact set of formulas and read carefully what is being asked, this becomes a generous source of quick marks. Let's build that foundation thoroughly.
Geometry is the study of shapes; mensuration is the part of geometry that measures them — their perimeter (distance around), area (surface covered), volume (space filled), and surface area (outer covering). Exam questions describe a shape (a room, a garden, a tank, a ladder against a wall) and ask for one of these measurements.
On the Civil Service Exam, geometry and mensuration are a steady part of Numerical Reasoning. The questions reward memorized formulas applied to a clearly identified shape, plus the closely related Pythagorean theorem for right triangles. Altogether the topic touches 4 to 7 items on a typical exam.
The difficulty is beginner to intermediate. The arithmetic is rarely hard; the exam tests whether you (1) recall the right formula, (2) tell area from perimeter from volume from surface area, and (3) use the radius where a formula wants a radius (not the diameter). Nail those three habits and this topic pays off reliably.
After completing this lesson you will be able to:
You should be comfortable with:
A quick refresher on the two ideas we lean on most. First, squaring: 7² means 7 × 7 = 49 (not 7 × 2). Circles and the Pythagorean theorem square things constantly. Second, radius vs. diameter: the radius is the distance from a circle's center to its edge; the diameter goes all the way across through the center, so diameter = 2 × radius. Almost every circle formula uses the radius — if you're given the diameter, halve it first. Keep both of these front of mind.
Mensuration is the math of building, buying, and making things:
Knowing mensuration means you can estimate materials, costs, and capacities in real life — precisely the practical reasoning the exam is checking.
We'll move from flat shapes to solids, one family at a time. Keep a notebook and sketch each shape as you read — geometry sticks far better when you draw it.
Before any formula, understand the four measurements, because choosing the wrong one is the #1 mistake here.
Analogy: For a swimming pool: the perimeter is the tiles you'd lay around the edge; the area is the cover you'd stretch over the top; the volume is the water that fills it; the surface area is all the paint to coat every inner wall and the floor. Same pool, four completely different questions.
A room 8 m long and 5 m wide: Area = 8 × 5 = 40 m² (flooring); Perimeter = 2(8 + 5) = 26 m (baseboard).
A triangle with base 10 cm and height 6 cm has area ½ × 10 × 6 = 30 cm².
For a triangle where you know all three sides but not the height, Heron's formula finds the area: let s = (a + b + c) ÷ 2 (the "semi-perimeter"), then Area = √(s(s − a)(s − b)(s − c)).
For sides 3, 4, 5: s = 6, Area = √(6 × 3 × 2 × 1) = √36 = 6. (This matches ½ × 3 × 4 = 6, since 3-4-5 is a right triangle.)
A trapezoid with parallel sides 8 cm and 12 cm and height 5 cm: Area = ½ × (8 + 12) × 5 = ½ × 20 × 5 = 50 cm².
About π: it's the constant ≈ 3.14159, the ratio of any circle's circumference to its diameter. For the exam, use 3.14 for a decimal answer, or 22/7 when the radius is a multiple of 7 (it cancels beautifully).
A circle of radius 7 cm: Area = 22/7 × 7 × 7 = 154 cm²; Circumference = 2 × 22/7 × 7 = 44 cm.
Common misconception: the area is π r squared, not π times 2r. And if you're given the diameter (say 14 cm), halve it to the radius (7 cm) before using any formula.
For any right triangle, the square of the hypotenuse (the longest side, opposite the right angle) equals the sum of the squares of the other two sides:
a² + b² = c², where c is the hypotenuse.
A ladder 13 m long rests against a wall with its base 5 m out. How high does it reach? 5² + b² = 13² → 25 + b² = 169 → b² = 144 → b = 12 m.
Pythagorean triples are whole-number side sets worth memorizing, because they let you skip the square roots: 3-4-5, 5-12-13, 8-15-17, 7-24-25, and any multiple (6-8-10, 9-12-15, 10-24-26…). If you spot two sides that fit a triple, the third is instant.
Analogy: the theorem is why a "3-4-5" corner is used by carpenters and masons to make a perfect right angle — measure 3 units along one edge, 4 along the other, and if the diagonal is exactly 5, the corner is square.
Volume measures how much space a solid fills (or holds):
A box 5 × 4 × 3 cm has volume 5 × 4 × 3 = 60 cm³. A cylinder of radius 7 cm and height 10 cm holds 22/7 × 7² × 10 = 1,540 cm³.
Surface area totals the outside faces:
A cube of side 4 cm has surface area 6 × 4² = 6 × 16 = 96 cm². That box (5 × 4 × 3) has surface area 2(5·4 + 5·3 + 4·3) = 2(20 + 15 + 12) = 2 × 47 = 94 cm².
A rhombus is a "pushed-over square" — four equal sides but slanted. It has two useful area formulas:
Its perimeter is 4 × side. And because the diagonals meet at right angles and bisect each other, each side is the hypotenuse of a right triangle formed by half of each diagonal.
A rhombus has diagonals 6 cm and 8 cm. Area = ½ × 6 × 8 = 24 cm². Each side = √(3² + 4²) = 5 cm, so the perimeter = 20 cm.
An equilateral triangle (all three sides equal) has a handy dedicated area formula:
Area = (√3 ÷ 4) × side².
For a side of 4 cm: Area = (√3 ÷ 4) × 16 = 4√3 ≈ 6.93 cm². (√3 ≈ 1.732.)
A sector is a "slice of pizza" — a fraction of a circle defined by a central angle θ (theta). An arc is the curved edge of that slice. Both are just the fraction θ/360 of the whole circle:
A sector with radius 6 cm and central angle 60°: sector area = (60 ÷ 360) × 3.14 × 6² = (1/6) × 3.14 × 36 = 18.84 cm².
Two more solids appear occasionally:
Notice cones and pyramids both hold one-third of the "box or cylinder" that would surround them — a cone is a third of the cylinder with the same base and height.
A cone with radius 3 cm and height 4 cm: slant l = 5 (the 3-4-5 triple). Volume = (1/3) × 3.14 × 9 × 4 ≈ 37.68 cm³; total surface = 3.14 × 3 × (3 + 5) ≈ 75.36 cm².
Real exam questions often combine shapes — a rectangle with a semicircle on top, or a path around a garden. The trick is to add or subtract areas:
A rectangular garden 20 m × 15 m has a 2-m-wide path running all around the outside. Find the path's area. Outer rectangle = (20 + 4) × (15 + 4) = 24 × 19 = 456 m². Garden = 300 m². Path = 456 − 300 = 156 m².
Keep units consistent: don't mix meters and centimeters in one calculation. And note that area units are squared (100 cm = 1 m, but 10,000 cm² = 1 m²) — a frequent trap in "how many tiles" problems.
With every shape and measure covered, let's picture, tabulate, and drill.
| Shape | Measure | Formula |
|---|---|---|
| Rectangle | Area / Perimeter | l × w / 2(l + w) |
| Square | Area / Perimeter | s² / 4s |
| Triangle | Area | ½ × base × height |
| Triangle (3 sides) | Area (Heron) | √(s(s−a)(s−b)(s−c)), s = (a+b+c)/2 |
| Parallelogram | Area | base × height |
| Trapezoid | Area | ½ × (sum of parallel sides) × height |
| Circle | Area / Circumference | πr² / 2πr |
| Right triangle | Sides | a² + b² = c² |
| Cube | Volume / Surface | s³ / 6s² |
| Box (prism) | Volume / Surface | lwh / 2(lw + lh + wh) |
| Cylinder | Volume / Surface | πr²h / 2πr(r + h) |
| Sphere | Volume / Surface | (4/3)πr³ / 4πr² |
| Rhombus | Area | ½ × d₁ × d₂ |
| Equilateral triangle | Area | (√3/4) × side² |
| Sector | Area / Arc | (θ/360)πr² / (θ/360)2πr |
| Cone | Volume / Surface | (1/3)πr²h / πr(r + l) |
| Pyramid | Volume | (1/3) × base area × height |
Two notes. (1) Every circle/cylinder/sphere formula uses the radius — halve the diameter if that's what you're given. (2) The cylinder's surface area 2πr(r + h) is just the two circle ends (2 × πr²) plus the curved side (circumference 2πr times height h), factored neatly.
| When the question says… | It wants… |
|---|---|
| "fencing," "border," "ribbon around," "how far around" | perimeter (or circumference) |
| "flooring," "paint the wall," "carpet," "how much surface" | area |
| "how much it holds," "capacity," "water," "fill" | volume |
| "wrap," "cover the outside," "material to make the box" | surface area |
| "ladder against a wall," "diagonal," "right angle" | Pythagorean theorem |
| "path around," "shaded region," "frame" | subtract inner area from outer |
| a circle with a diameter given | halve it to the radius first |
Step 1 — Identify the shape (or shapes, for a composite figure) and sketch it. ↓ Step 2 — Decide which measure is asked: perimeter, area, volume, or surface area. Underline the keyword. ↓ Step 3 — Gather the right inputs. Convert diameter to radius; make sure all lengths share one unit. ↓ Step 4 — Apply the formula, squaring or cubing carefully, and choosing π = 3.14 or 22/7 wisely. ↓ Step 5 — Check the units and reasonableness. Area comes out in square units, volume in cubic units; a tiny room shouldn't yield a huge area.
Why Step 2 matters most: the arithmetic here is easy — the marks are lost by computing area when perimeter was wanted, or volume when surface area was wanted. Name the measure before you compute.
Example 1. Find the area and perimeter of a rectangle 12 cm by 7 cm. Solution: Area = 12 × 7 = 84 cm²; Perimeter = 2(12 + 7) = 38 cm. Difficulty: ★☆☆☆☆
Example 2. Find the area of a triangle with base 8 cm and height 5 cm. Solution: ½ × 8 × 5 = 20 cm². Difficulty: ★☆☆☆☆
Example 3 (circle). Find the area and circumference of a circle with radius 7 cm (use π = 22/7). Solution: Area = 22/7 × 7 × 7 = 154 cm²; Circumference = 2 × 22/7 × 7 = 44 cm. Difficulty: ★★☆☆☆
Example 4 (diameter trap). A circular table has a diameter of 14 dm. Find its area (π = 22/7). Thinking: The formula needs the radius, so halve the diameter: r = 7. Solution: Area = 22/7 × 7² = 154 dm². Common mistake: using 14 as the radius, which quadruples the answer. Difficulty: ★★☆☆☆
Example 5 (Pythagorean triple). A right triangle has legs 8 cm and 15 cm. Find the hypotenuse. Solution: This is the 8-15-17 triple, so the hypotenuse is 17 cm. (Check: 8² + 15² = 64 + 225 = 289 = 17².) Difficulty: ★★☆☆☆
Example 6 (trapezoid). A trapezoidal lot has parallel sides 30 m and 50 m, and the perpendicular distance between them is 20 m. Find its area. Solution: ½ × (30 + 50) × 20 = ½ × 80 × 20 = 800 m². Difficulty: ★★★☆☆
Example 7 (box volume and surface). A closed box measures 10 × 6 × 4 cm. Find its volume and surface area. Solution: Volume = 10 × 6 × 4 = 240 cm³. Surface area = 2(10·6 + 10·4 + 6·4) = 2(60 + 40 + 24) = 2 × 124 = 248 cm². Difficulty: ★★★☆☆
Example 8 (cylinder). A cylindrical tank has radius 7 m and height 10 m. Find its volume and total surface area (π = 22/7). Solution: Volume = 22/7 × 7² × 10 = 22 × 7 × 10 = 1,540 m³. Surface area = 2 × 22/7 × 7 × (7 + 10) = 2 × 22 × 17 = 748 m². Difficulty: ★★★☆☆
Example 9 (tiles — unit trap). How many 50 cm × 50 cm tiles are needed to cover a floor 4 m by 3 m? Thinking: Work in the same units. Floor area = 4 × 3 = 12 m². Each tile is 0.5 × 0.5 = 0.25 m². Solution: 12 ÷ 0.25 = 48 tiles. Common mistake: forgetting 50 cm = 0.5 m and mixing units. Difficulty: ★★★★☆
Example 10 (path around a garden). A rectangular garden 20 m × 12 m is bordered by a 1-m-wide path all around the outside. Find the path's area. Solution: Outer rectangle = (20 + 2) × (12 + 2) = 22 × 14 = 308 m². Garden = 240 m². Path = 308 − 240 = 68 m². Difficulty: ★★★★☆
Example 11 (Pythagoras in a rectangle). Find the length of the diagonal of a rectangle 24 cm long and 7 cm wide. Thinking: The diagonal is the hypotenuse of a right triangle with legs 24 and 7. Solution: √(24² + 7²) = √(576 + 49) = √625 = 25 cm (the 7-24-25 triple). Difficulty: ★★★★☆
Example 12 (composite figure). A shape is a 10 cm × 6 cm rectangle with a semicircle (diameter 6 cm) attached to one short side. Find the total area (π = 3.14). Thinking: Add the rectangle's area and the semicircle's area. The semicircle's radius is 3 cm. Solution: Rectangle = 60 cm². Semicircle = ½ × 3.14 × 3² = ½ × 3.14 × 9 = 14.13 cm². Total ≈ 74.13 cm². Difficulty: ★★★★★
Example 13 (fencing cost). A square lot has an area of 400 m². Fencing costs ₱250 per meter. Find the total cost to fence the lot. Thinking: From area to side to perimeter to cost. Solution: Side = √400 = 20 m. Perimeter = 4 × 20 = 80 m. Cost = 80 × 250 = ₱20,000. Difficulty: ★★★★☆
Example 14 (sphere). Find the volume and surface area of a sphere with radius 3 cm (π = 3.14). Solution: Volume = (4/3) × 3.14 × 3³ = (4/3) × 3.14 × 27 = 4 × 3.14 × 9 = 113.04 cm³. Surface area = 4 × 3.14 × 3² = 4 × 3.14 × 9 = 113.04 cm². Difficulty: ★★★★☆
Example 15 (rhombus). A rhombus-shaped tile has diagonals 10 cm and 24 cm. Find its area and perimeter. Thinking: Area is half the product of diagonals; each side is the hypotenuse of a right triangle with legs half of each diagonal (5 and 12). Solution: Area = ½ × 10 × 24 = 120 cm². Side = √(5² + 12²) = √169 = 13 cm (the 5-12-13 triple), so perimeter = 4 × 13 = 52 cm. Difficulty: ★★★★☆
Example 16 (sector). A sprinkler waters a sector of a lawn with radius 14 m and central angle 90°. Find the watered area (π = 22/7). Thinking: 90° is a quarter of the circle. Solution: (90 ÷ 360) × 22/7 × 14² = ¼ × 22/7 × 196 = ¼ × 616 = 154 m². Difficulty: ★★★★☆
Example 17 (equilateral triangle). Find the area of an equilateral triangle with side 6 cm (√3 ≈ 1.732). Solution: (√3 ÷ 4) × 6² = (1.732 ÷ 4) × 36 = 0.433 × 36 ≈ 15.59 cm². Difficulty: ★★★☆☆
Example 18 (cone volume). A conical container has radius 7 cm and height 12 cm. Find its volume (π = 22/7). Solution: (1/3) × 22/7 × 7² × 12 = (1/3) × 22 × 7 × 12 = (1/3) × 1,848 = 616 cm³. Difficulty: ★★★★☆
Example 19 (pyramid). A square-based pyramid has a base edge of 6 m and a height of 10 m. Find its volume. Thinking: Base area = 6² = 36 m²; volume = one-third of base area × height. Solution: (1/3) × 36 × 10 = 120 m³. Difficulty: ★★★★☆
Example 20 (shaded region: circle in a square). A circle is inscribed in a square of side 14 cm (the circle just touches all four sides). Find the area left over inside the square but outside the circle (π = 22/7). Thinking: The circle's diameter equals the square's side, so radius = 7. Solution: Square = 14² = 196 cm²; circle = 22/7 × 7² = 154 cm²; leftover = 196 − 154 = 42 cm². Difficulty: ★★★★★
Example 21 (volume to capacity). A rectangular water tank measures 2 m × 1 m × 0.5 m. How many liters of water does it hold? (1 m³ = 1,000 liters.) Solution: Volume = 2 × 1 × 0.5 = 1 m³ = 1,000 liters. Difficulty: ★★★☆☆
Example 22 (increase in area). If the side of a square is doubled, by what factor does its area increase? Thinking: Area depends on side², so doubling the side multiplies area by 2² = 4. Solution: The area becomes 4 times as large. (This "square the scale factor" idea is a favorite exam concept: triple the side → 9× the area.) Difficulty: ★★★★☆
Geometry and mensuration reward memorized formulas applied to the right measure. First decide what's asked — perimeter (around), area (surface, squared units), volume (space, cubed units), or surface area (outside covering). Then apply the formula, always using the radius for circles/cylinders/spheres (halve any diameter), the perpendicular height for triangles and parallelograms, and consistent units throughout. The Pythagorean theorem (a² + b² = c²) and its triples handle right-triangle sides quickly, and composite figures yield to adding or subtracting familiar areas. Recall the formulas, read the keyword, mind the radius and the units — that's the whole game.
| Shape | Key formula |
|---|---|
| Rectangle | A = lw, P = 2(l + w) |
| Square | A = s², P = 4s |
| Triangle | A = ½ bh |
| Trapezoid | A = ½(a + b)h |
| Circle | A = πr², C = 2πr |
| Right triangle | a² + b² = c² |
| Cube | V = s³, SA = 6s² |
| Box | V = lwh, SA = 2(lw + lh + wh) |
| Cylinder | V = πr²h, SA = 2πr(r + h) |
| Sphere | V = (4/3)πr³, SA = 4πr² |
Triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25 (and multiples). π: 3.14, or 22/7 when r is a multiple of 7. Remember: radius (not diameter); perpendicular height (not slant); square units for area, cubic for volume.
When do I use 3.14 versus 22/7 for π? Both approximate π. Use 22/7 when the radius is a multiple of 7 (the 7s cancel to give clean numbers); use 3.14 otherwise, or whichever the problem specifies.
How do I know if a problem wants area or perimeter? Look at the action. Covering a surface (paint, tiles, carpet) is area; going around an edge (fence, border, ribbon) is perimeter; filling a space (water, sand) is volume; wrapping the outside is surface area.
What's the difference between the radius and the diameter? The radius runs from the center to the edge; the diameter runs all the way across through the center, so diameter = 2 × radius. Circle formulas use the radius — halve the diameter first.
Do I have to memorize Heron's formula? It's helpful when you know all three sides of a triangle but not the height. If you know the base and perpendicular height, ½ × base × height is simpler.
Why are area units squared and volume units cubed? Area covers a two-dimensional surface (length × width → square units), while volume fills three dimensions (length × width × height → cubic units). This is also why 1 m² = 10,000 cm², not 100.
Tick them all and geometry and mensuration becomes a dependable, formula-driven scoring area — and you'll estimate real-world materials and costs with confidence.
Put it to the test with 1,563 practice questions on this topic.