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Numerical Aptitude Syllabus / Module 15 · Geometry & Mensuration / Week 15

Week 15: Geometry & Mensuration

Fewer clever shortcuts this week, and more formulas worth having cold — geometry questions are usually fast once you recognize the shape and recall the right relationship, so the actual skill is building instant recall for a compact set of facts, not deriving them under time pressure.

Module 15 of 20 Week 15 of 20 ~2–3 Hours Practice Problems Included

By the end of this week, you'll be able to

  • Recall the key triangle and circle properties that recur across questions
  • Compute area and perimeter for every standard 2D shape
  • Compute volume and surface area for the standard 3D solids
1 Triangle & Circle Properties 2 Area & Perimeter — 2D Shapes 3 Volume & Surface Area — 3D Solids 4 Pythagorean Triples — A Speed Shortcut 5 Practice Problems 6 Knowledge Check

1. Triangle & Circle Properties

A short list of relationships that appear inside a large fraction of geometry questions, often as one step within a longer problem rather than the whole question itself.

triangle facts
Angle sum: the three interior angles always sum to 180 deg

Exterior angle theorem: an exterior angle equals the sum of
the two REMOTE (non-adjacent) interior angles

Pythagorean theorem (right triangles only):
  (leg1)^2 + (leg2)^2 = (hypotenuse)^2

Similar triangles: if two triangles are similar, their
corresponding SIDES are in the same ratio, and their AREAS
are in the ratio of that side-ratio SQUARED
circle facts
Angle in a semicircle: any angle inscribed in a semicircle
(subtended by the diameter) is exactly 90 degrees

Angle at the center vs. circumference: the angle at the
center is TWICE the angle at the circumference, for angles
subtending the SAME arc

Tangent-radius: a tangent line to a circle is always
PERPENDICULAR to the radius at the point of contact

The similar-triangles area rule (ratio squared, not the same ratio as the sides) is worth flagging specifically — it's a common trap, since the intuitive guess is that area scales the same way length does, when in fact area is a two-dimensional quantity and scales with the square of any linear scale factor.

2. Area & Perimeter — 2D Shapes

standard 2D formulas
Rectangle:   Area = l x w         Perimeter = 2(l+w)
Square:      Area = s^2           Perimeter = 4s
Triangle:    Area = (1/2) x base x height
Parallelogram: Area = base x height
Trapezoid:   Area = (1/2) x (sum of parallel sides) x height
Circle:      Area = pi x r^2      Circumference = 2 x pi x r

Worked: find the area of a trapezoid with parallel sides
10cm and 16cm, and height 8cm.
  Area = (1/2) x (10+16) x 8 = (1/2) x 26 x 8 = 104 cm^2

Every one of these is worth being able to state instantly — the actual test-question difficulty almost always lives in figuring out WHICH shape (or combination of shapes) a word problem describes, not in recalling the formula once that's clear.

3. Volume & Surface Area — 3D Solids

standard 3D formulas
Cube:      Volume = s^3         Surface area = 6s^2
Cuboid:    Volume = lxwxh       Surface area = 2(lw+wh+hl)
Cylinder:  Volume = pi r^2 h    Surface area = 2 pi r h + 2 pi r^2
                                 (curved + two circular ends)
Cone:      Volume = (1/3) pi r^2 h
                                 Curved surface area = pi r l
                                 (l = slant height, NOT h)
Sphere:    Volume = (4/3) pi r^3
                                 Surface area = 4 pi r^2

Worked: find the volume of a cylinder with radius 7cm and
height 10cm. (use pi = 22/7)
  Volume = (22/7) x 7^2 x 10 = (22/7) x 49 x 10 = 1,540 cm^3

The cone's curved surface area using slant height l (not the perpendicular height h) is the detail most often mixed up — the slant height is the hypotenuse of the right triangle formed by the height and radius: l = √(r² + h²), Section 1's Pythagorean theorem applied to the cone's own cross-section.

4. Pythagorean Triples — A Speed Shortcut

Certain integer side-length combinations satisfy the Pythagorean theorem exactly — recognizing these instantly, rather than computing square roots, is a real speed advantage on right-triangle questions.

common triples worth recognizing on sight
3-4-5      (and its multiples: 6-8-10, 9-12-15, 30-40-50...)
5-12-13
8-15-17
7-24-25

Worked: a right triangle has legs 9 and 12. Find the
hypotenuse WITHOUT computing sqrt(9^2+12^2) by hand.

Recognize: 9 = 3x3, 12 = 3x4 -- this is a SCALED 3-4-5 triple
  Hypotenuse = 3 x 5 = 15   (scaled by the same factor of 3)

Scanning for a recognizable triple (or a scaled multiple of one) before reaching for the full Pythagorean calculation saves real time on any question where the numbers happen to fit — worth a quick mental check as the very first move on any right- triangle problem.

5. Practice Problems

Practice

30 problems, no calculator — Easy, Medium & Tough

Check for a recognizable Pythagorean triple before computing any square root by hand. Easy problems should take under 30 seconds each; Medium under 60 seconds; Tough under 90 seconds.

Easy (1–10)

  1. Find the area of a rectangle with length 10cm and width 6cm.
  2. Find the perimeter of a square with side 8cm.
  3. Find the area of a square with side 7cm.
  4. Find the area of a triangle with base 10cm and height 6cm.
  5. Find the circumference of a circle with radius 7cm (use π=22/7).
  6. Find the area of a circle with radius 14cm (use π=22/7).
  7. Find the sum of the angles of a triangle.
  8. A right triangle has legs 3cm and 4cm. Find the hypotenuse.
  9. Find the volume of a cube with side 4cm.
  10. Find the perimeter of a rectangle with length 12cm and width 5cm.
Easy — Answers

1) 60cm². 2) 32cm. 3) 49cm². 4) 30cm². 5) 44cm. 6) 616cm². 7) 180°. 8) 5cm. 9) 64cm³. 10) 34cm.

Medium (11–20)

  1. Find the area of a trapezoid with parallel sides 8cm and 12cm, and height 5cm.
  2. A right triangle has legs 6cm and 8cm. Find its area.
  3. Find the volume of a cuboid with dimensions 5cm × 4cm × 3cm.
  4. Find the total surface area of a cube with side 5cm.
  5. Find the volume of a cylinder with radius 7cm and height 10cm (use π=22/7).
  6. Find the curved surface area of a cylinder with radius 7cm and height 5cm (use π=22/7).
  7. A right triangle has legs 9cm and 12cm. Find its hypotenuse.
  8. Find the area of a parallelogram with base 15cm and height 8cm.
  9. Two similar triangles have sides in the ratio 3:5. Find the ratio of their areas.
  10. Find the volume of a cone with radius 6cm and height 7cm (use π=22/7).
Medium — Answers

11) 50cm². 12) 24cm². 13) 60cm³. 14) 150cm². 15) 1,540cm³. 16) 220cm². 17) 9=3×3, 12=3×4 → hypotenuse=3×5=15cm. 18) 120cm². 19) (3/5)²=9/25. 20) 264cm³.

Tough (21–30)

  1. Find the curved surface area of a cone with radius 7cm and height 24cm (use π=22/7; find the slant height first).
  2. Find the total surface area of a cylinder with radius 7cm and height 10cm (use π=22/7).
  3. Find the volume of a sphere with radius 21cm (use π=22/7).
  4. The area of a circle is 154 sq.cm. Find its radius (use π=22/7).
  5. A wire is bent into a circle of radius 21cm. If the same wire is bent into a square, find the side of the square (use π=22/7).
  6. Two similar triangles have areas 50 sq.cm and 72 sq.cm. If the smaller triangle's base is 5cm, find the larger triangle's base.
  7. Find the length of the diagonal of a rectangle with length 16cm and width 12cm.
  8. A cone and a cylinder have the same base radius and height. Find the ratio of their volumes.
  9. The perimeter of a right triangle is 30cm and its hypotenuse is 13cm, following a scaled 5-12-13 triple. Find its area.
  10. Find the number of diagonals in a polygon with 10 sides using the formula n(n-3)/2.
Tough — Answers

21) l=√(7²+24²)=25cm; CSA=(22/7)×7×25=550cm². 22) TSA=2πr(h+r)=2×(22/7)×7×17=748cm². 23) V=(4/3)×(22/7)×21³=38,808cm³. 24) πr²=154 → r²=49 → r=7cm. 25) Circumference=132cm; square side=132/4=33cm. 26) Area ratio 50:72=25:36=(5:6)²; side ratio 5:6; larger base=5×6/5=6cm. 27) √(16²+12²)=√400=20cm. 28) Cone:cylinder volume ratio=1:3. 29) 30=5k+12k+13k=30k → k=1, so it's exactly 5,12,13; area=(1/2)×5×12=30cm². 30) 10×7/2=35.

6. Knowledge Check

Four quick questions. Expand each to check your answer.

Q1

If two similar triangles have sides in a 2:3 ratio, why is their area ratio 4:9, not 2:3?

Area is a two-dimensional quantity, built from a length multiplied by another length, so scaling every linear dimension by a factor scales the area by that factor SQUARED. A 2:3 side ratio therefore produces a (2²):(3²) = 4:9 area ratio, not a direct 2:3 match.

Q2

Why does a cone's curved surface area formula use the slant height, not the perpendicular (vertical) height?

The curved surface of a cone, when "unrolled" flat, forms a sector of a circle whose radius is the distance along the cone's actual sloped surface — the slant height — not the straight-down vertical height. The slant height is the physically relevant distance covered by that curved surface, which is why it appears in the formula instead of the vertical height.

Q3

Why does recognizing legs 15 and 20 as a scaled 3-4-5 triple save time compared to computing √(15²+20²) directly?

Once 15 and 20 are recognized as 5 times the base triple 3 and 4, the hypotenuse is immediately known to be 5 times the base triple's hypotenuse (5), giving 25 — a quick multiplication instead of squaring two two-digit numbers, adding them, and extracting a square root by hand.

Q4

Why is any angle inscribed in a semicircle (subtending the diameter) always exactly 90 degrees, regardless of where on the arc the vertex sits?

This is a direct consequence of the angle-at-center-vs-circumference rule: the diameter subtends a straight angle (180°) at the center, and any inscribed angle subtending the same arc is exactly half the central angle — half of 180° is 90°, regardless of exactly where the inscribed angle's vertex sits on the remaining arc.

← Back to Week 14: Logarithms, Surds & Indices Up next Week 16: Clocks, Calendars & Age-Based Problems →