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Veda Skills Quantitative Aptitude Unit 15: Mensuration

Mensuration is Unit 15 of Veda Skills Quantitative Aptitude. This page has the complete chapter notes across 2 topics, 139 practice MCQs with answers and explanations, 10 important short and long questions and 31 flashcards, written for SSC, Bank, Railway and placement exams and free to read.
  • 2 topics
  • 139 MCQs
  • 8 short answers
  • 2 long answers
  • 31 flashcards
  • 2 games
  • ~130 min read
  • English

Checked before publishing: 23 automated checks, including a plagiarism check against the textbook's own words. How we check our content

Area and Perimeter of Plane Figures

Areas and perimeters of rectangles, squares, triangles (including Heron), parallelograms, rhombuses, trapeziums, circles, sectors and rings, with paths, roads, costs and shaded-region questions.

What you will be able to do

  • I can find the area and perimeter of rectangles, squares and triangles, including by Heron's formula.
  • I can find areas of parallelograms, rhombuses and trapeziums.
  • I can work with circles, semicircles, sectors and rings.
  • I can find shaded areas, path areas and costs of fencing or flooring.
  • I can find the percentage change in area when the sides change.

Key points

  • Rectangle: area l × b, perimeter 2(l + b). Square: area a² or d²/2, perimeter 4a.

  • Triangle: ½ × base × height. With three sides use Heron: s = (a + b + c)/2, area = √(s(s − a)(s − b)(s − c)).

  • Equilateral triangle of side a: area (√3/4)a², height (√3/2)a.

  • Parallelogram: base × height. Rhombus: ½ × d₁ × d₂. Trapezium: ½ × (sum of parallel sides) × height.

  • Circle: area πr², circumference 2πr. Semicircle perimeter: πr + 2r.

  • Sector of angle θ: area (θ/360) × πr², arc length (θ/360) × 2πr.

  • Ring between radii R and r: π(R² − r²) = π(R + r)(R − r).

  • Shaded or path area = outer area − inner area.

  • If two sides change by a% and b%, the area changes by a + b + ab/100 percent.

Full explanation

Sunita's school garden

A farmer measuring a green field with a rope beside a cylindrical water tank and a brick house.

Sunita's school has a rectangular garden 40 m by 25 m. The principal wants a 3 m wide brick path all around it, a round fountain of radius 7 m in the middle, and a fence on the outside. How many square metres of bricks, and how many metres of fence?

Every part of that plan is a mensuration question. Let us build the toolkit one shape at a time.

Perimeter and area

A family fencing a garden plot.

Perimeter is the distance around a shape, measured in m or cm. Area is the surface inside, measured in m² or cm².

  • Rectangle: $A = l \times b$, $P = 2(l + b)$
  • Square: $A = a^2 = \frac{d^2}{2}$, $P = 4a$

Cost questions: fencing cost = perimeter × rate per metre; flooring cost = area × rate per m².

Four formula cards: rectangle and square, triangle with Heron and equilateral, quadrilaterals, and circle with sector and ring.

Paths and roads

A path of width $w$ outside an $l \times b$ field makes an outer rectangle $(l + 2w) \times (b + 2w)$. The path area is outer minus inner. For a path inside, the inner rectangle is $(l - 2w) \times (b - 2w)$.

Two roads of width $w$ crossing through the middle cover $w(l + b) - w^2$, because the crossing square is counted twice.

A 40 by 25 field with a shaded path 3 wide all around it; the outer rectangle is 46 by 31 and the path area is 426.

Triangles

  • Base and height known: $A = \frac{1}{2} b h$.
  • Three sides known: Heron's formula with $s = \frac{a + b + c}{2}$:

$$A = \sqrt{s(s - a)(s - b)(s - c)}$$

  • Equilateral of side $a$: $A = \frac{\sqrt{3}}{4} a^2$.

Triangle ABC with sides 13, 14 and 15; s = 21 and the area is 84.

Four-sided figures

  • Parallelogram: base × perpendicular height.
  • Rhombus: $\frac{1}{2} d_1 d_2$.
  • Trapezium: $\frac{1}{2}(a + b) h$, where $a$, $b$ are the parallel sides.

Circles, semicircles and rings

$A = \pi r^2$ and $C = 2\pi r$. With $\pi = \frac{22}{7}$, choose $r$ as a multiple of 7: $r = 7$ gives $A = 154$ and $C = 44$.

  • Semicircle perimeter $= \pi r + 2r$.
  • Ring: $\pi(R^2 - r^2)$.
  • A wheel turns once for every $2\pi r$ of road.

Sectors and arcs

A sector of angle $\theta$ is $\frac{\theta}{360}$ of the circle:

$$\text{Area} = \frac{\theta}{360} \pi r^2, \qquad \text{Arc} = \frac{\theta}{360} \times 2\pi r$$

A circle of radius 21 with a shaded 60° sector; its area is 231 and its arc is 22.

Shaded regions

Shaded area = whole shape − the white parts. A square of side 14 with a circle of radius 7 inside has corner pieces of $196 - 154 = 42$. Four quarter circles of radius 7 at the corners of the same square also cover one full circle, so the middle piece is also 42.

A square of side 14 with its inscribed circle; the four corner pieces are shaded, area 42.

Percentage change in area

If length rises $a%$ and breadth $b%$, area changes by $a + b + \frac{ab}{100}$ percent. A square's side up 10%: $10 + 10 + 1 = 21%$. Length +20%, breadth −20%: $0 - 4 = -4%$.

Sunita's answers

Path: $46 \times 31 - 40 \times 25 = 426$ m² of bricks. Fence: $2(46 + 31) = 154$ m. A fountain of radius 7 m needs $154$ m² of the lawn. Three questions, three formulas, and the subtraction idea tying them together.

Mind map: rectangle, triangle, circle, sector and ring, shaded areas, and paths and roads around Area and perimeter.

Concepts and terms

Area vs perimeter

Perimeter is the length around a shape (m); area is the surface it covers (m²). Fencing uses perimeter; flooring uses area.

Heron's formula

Gives a triangle's area from its three sides, no height needed.

Sector

A slice of a circle between two radii. It is θ/360 of the whole circle.

Subtraction method

For any shaded region, find the area of the full shape and subtract the unshaded parts.

Did you know

Of all shapes with the same perimeter, the circle encloses the largest area; that is why many storage tanks are round.

Aryabhata (499 CE) gave π ≈ 3.1416, correct to four decimal places.

Memory tricks

Circle formulas — Area is squared, circumference is doubled — Area πr² has r squared; circumference 2πr has a 2 in front.

Area change — a + b + ab/100 — +20% and −10% give 20 − 10 − 2 = +8%.

Practice MCQs with answers (60)

Q1 A rectangular plot is 18 m long and 12 m wide. Find its area (in m²). beginner
  • A 60
  • B 234
  • C 216
  • D 232
Show answer
Answer: 216
Why:

18 × 12 = 216 m².

Q2 The area of a rectangle is 400 m² and its length is 25 m. Find its perimeter (in m). beginner
  • A 66
  • B 16
  • C 41
  • D 82
Show answer
Answer: 82
Why:

Breadth = 400 ÷ 25 = 16, so perimeter = 2(25 + 16) = 82 m.

Q3 A field is 60 m by 40 m. Fencing costs ₹15 per metre. What is the cost of fencing it (in ₹)? beginner
  • A 3000
  • B 1500
  • C 3300
  • D 36,000
Show answer
Answer: 3000
Why:

Perimeter = 2(60 + 40) = 200 m; cost = 200 × 15 = ₹3000.

Q4 A hall floor is 12 m by 9 m. Tiling costs ₹80 per m². Find the cost of tiling (in ₹). intermediate
  • A 3360
  • B 8640
  • C 9600
  • D 17,280
Show answer
Answer: 8640
Why:

Area = 12 × 9 = 108 m²; cost = 108 × 80 = ₹8640.

Q5 A field is 75 m by 45 m. Fencing costs ₹22 per metre. What is the cost of fencing it (in ₹)? intermediate
  • A 74,250
  • B 2640
  • C 5280
  • D 5720
Show answer
Answer: 5280
Why:

Perimeter = 2(75 + 45) = 240 m; cost = 240 × 22 = ₹5280.

Q6 The diagonal of a square is 12 cm. Find its area (in cm²). beginner
  • A 144
  • B 36
  • C 24
  • D 72
Show answer
Answer: 72
Why:

Area = d²/2 = 144/2 = 72.

Q7 The diagonal of a square is 9 cm. Find its area (in cm²). intermediate
  • A 40.5
  • B 81
  • C 20.25
  • D 18
Show answer
Answer: 40.5
Why:

Area = d²/2 = 81/2 = 40.5.

Q8 The length of a rectangle is increased by 20% and its breadth is decreased by 10%. What is the percentage change in its area? intermediate
  • A +10%
  • B +8%
  • C -8%
  • D -2%
Show answer
Answer: +8%
Why:

a + b + ab/100 = 20 + (-10) + (20 × -10)/100 = +8%.

Q9 The side of a square is increased by 10%. By what percent does its area change? intermediate
  • A -21%
  • B +20%
  • C +21%
  • D +23%
Show answer
Answer: +21%
Why:

a + b + ab/100 = 10 + (10) + (10 × 10)/100 = +21%.

Q10 The length of a rectangle is increased by 25% and its breadth is decreased by 20%. What is the percentage change in its area? advanced
  • A +1%
  • B +5%
  • C +2%
  • D 0%
Show answer
Answer: 0%
Why:

a + b + ab/100 = 25 + (-20) + (25 × -20)/100 = 0%.

Q11 How many square tiles of side 25 cm are needed to cover a floor 5 m by 4 m? intermediate
  • A 320
  • B 3200
  • C 80
  • D 32
Show answer
Answer: 320
Why:

Floor = 500 × 400 = 2,00,000 cm²; tile = 625 cm²; 2,00,000 ÷ 625 = 320.

Q12 A 40 m × 25 m garden has a path 3 m wide running outside along its edges (shaded). Find the area of the path (in m²). ![A 40 by 25 rectangle with a shaded path 3 wide outside it.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-path-1.svg) intermediate
  • A 204
  • B 426
  • C 390
  • D 462
Show answer
Answer: 426
Why:

Outer = 46 × 31 = 1426; path = that − 40 × 25 = 426 m².

Q13 A 30 m × 20 m garden has a path 2.5 m wide running inside along its edges (shaded). Find the area of the path (in m²). ![A 30 by 20 rectangle with a shaded path 2.5 wide inside it.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-path-2.svg) intermediate
  • A 250
  • B 220
  • C 225
  • D 118.75
Show answer
Answer: 225
Why:

Inner = 25 × 15 = 375; path = 600 − that = 225 m².

Q14 A 50 m × 35 m garden has a path 2 m wide running outside along its edges (shaded). Find the area of the path (in m²). ![A 50 by 35 rectangle with a shaded path 2 wide outside it.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-path-3.svg) beginner
  • A 174
  • B 340
  • C 372
  • D 356
Show answer
Answer: 356
Why:

Outer = 54 × 39 = 2106; path = that − 50 × 35 = 356 m².

Q15 A 60 m × 40 m park has two roads, each 3 m wide, crossing at the centre parallel to its sides (shaded). Find the area of the roads (in m²). ![A 60 by 40 rectangle with two shaded roads 3 wide crossing in the middle.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-roads-1.svg) intermediate
  • A 291
  • B 309
  • C 180
  • D 300
Show answer
Answer: 291
Why:

Roads = 3 × 60 + 3 × 40 − 3 × 3 = 291 m².

Q16 A 80 m × 50 m park has two roads, each 4 m wide, crossing at the centre (shaded). Gravelling costs ₹12 per m². Find the cost (in ₹). ![A 80 by 50 rectangle with two shaded roads 4 wide crossing in the middle.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-roads-2.svg) advanced
  • A 6240
  • B 6048
  • C 6432
  • D 3840
Show answer
Answer: 6048
Why:

Roads = 4 × 80 + 4 × 50 − 4 × 4 = 504 m²; cost = 504 × 12 = ₹6048.

Q17 Find the area of the triangle (in cm²). All lengths are in cm. ![A triangle with base 12 and perpendicular height 8 drawn dashed.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-tri-1.svg) beginner
  • A 20
  • B 60
  • C 48
  • D 96
Show answer
Answer: 48
Why:

½ × 12 × 8 = 48 cm².

Q18 Find the area of the triangle (in cm²). All lengths are in cm. ![A triangle with base 15 and perpendicular height 9 drawn dashed.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-tri-2.svg) beginner
  • A 135
  • B 24
  • C 82.5
  • D 67.5
Show answer
Answer: 67.5
Why:

½ × 15 × 9 = 67.5 cm².

Q19 Find the area of the triangle (in cm²). All lengths are in cm. ![Triangle ABC with sides 14, 15 and 13.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-heron-1.svg) intermediate
  • A 84
  • B 105
  • C 98
  • D 168
Show answer
Answer: 84
Why:

s = (14 + 15 + 13) ÷ 2 = 21; area = √(21 × 7 × 6 × 8) = 84.

Q20 Find the area of the triangle (in cm²). All lengths are in cm. ![Triangle ABC with sides 10, 17 and 9.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-heron-2.svg) intermediate
  • A 46
  • B 36
  • C 72
  • D 85
Show answer
Answer: 36
Why:

s = (10 + 17 + 9) ÷ 2 = 18; area = √(18 × 8 × 1 × 9) = 36.

Q21 Find the area of the triangle (in cm²). All lengths are in cm. ![Triangle ABC with sides 21, 20 and 13.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-heron-3.svg) advanced
  • A 210
  • B 147
  • C 126
  • D 252
Show answer
Answer: 126
Why:

s = (21 + 20 + 13) ÷ 2 = 27; area = √(27 × 6 × 7 × 14) = 126.

Q22 The sides of a triangle are 6 m, 5 m and 5 m. Find its area (in m²). intermediate
  • A 15
  • B 18
  • C 24
  • D 12
Show answer
Answer: 12
Why:

s = (6 + 5 + 5) ÷ 2 = 8; area = √(8 × 2 × 3 × 3) = 12.

Q23 Find the area of an equilateral triangle of side 8 cm. beginner
  • A 16√3 cm²
  • B 32√3 cm²
  • C 64 cm²
  • D 8√3 cm²
Show answer
Answer: 16√3 cm²
Why:

(√3/4) × 8² = (√3/4) × 64 = 16√3.

Q24 The area of an equilateral triangle is 36√3 cm². Find its perimeter (in cm). advanced
  • A 108
  • B 36
  • C 12
  • D 18
Show answer
Answer: 36
Why:

(√3/4)a² = 36√3 gives a² = 144, a = 12, perimeter = 36.

Q25 The diagonals of rhombus ABCD are AC = 16 cm and BD = 12 cm. Find its area (in cm²). ![A rhombus ABCD with diagonals AC = 16 and BD = 12.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-rh-1.svg) beginner
  • A 112
  • B 192
  • C 96
  • D 56
Show answer
Answer: 96
Why:

½ × 16 × 12 = 96.

Q26 The diagonals of rhombus ABCD are AC = 24 cm and BD = 10 cm. Find its area (in cm²). ![A rhombus ABCD with diagonals AC = 24 and BD = 10.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-rh-2.svg) beginner
  • A 240
  • B 68
  • C 144
  • D 120
Show answer
Answer: 120
Why:

½ × 24 × 10 = 120.

Q27 AB ∥ DC. Find the area of the trapezium (in cm²). All lengths are in cm. ![Trapezium ABCD with parallel sides AB = 14 and DC = 8 and height 6.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-trap-1.svg) intermediate
  • A 66
  • B 132
  • C 84
  • D 72
Show answer
Answer: 66
Why:

½ × (14 + 8) × 6 = 66.

Q28 AB ∥ DC. Find the area of the trapezium (in cm²). All lengths are in cm. ![Trapezium ABCD with parallel sides AB = 20 and DC = 12 and height 9.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-trap-2.svg) intermediate
  • A 180
  • B 144
  • C 153
  • D 288
Show answer
Answer: 144
Why:

½ × (20 + 12) × 9 = 144.

Q29 Find the area of the parallelogram (in cm²). All lengths are in cm. ![A parallelogram with base 10 and dashed perpendicular height 6.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-pg-1.svg) beginner
  • A 66
  • B 30
  • C 60
  • D 32
Show answer
Answer: 60
Why:

Base × height = 10 × 6 = 60.

Q30 Find the area of a circle of radius 7 cm (in cm²). Take π = 22/7. beginner
  • A 44
  • B 616
  • C 168
  • D 154
Show answer
Answer: 154
Why:

(22/7) × 7² = 154 cm².

Q31 Find the circumference of a circle of radius 14 cm (in cm). Take π = 22/7. beginner
  • A 88
  • B 616
  • C 44
  • D 95
Show answer
Answer: 88
Why:

2 × (22/7) × 14 = 88 cm.

Q32 A circular plot has diameter 42 m. Find its area (in m²). Take π = 22/7. intermediate
  • A 1407
  • B 1386
  • C 5544
  • D 132
Show answer
Answer: 1386
Why:

r = 21; area = (22/7) × 21² = 1386 m².

Q33 The circumference of a circle is 220 cm. Find its area (in cm²). Take π = 22/7. intermediate
  • A 7700
  • B 4070
  • C 3850
  • D 15,400
Show answer
Answer: 3850
Why:

2πr = 220 gives r = 35; area = (22/7) × 35² = 3850.

Q34 A wheel has diameter 70 cm. How many complete turns does it make in covering 1.1 km? Take π = 22/7. intermediate
  • A 5000
  • B 1000
  • C 250
  • D 500
Show answer
Answer: 500
Why:

One turn = π × 70 = 220 cm; 110000 ÷ 220 = 500.

Q35 A wheel has diameter 56 cm. How many complete turns does it make in covering 2.2 km? Take π = 22/7. advanced
  • A 1250
  • B 625
  • C 12,500
  • D 2500
Show answer
Answer: 1250
Why:

One turn = π × 56 = 176 cm; 220000 ÷ 176 = 1250.

Q36 A wire bent into a square of side 22 cm is reshaped into a circle. Find the area of the circle (in cm²). Take π = 22/7. intermediate
  • A 88
  • B 616
  • C 484
  • D 154
Show answer
Answer: 616
Why:

Wire = 4 × 22 = 88 = 2πr, so r = 14 and area = 616.

Q37 The figure is a semicircle with the diameter shown. Find its perimeter (in cm). All lengths are in cm. Take π = 22/7. ![A semicircle with diameter 14.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-semi-1.svg) intermediate
  • A 44
  • B 29
  • C 36
  • D 22
Show answer
Answer: 36
Why:

πr + 2r = (22/7) × 7 + 14 = 22 + 14 = 36.

Q38 The figure is a semicircle with the diameter shown. Find its perimeter (in cm). All lengths are in cm. Take π = 22/7. ![A semicircle with diameter 28.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-semi-2.svg) beginner
  • A 58
  • B 44
  • C 88
  • D 72
Show answer
Answer: 72
Why:

πr + 2r = (22/7) × 14 + 28 = 44 + 28 = 72.

Q39 Find the area of the shaded sector (in cm²). All lengths are in cm. Take π = 22/7. ![A circle of radius 21 with a shaded sector of angle 60°.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-sector-1.svg) intermediate
  • A 231
  • B 462
  • C 22
  • D 1386
Show answer
Answer: 231
Why:

(60/360) × (22/7) × 21² = 231.

Q40 Find the area of the shaded sector (in cm²). All lengths are in cm. Take π = 22/7. ![A circle of radius 14 with a shaded sector of angle 90°.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-sector-2.svg) beginner
  • A 308
  • B 154
  • C 22
  • D 616
Show answer
Answer: 154
Why:

(90/360) × (22/7) × 14² = 154.

Q41 Find the length of the arc of the shaded sector (in cm). All lengths are in cm. Take π = 22/7. ![A circle of radius 21 with a shaded sector of angle 120°.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-sector-3.svg) intermediate
  • A 462
  • B 132
  • C 44
  • D 88
Show answer
Answer: 44
Why:

Arc = (120/360) × 2 × (22/7) × 21 = 44.

Q42 Find the length of the arc of the shaded sector (in cm). All lengths are in cm. Take π = 22/7. ![A circle of radius 7 with a shaded sector of angle 72°.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-sector-4.svg) advanced
  • A 17.6
  • B 30.8
  • C 44
  • D 8.8
Show answer
Answer: 8.8
Why:

Arc = (72/360) × 2 × (22/7) × 7 = 8.8.

Q43 The minute hand of a clock is 14 cm long. What area does it sweep in 15 minutes (in cm²)? Take π = 22/7. intermediate
  • A 154
  • B 22
  • C 308
  • D 616
Show answer
Answer: 154
Why:

15 minutes = 90°, a quarter circle: ¼ × 616 = 154.

Q44 Find the area of the shaded ring (in cm²). All lengths are in cm. Take π = 22/7. ![Two circles with the same centre, radii 14 and 7; the ring between them is shaded.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-ring-1.svg) intermediate
  • A 44
  • B 462
  • C 154
  • D 616
Show answer
Answer: 462
Why:

π(R² − r²) = (22/7) × (196 − 49) = 462.

Q45 A circular park of radius 14 m has a track 7 m wide all around it, outside. Find the area of the track (in m²). Take π = 22/7. intermediate
  • A 154
  • B 1386
  • C 770
  • D 44
Show answer
Answer: 770
Why:

R = 14 + 7 = 21; π(R² − r²) = (22/7) × (441 − 196) = 770.

Q46 A circle is drawn inside a square of side 28 cm, touching all four sides. Find the shaded area (in cm²). Take π = 22/7. ![A square of side 28 with its inscribed circle; the four corner pieces are shaded.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-shade-1.svg) intermediate
  • A 696
  • B 616
  • C 42
  • D 168
Show answer
Answer: 168
Why:

784 − (22/7) × (14)² = 784 − 616 = 168.

Q47 A circle is drawn inside a square of side 21 cm, touching all four sides. Find the shaded area (in cm²). Take π = 22/7. ![A square of side 21 with its inscribed circle; the four corner pieces are shaded.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-shade-2.svg) advanced
  • A 94.5
  • B 24
  • C 375
  • D 346.5
Show answer
Answer: 94.5
Why:

441 − (22/7) × (10.5)² = 441 − 346.5 = 94.5.

Q48 Four quarter circles of radius 7 cm are drawn at the corners of a square of side 14 cm. Find the shaded area (in cm²). Take π = 22/7. ![A square of side 14 with a white quarter circle at each corner; the curved middle part is shaded.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-shade-3.svg) intermediate
  • A 154
  • B 42
  • C 157.5
  • D 168
Show answer
Answer: 42
Why:

The four quarters make one full circle: 196 − (22/7) × (7)² = 42.

Q49 Four quarter circles of radius 14 cm are drawn at the corners of a square of side 28 cm. Find the shaded area (in cm²). Take π = 22/7. ![A square of side 28 with a white quarter circle at each corner; the curved middle part is shaded.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-shade-4.svg) advanced
  • A 630
  • B 672
  • C 168
  • D 616
Show answer
Answer: 168
Why:

The four quarters make one full circle: 784 − (22/7) × (14)² = 168.

Q50 A square is drawn inside a circle of radius 7 cm with its corners on the circle. Find the shaded area (in cm²). Take π = 22/7. ![A circle of radius 7 with a square inside it touching the circle at its corners; the four segments outside the square are shaded.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-shade-5.svg) intermediate
  • A 28
  • B 105
  • C 98
  • D 56
Show answer
Answer: 56
Why:

Square diagonal = 14, so square = 14²/2 = 98. Shaded = 154 − 98 = 56.

Q51 A square is drawn inside a circle of radius 14 cm with its corners on the circle. Find the shaded area (in cm²). Take π = 22/7. ![A circle of radius 14 with a square inside it touching the circle at its corners; the four segments outside the square are shaded.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-shade-6.svg) advanced
  • A 224
  • B 392
  • C 112
  • D 420
Show answer
Answer: 224
Why:

Square diagonal = 28, so square = 28²/2 = 392. Shaded = 616 − 392 = 224.

Q52 A semicircle with diameter 14 cm is added to the end of a 20 cm × 14 cm rectangle, as shown. Find the total shaded area (in cm²). Take π = 22/7. ![A 20 by 14 rectangle with a semicircle of diameter 14 attached to one end; the whole shape is shaded.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-shade-7.svg) intermediate
  • A 280
  • B 357
  • C 203
  • D 434
Show answer
Answer: 357
Why:

20 × 14 + ½ × (22/7) × (7)² = 280 + 77 = 357.

Q53 If the radius of a circle is doubled, its area becomes beginner
  • A 8 times
  • B unchanged
  • C 4 times
  • D 2 times
Show answer
Answer: 4 times
Why:

Area goes as r², and 2² = 4.

Q54 A square and a circle have the same perimeter. Which has the larger area? intermediate
  • A It depends on the size
  • B The square
  • C Both equal
  • D The circle
Show answer
Answer: The circle
Why:

For perimeter 88: circle 616, square 484. The circle always wins.

Q55 What is the ratio of the areas of the circle inscribed in a square to the circle drawn around it? advanced
  • A 1 : 2
  • B 1 : 4
  • C 1 : √2
  • D 2 : 3
Show answer
Answer: 1 : 2
Why:

Radii are a/2 and a/√2; squares give a²/4 : a²/2 = 1 : 2.

Q56 The ratio of the areas of two circles is 16 : 25. What is the ratio of their circumferences? intermediate
  • A 16 : 25
  • B 4 : 5
  • C 2 : 5
  • D 8 : 5
Show answer
Answer: 4 : 5
Why:

Radii ratio = √(16/25) = 4/5, and circumference goes as r.

Q57 Assertion (A): A 10% increase in each side of a square increases its area by 21%. Reason (R): Area of a square depends on the square of its side. intermediate
  • A Both A and R are true, and R is the correct explanation of A
  • B Both A and R are true, but R is not the correct explanation of A
  • C A is true but R is false
  • D A is false but R is true
Show answer
Answer: Both A and R are true, and R is the correct explanation of A
Why:

1.1² = 1.21, so A is true and follows from R.

Q58 Assertion (A): The perimeter of a semicircle of radius 7 cm is 22 cm. Reason (R): The curved part of a semicircle of radius 7 cm is 22 cm long. intermediate
  • A Both A and R are true, and R is the correct explanation of A
  • B Both A and R are true, but R is not the correct explanation of A
  • C A is true but R is false
  • D A is false but R is true
Show answer
Answer: A is false but R is true
Why:

R is true (π × 7 = 22), but the perimeter also includes the diameter: 22 + 14 = 36. So A is false.

Q59 Consider the statements: 1. A rhombus with diagonals 10 cm and 24 cm has area 120 cm². 2. A triangle with sides 5, 12 and 13 cm has area 60 cm². intermediate
  • A 1 only
  • B 2 only
  • C Both 1 and 2
  • D Neither 1 nor 2
Show answer
Answer: 1 only
Why:

½ × 10 × 24 = 120, true. The 5-12-13 triangle is right-angled: ½ × 5 × 12 = 30, so 2 is false.

Q60 Which of these statements is/are true? 1. Doubling the length and breadth of a rectangle doubles its area. 2. A 90° sector is one quarter of the circle. beginner
  • A 1 only
  • B 2 only
  • C Both 1 and 2
  • D Neither 1 nor 2
Show answer
Answer: 2 only
Why:

Doubling both makes the area 4 times, so 1 is false. 90/360 = ¼, so 2 is true.

Important short questions (4)

Find the area of a triangle with sides 9 cm, 10 cm and 17 cm.

s = 18; area = √(18 × 9 × 8 × 1) = √1296 = 36 cm².

A circular field has circumference 132 m. Find its area (π = 22/7).

2πr = 132 gives r = 21; area = (22/7) × 441 = 1386 m².

Two roads 5 m wide cross at the centre of a 100 m × 80 m field, parallel to the sides. Find their area.

5 × 100 + 5 × 80 − 5 × 5 = 875 m².

Find the area of a sector of angle 45° in a circle of radius 28 cm.

(45/360) × (22/7) × 784 = ⅛ × 2464 = 308 cm².

Long answer questions (1)

A rectangular lawn is 50 m by 35 m. A path 2.5 m wide runs all round it outside, and a circular flower bed of radius 7 m is made in the middle of the lawn. (a) Find the area of the path. (b) Find the cost of laying the path at ₹40 per m². (c) Find the area of the lawn left for grass.

(a) Outer 55 × 40 = 2200; path = 2200 − 1750 = 450 m².

(b) 450 × 40 = ₹18,000.

(c) Bed = (22/7) × 49 = 154; grass = 1750 − 154 = 1596 m².

Flashcards (15)

Area of a rectangle 12 × 7?
84
Perimeter of a rectangle 12 × 7?
38
Area of a square with diagonal 10?
50
Area of a circle of radius 7 (π = 22/7)?
154
Circumference for radius 14?
88
Perimeter of a semicircle of radius 7?
36
Heron's s for sides 5, 12, 13?
15
Area of an equilateral triangle of side 4?
4√3
Rhombus with diagonals 10 and 8: area?
40
Trapezium sides 10 and 6, height 4: area?
32
Sector of 90° in a circle of radius 14: area?
154
Ring with R = 14, r = 7: area?
462
Square side +20%: area change?
+44%
Radius doubled: area becomes?
4 times
Distance in one turn of a wheel of radius r?
2πr

Play this topic

  • Area check

Volume and Surface Area of Solids

Volume, curved and total surface area of cuboids, cubes, cylinders, cones, spheres and hemispheres; melting and recasting, water-level rise, percentage change in volume, composite solids, and the painted-cube and frustum questions of harder exams.

What you will be able to do

  • I can find the volume, surface area and diagonal of a cuboid and a cube.
  • I can work with cylinders, cones, spheres and hemispheres, including slant height.
  • I can solve melting, recasting and water-level questions by equating volumes.
  • I can find surface areas and volumes of composite solids such as tents and capsules.
  • I can find the percentage change and ratio of volumes when dimensions change.

Key points

  • Cuboid: V = lbh, TSA = 2(lb + bh + hl), four walls = 2h(l + b), diagonal = √(l² + b² + h²).

  • Cube of side a: V = a³, TSA = 6a², diagonal = a√3.

  • Cylinder: V = πr²h, curved surface 2πrh, total 2πr(r + h).

  • Cone: V = ⅓πr²h, slant l = √(r² + h²), curved πrl, total πr(l + r).

  • Sphere: V = ⁴⁄₃πr³, S = 4πr². Hemisphere: V = ⅔πr³, curved 2πr², total 3πr².

  • Melting and recasting keeps the volume the same; the surface area changes.

  • Rise in water level = volume put in ÷ base area of the vessel.

  • 1 m³ = 1000 litres and 1 litre = 1000 cm³.

  • If every length is multiplied by k, area becomes k² times and volume k³ times.

  • Cone, hemisphere and cylinder of the same radius r and height r have volumes in the ratio 1 : 2 : 3.

Full explanation

Raju's ice-cream cart

Gift boxes, a can and a ball arranged on a table.

Raju sells kulfi in cones and ice cream from a cylindrical tub. He wants to know how many cones one tub fills, how much paper wraps each cone, and what happens to his cost if he makes the tub 10% wider.

Each question is about a solid: how much it holds (volume) and how much covers it (surface area).

Four formula cards: cuboid and cube, cylinder, cone, and sphere with hemisphere.

Cuboid and cube

A cuboid of length $l$, breadth $b$, height $h$:

  • Volume $= lbh$
  • Total surface $= 2(lb + bh + hl)$
  • Four walls $= 2h(l + b)$
  • Longest diagonal $= \sqrt{l^2 + b^2 + h^2}$

A cube of side $a$: volume $a^3$, surface $6a^2$, diagonal $a\sqrt{3}$.

A 12 by 4 by 3 cuboid with its longest diagonal drawn in red; the diagonal is 13, the volume 144 and the total surface 192.

Cylinder

A cylinder is a stack of circles: volume $= \pi r^2 h$. Unroll its side and you get a rectangle $2\pi r$ by $h$, so the curved surface is $2\pi r h$. Add two circular ends for the total: $2\pi r(r + h)$.

A cylinder of radius 7 and height 10; its volume is 1540, curved surface 440 and total surface 748.

Cone

A cone holds one-third of the cylinder with the same base and height: $V = \frac{1}{3}\pi r^2 h$. Its slant height comes from the right triangle inside it: $l = \sqrt{r^2 + h^2}$. Curved surface $= \pi r l$; total $= \pi r(l + r)$.

A cone with radius 7, height 24 and slant height 25; volume 1232 and curved surface 550.

Sphere and hemisphere

  • Sphere: $V = \frac{4}{3}\pi r^3$, $S = 4\pi r^2$.
  • Hemisphere: $V = \frac{2}{3}\pi r^3$, curved $2\pi r^2$, total $3\pi r^2$ (curved plus the flat circle).

A sphere of radius 21; volume 38,808 and surface 5544.

Melting and recasting

Metal melted into a new shape keeps its volume. A sphere of radius 6 recast into a cylinder of radius 4:

$$\tfrac{4}{3}\pi \cdot 6^3 = \pi \cdot 4^2 \cdot h \Rightarrow h = 18$$

A ball of radius 12 makes $(12/3)^3 = 64$ balls of radius 3. Volume ratio always goes as the cube of the length ratio.

Water level rise

Drop a solid into a vessel and the water rises by (volume of the solid) ÷ (base area of the vessel). A cube of side 20 cm in a 50 cm × 40 cm tank raises the water by $8000 \div 2000 = 4$ cm.

Composite solids

Add the volumes of the parts. For the surface, add only the faces you can touch from outside. A tent (cylinder plus cone) uses $2\pi r h + \pi r l$ of canvas: no base and no joining circle.

A tent made of a cylinder of radius 6 and height 9 with a cone of height 8 on top; slant 10, canvas 528.

Percentage change in volume

Scale every length by $k$ and volume scales by $k^3$. Cube side up 10%: $1.1^3 = 1.331$, so volume up 33.1%. Cylinder radius up 10% with the same height: $1.1^2 = 1.21$, volume up 21%.

Two harder ideas

Painted cube. A cube painted outside and cut into $n^3$ small cubes has 8 cubes with three painted faces, $12(n - 2)$ with two, $6(n - 2)^2$ with one and $(n - 2)^3$ with none.

Frustum. A cone with its top cut off: $V = \frac{\pi h}{3}(R^2 + r^2 + Rr)$, curved surface $\pi(R + r)l$ with $l = \sqrt{h^2 + (R - r)^2}$.

Raju's answers

A tub of radius 14 cm and height 21 cm holds $\frac{22}{7} \cdot 196 \cdot 21 = 12{,}936$ cm³; each cone of radius 3.5 cm and height 12 cm holds 154 cm³, so one tub fills 84 cones. Each cone has slant height $\sqrt{3.5^2 + 12^2} = 12.5$ cm, so its paper wrap is $\pi r l = \frac{22}{7} \times 3.5 \times 12.5 = 137.5$ cm². A tub with a 10% larger radius holds 21% more. Volume questions are about one idea: find the space, then compare.

Mind map: cuboid and cube, cylinder, cone, sphere, recasting and composites around Solids.

Concepts and terms

Volume

The space a solid fills, in cubic units. For any prism or cylinder it is base area × height.

Curved vs total surface

Curved surface leaves out the flat ends; total surface adds them. A pipe or a tent uses only the curved part.

Slant height

In a cone, the radius, height and slant height form a right triangle: l² = r² + h².

Conservation of volume

When a solid is melted and recast, or water is displaced, the volume before equals the volume after.

Scaling

Lengths scale by k, areas by k², volumes by k³.

Did you know

Archimedes proved that a sphere has exactly two-thirds the volume of the cylinder that just encloses it, and asked for this to be carved on his tomb.

A 1-litre carton holds exactly 1000 cm³, a cube 10 cm on each side.

Memory tricks

Cone vs cylinder — A cone is one-third of its cylinder — Same base and height: three cones of water fill one cylinder.

Sphere — Four-thirds pi r cubed, four pi r squared — Volume has the cube, surface has the square, both carry a 4.

Cone, hemisphere, cylinder — 1 : 2 : 3 — With radius r and height r, the volumes are ⅓, ⅔ and 1 times πr³.

Practice MCQs with answers (79)

Q1 The cuboid shown is 12 cm long, 8 cm wide and 5 cm high. Find its volume (in cm³). ![A cuboid with length 12, breadth 8 and height 5 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cub-1.svg) beginner
  • A 480
  • B 392
  • C 576
  • D 960
Show answer
Answer: 480
Why:

V = 12 × 8 × 5 = 480 cm³.

Q2 The cuboid shown is 10 cm long, 6 cm wide and 4 cm high. Find its total surface area (in cm²). ![A cuboid with length 10, breadth 6 and height 4 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cub-2.svg) beginner
  • A 240
  • B 248
  • C 128
  • D 124
Show answer
Answer: 248
Why:

2(60 + 24 + 40) = 2 × 124 = 248 cm².

Q3 The cuboid shown is 15 cm long, 10 cm wide and 6 cm high. Find the area of its four side faces, leaving out the top and bottom (in cm²). ![A cuboid with length 15, breadth 10 and height 6 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cub-3.svg) intermediate
  • A 450
  • B 600
  • C 300
  • D 150
Show answer
Answer: 300
Why:

Four walls = 2h(l + b) = 2 × 6 × 25 = 300 cm².

Q4 The cuboid shown is 12 cm long, 4 cm wide and 3 cm high. Find the length of the diagonal shown in red (in cm). ![A cuboid with length 12, breadth 4 and height 3 marked; the space diagonal is drawn in red and marked ?.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cub-4.svg) beginner
  • A 19
  • B 15
  • C 12
  • D 13
Show answer
Answer: 13
Why:

d = √(12² + 4² + 3²) = √169 = 13 cm.

Q5 The cuboid shown is 9 cm long, 6 cm wide and 2 cm high. Find the length of the diagonal shown in red (in cm). ![A cuboid with length 9, breadth 6 and height 2 marked; the space diagonal is drawn in red and marked ?.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cub-5.svg) intermediate
  • A 11
  • B 13
  • C 10
  • D 17
Show answer
Answer: 11
Why:

d = √(9² + 6² + 2²) = √121 = 11 cm.

Q6 The cuboid shown is 12 cm long, 9 cm wide and 8 cm high. Find the length of the diagonal shown in red (in cm). ![A cuboid with length 12, breadth 9 and height 8 marked; the space diagonal is drawn in red and marked ?.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cub-6.svg) intermediate
  • A 16
  • B 17
  • C 29
  • D 19
Show answer
Answer: 17
Why:

d = √(12² + 9² + 8²) = √289 = 17 cm.

Q7 The cuboid shown is 20 cm long, 14 cm wide and 7 cm high. Find its total surface area (in cm²). ![A cuboid with length 20, breadth 14 and height 7 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cub-7.svg) intermediate
  • A 1960
  • B 476
  • C 1036
  • D 518
Show answer
Answer: 1036
Why:

2(280 + 98 + 140) = 2 × 518 = 1036 cm².

Q8 Find the volume of the cube shown, of side 7 cm (in cm³). ![A cube with edge 7 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cube-1.svg) beginner
  • A 294
  • B 147
  • C 49
  • D 343
Show answer
Answer: 343
Why:

a³ = 7³ = 343 cm³.

Q9 Find the total surface area of the cube shown, of side 9 cm (in cm²). ![A cube with edge 9 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cube-2.svg) beginner
  • A 486
  • B 81
  • C 324
  • D 729
Show answer
Answer: 486
Why:

6a² = 6 × 81 = 486 cm².

Q10 The total surface area of the cube shown is 216 cm². Find its volume (in cm³). ![A cube with its edge marked a.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cube-3.svg) intermediate
  • A 343
  • B 216
  • C 36
  • D 252
Show answer
Answer: 216
Why:

6a² = 216 gives a² = 36, a = 6; volume = 216 cm³.

Q11 The total surface area of the cube shown is 726 cm². Find its volume (in cm³). ![A cube with its edge marked a.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cube-4.svg) intermediate
  • A 1452
  • B 1728
  • C 1331
  • D 726
Show answer
Answer: 1331
Why:

6a² = 726 gives a² = 121, a = 11; volume = 1331 cm³.

Q12 Find the length of the longest diagonal of a cube of side 4 cm. beginner
  • A 4√2 cm
  • B 12 cm
  • C 4√6 cm
  • D 4√3 cm
Show answer
Answer: 4√3 cm
Why:

d = a√3 = 4√3 cm.

Q13 A room is 6 m long, 4 m wide and 3 m high. Find the cost of painting its four walls at ₹25 per m² (in ₹). ![A room drawn as a cuboid with length 6, breadth 4 and height 3.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-room-1.svg) intermediate
  • A 1500
  • B 2700
  • C 750
  • D 2100
Show answer
Answer: 1500
Why:

Walls = 2 × 3 × (6 + 4) = 60 m²; cost = 60 × 25 = ₹1500.

Q14 A room is 8 m long, 5 m wide and 4 m high. Find the cost of painting its four walls at ₹30 per m² (in ₹). ![A room drawn as a cuboid with length 8, breadth 5 and height 4.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-room-2.svg) intermediate
  • A 4320
  • B 3120
  • C 5520
  • D 1560
Show answer
Answer: 3120
Why:

Walls = 2 × 4 × (8 + 5) = 104 m²; cost = 104 × 30 = ₹3120.

Q15 How many bricks of size 25 × 12 × 8 cm are needed to build a wall 1000 × 30 × 240 cm, ignoring mortar? intermediate
  • A 1500
  • B 6000
  • C 3000
  • D 3100
Show answer
Answer: 3000
Why:

Wall = 72,00,000 cm³; brick = 2400 cm³; 72,00,000 ÷ 2400 = 3000.

Q16 How many bricks of size 20 × 10 × 7.5 cm are needed to build a wall 800 × 50 × 600 cm, ignoring mortar? advanced
  • A 16,100
  • B 8000
  • C 32,000
  • D 16,000
Show answer
Answer: 16,000
Why:

Wall = 2,40,00,000 cm³; brick = 1500 cm³; 2,40,00,000 ÷ 1500 = 16,000.

Q17 A cuboidal tank is 2 m × 1.5 m × 1 m. How many litres of water can it hold? beginner
  • A 3000
  • B 300
  • C 30,000
  • D 3
Show answer
Answer: 3000
Why:

2 × 1.5 × 1 = 3 m³; × 1000 = 3000 litres.

Q18 A cylindrical tank has radius 0.7 m and height 2 m. How many litres of water can it hold? (π = 22/7) intermediate
  • A 308
  • B 3080
  • C 30,800
  • D 3.08
Show answer
Answer: 3080
Why:

(22/7) × 0.7² × 2 = 3.08 m³; × 1000 = 3080 litres.

Q19 A cylinder has radius 7 cm and height 10 cm, as shown. Find its volume (in cm³). Take π = 22/7. ![A cylinder with radius 7 and height 10 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cyl-1.svg) beginner
  • A 770
  • B 440
  • C 1540
  • D 3080
Show answer
Answer: 1540
Why:

πr²h = (22/7) × 49 × 10 = 1540 cm³.

Q20 A cylinder has radius 14 cm and height 20 cm, as shown. Find its curved surface area (in cm²). Take π = 22/7. ![A cylinder with radius 14 and height 20 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cyl-2.svg) beginner
  • A 2992
  • B 880
  • C 12,320
  • D 1760
Show answer
Answer: 1760
Why:

2πrh = 2 × (22/7) × 14 × 20 = 1760 cm².

Q21 A cylinder has radius 7 cm and height 15 cm, as shown. Find its total surface area (in cm²). Take π = 22/7. ![A cylinder with radius 7 and height 15 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cyl-3.svg) intermediate
  • A 968
  • B 814
  • C 1276
  • D 660
Show answer
Answer: 968
Why:

2πr(r + h) = 2 × (22/7) × 7 × 22 = 968 cm².

Q22 A cylinder has radius 3.5 cm and height 8 cm, as shown. Find its volume (in cm³). Take π = 22/7. ![A cylinder with radius 3.5 and height 8 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cyl-4.svg) intermediate
  • A 616
  • B 308
  • C 154
  • D 176
Show answer
Answer: 308
Why:

πr²h = (22/7) × 12.25 × 8 = 308 cm³.

Q23 A cylinder has radius 21 cm and height 10 cm, as shown. Find its curved surface area (in cm²). Take π = 22/7. ![A cylinder with radius 21 and height 10 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cyl-5.svg) beginner
  • A 13,860
  • B 4092
  • C 1320
  • D 660
Show answer
Answer: 1320
Why:

2πrh = 2 × (22/7) × 21 × 10 = 1320 cm².

Q24 A cylinder has radius 14 cm and height 6 cm, as shown. Find its total surface area (in cm²). Take π = 22/7. ![A cylinder with radius 14 and height 6 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cyl-6.svg) advanced
  • A 2992
  • B 528
  • C 1144
  • D 1760
Show answer
Answer: 1760
Why:

2πr(r + h) = 2 × (22/7) × 14 × 20 = 1760 cm².

Q25 A cylinder of height 9 cm has volume 1386 cm³. Find its radius (in cm). Take π = 22/7. intermediate
  • A 7
  • B 49
  • C 9
  • D 14
Show answer
Answer: 7
Why:

r² = 1386 ÷ ((22/7) × 9) = 49, so r = 7 cm.

Q26 A cone has radius 7 cm and height 24 cm, as shown. Find its slant height (in cm). ![A cone with radius 7 and height 24 marked; the slant height is marked ?.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cone-1.svg) beginner
  • A 26
  • B 25
  • C 23
  • D 31
Show answer
Answer: 25
Why:

l = √(7² + 24²) = √625 = 25 cm.

Q27 A cone has radius 21 cm and height 20 cm, as shown. Find its slant height (in cm). ![A cone with radius 21 and height 20 marked; the slant height is marked ?.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cone-2.svg) beginner
  • A 30
  • B 21
  • C 29
  • D 41
Show answer
Answer: 29
Why:

l = √(21² + 20²) = √841 = 29 cm.

Q28 A cone has radius 7 cm and height 24 cm, as shown. Find its volume (in cm³). Take π = 22/7. ![A cone with radius 7 and height 24 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cone-3.svg) intermediate
  • A 1283
  • B 550
  • C 3696
  • D 1232
Show answer
Answer: 1232
Why:

⅓πr²h = ⅓ × (22/7) × 49 × 24 = 1232 cm³.

Q29 A cone has radius 21 cm and slant height 29 cm, as shown. Find its curved surface area (in cm²). Take π = 22/7. ![A cone with radius 21 and slant height 29 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cone-4.svg) intermediate
  • A 1914
  • B 3828
  • C 3300
  • D 1320
Show answer
Answer: 1914
Why:

πrl = (22/7) × 21 × 29 = 1914 cm².

Q30 A cone has radius 14 cm and height 48 cm, as shown. Find its volume (in cm³). Take π = 22/7. ![A cone with radius 14 and height 48 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cone-5.svg) advanced
  • A 29,568
  • B 9856
  • C 10,267
  • D 2200
Show answer
Answer: 9856
Why:

⅓πr²h = ⅓ × (22/7) × 196 × 48 = 9856 cm³.

Q31 A cone has radius 7 cm and height 24 cm, as shown. Find its total surface area (in cm²). Take π = 22/7. ![A cone with radius 7 and height 24 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cone-6.svg) advanced
  • A 858
  • B 550
  • C 704
  • D 682
Show answer
Answer: 704
Why:

l = √(7² + 24²) = 25; πr(l + r) = (22/7) × 7 × 32 = 704 cm².

Q32 A cone has radius 28 cm and slant height 35 cm, as shown. Find its curved surface area (in cm²). Take π = 22/7. ![A cone with radius 28 and slant height 35 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-cone-7.svg) intermediate
  • A 5544
  • B 1848
  • C 6160
  • D 3080
Show answer
Answer: 3080
Why:

πrl = (22/7) × 28 × 35 = 3080 cm².

Q33 A sphere has radius 21 cm, as shown. Find its volume (in cm³). Take π = 22/7. ![A sphere with radius 21 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-sph-1.svg) intermediate
  • A 38,808
  • B 5544
  • C 19,404
  • D 29,106
Show answer
Answer: 38,808
Why:

⁴⁄₃πr³ = ⁴⁄₃ × (22/7) × 9261 = 38,808 cm³.

Q34 A sphere has radius 7 cm, as shown. Find its surface area (in cm²). Take π = 22/7. ![A sphere with radius 7 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-sph-2.svg) beginner
  • A 462
  • B 616
  • C 308
  • D 154
Show answer
Answer: 616
Why:

4πr² = 4 × (22/7) × 49 = 616 cm².

Q35 A sphere has radius 3.5 cm, as shown. Find its surface area (in cm²). Take π = 22/7. ![A sphere with radius 3.5 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-sph-3.svg) intermediate
  • A 77
  • B 38.5
  • C 154
  • D 115.5
Show answer
Answer: 154
Why:

4πr² = 4 × (22/7) × 12.25 = 154 cm².

Q36 A solid hemisphere has radius 21 cm, as shown. Find its volume (in cm³). Take π = 22/7. ![A solid hemisphere with radius 21 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-sph-4.svg) intermediate
  • A 29,106
  • B 38,808
  • C 4158
  • D 19,404
Show answer
Answer: 19,404
Why:

⅔πr³ = ⅔ × (22/7) × 9261 = 19,404 cm³.

Q37 A solid hemisphere has radius 7 cm, as shown. Find its total surface area (in cm²). Take π = 22/7. ![A solid hemisphere with radius 7 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-sph-5.svg) intermediate
  • A 462
  • B 154
  • C 308
  • D 616
Show answer
Answer: 462
Why:

Curved 2πr² + flat πr² = 3πr² = 3 × (22/7) × 49 = 462 cm².

Q38 A solid hemisphere has radius 14 cm, as shown. Find its curved surface area (in cm²). Take π = 22/7. ![A solid hemisphere with radius 14 marked.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-sph-6.svg) beginner
  • A 616
  • B 1232
  • C 1848
  • D 2464
Show answer
Answer: 1232
Why:

2πr² = 2 × (22/7) × 196 = 1232 cm².

Q39 A metal ball of radius 12 cm is melted and made into small balls of radius 3 cm. How many small balls are made? intermediate
  • A 16
  • B 4
  • C 64
  • D 12
Show answer
Answer: 64
Why:

Number = (R/r)³ = (12/3)³ = 64.

Q40 A metal ball of radius 10 cm is melted and made into small balls of radius 2 cm. How many small balls are made? beginner
  • A 15
  • B 25
  • C 5
  • D 125
Show answer
Answer: 125
Why:

Number = (R/r)³ = (10/2)³ = 125.

Q41 A solid sphere of radius 6 cm is melted and recast into a cylinder of radius 4 cm. Find the height of the cylinder (in cm). intermediate
  • A 18
  • B 27
  • C 12
  • D 13.5
Show answer
Answer: 18
Why:

⁴⁄₃π × 6³ = π × 4² × h, so h = 4 × 216 ÷ (3 × 16) = 18 cm.

Q42 Three metal cubes of sides 3 cm, 4 cm and 5 cm are melted into one cube. Find the side of the new cube (in cm). intermediate
  • A 7
  • B 6
  • C 5
  • D 12
Show answer
Answer: 6
Why:

3³ + 4³ + 5³ = 27 + 64 + 125 = 216 = 6³, so the side is 6 cm.

Q43 Three metal cubes of sides 6 cm, 8 cm and 10 cm are melted into one cube. Find the side of the new cube (in cm). advanced
  • A 13
  • B 11
  • C 12
  • D 24
Show answer
Answer: 12
Why:

6³ + 8³ + 10³ = 216 + 512 + 1000 = 1728 = 12³, so the side is 12 cm.

Q44 A copper sphere of radius 3 cm is drawn into a wire of radius 1 mm. Find the length of the wire (in m). advanced
  • A 27
  • B 360
  • C 3.6
  • D 36
Show answer
Answer: 36
Why:

r = 0.1 cm. Length = ⁴⁄₃ × 27 ÷ 0.01 = 3600 cm = 36 m.

Q45 A solid cone of radius 6 cm and height 24 cm is melted and recast into a sphere. Find the radius of the sphere (in cm). advanced
  • A 6
  • B 8
  • C 12
  • D 7
Show answer
Answer: 6
Why:

⅓πr²h = ⁴⁄₃πR³ gives R³ = r²h/4 = 864/4 = 216, so R = 6 cm.

Q46 How many cones of radius 3 cm and height 5 cm can be made by melting a cylinder of radius 6 cm and height 15 cm? intermediate
  • A 24
  • B 36
  • C 12
  • D 72
Show answer
Answer: 36
Why:

Number = cylinder ÷ cone = π × 36 × 15 ÷ (⅓π × 9 × 5) = 3 × 36 × 15 ÷ (9 × 5) = 36.

Q47 2 iron balls, each of radius 3 cm, are dropped into a cylindrical jar of radius 6 cm partly filled with water. By how many cm does the water level rise? intermediate
  • A 1.5
  • B 6
  • C 2
  • D 1
Show answer
Answer: 2
Why:

Rise = 2 × ⁴⁄₃π3³ ÷ (π × 6²) = 72 ÷ 36 = 2 cm.

Q48 16 iron balls, each of radius 3 cm, are dropped into a cylindrical jar of radius 12 cm partly filled with water. By how many cm does the water level rise? advanced
  • A 2
  • B 3
  • C 12
  • D 4
Show answer
Answer: 4
Why:

Rise = 16 × ⁴⁄₃π3³ ÷ (π × 12²) = 576 ÷ 144 = 4 cm.

Q49 A cuboidal tank with base 50 cm × 40 cm contains water. A solid cube of side 20 cm is put in and is fully covered. By how many cm does the water rise? intermediate
  • A 4
  • B 16
  • C 2
  • D 8
Show answer
Answer: 4
Why:

Rise = 8000 ÷ (50 × 40) = 4 cm.

Q50 Each edge of a cube is increased by 10%. By what percent does its volume change? intermediate
  • A +30%
  • B +33.1%
  • C +10%
  • D +21%
Show answer
Answer: +33.1%
Why:

Volume factor = (1.1)³ = 1331/1000, a change of +33.1%.

Q51 The radius of a sphere is increased by 20%. By what percent does its volume change? advanced
  • A +44%
  • B +60%
  • C +72.8%
  • D +20%
Show answer
Answer: +72.8%
Why:

Volume factor = (1.2)³ = 216/125, a change of +72.8%.

Q52 The radius of a cylinder is increased by 10% and its height is unchanged. By what percent does its volume change? intermediate
  • A +10%
  • B +33.1%
  • C +20%
  • D +21%
Show answer
Answer: +21%
Why:

Volume factor = (1.1)² = 1.21, a change of +21%.

Q53 Each edge of a cube is decreased by 10%. By what percent does its volume change? advanced
  • A -27.1%
  • B -30%
  • C -10%
  • D -19%
Show answer
Answer: -27.1%
Why:

Volume factor = (0.9)³ = 729/1000, a change of -27.1%.

Q54 The radii of two cylinders are in the ratio 2 : 3 and their heights in the ratio 5 : 4. Find the ratio of their volumes. intermediate
  • A 4 : 9
  • B 5 : 9
  • C 16 : 45
  • D 5 : 6
Show answer
Answer: 5 : 9
Why:

V ∝ r²h: 4 × 5 : 9 × 4 = 5 : 9.

Q55 The radii of two cones are in the ratio 3 : 4 and their heights in the ratio 4 : 3. Find the ratio of their volumes. advanced
  • A 9 : 16
  • B 27 : 64
  • C 3 : 4
  • D 1 : 1
Show answer
Answer: 3 : 4
Why:

V ∝ r²h: 9 × 4 : 16 × 3 = 3 : 4.

Q56 A wooden cube is painted on all faces and then cut into 64 equal small cubes. How many small cubes have exactly three painted faces? advanced
  • A 56
  • B 14
  • C 24
  • D 8
Show answer
Answer: 8
Why:

n = 4: 8 corner cubes.

Q57 A wooden cube is painted on all faces and then cut into 125 equal small cubes. How many small cubes have exactly one painted face? advanced
  • A 54
  • B 27
  • C 8
  • D 36
Show answer
Answer: 54
Why:

n = 5: 6 faces × (n − 2)² = 6 × 9 = 54.

Q58 A wooden cube is painted on all faces and then cut into 216 equal small cubes. How many small cubes have no painted faces? advanced
  • A 96
  • B 64
  • C 8
  • D 48
Show answer
Answer: 64
Why:

n = 6: (n − 2)³ = 4³ = 64.

Q59 A wooden cube is painted on all faces and then cut into 125 equal small cubes. How many small cubes have exactly two painted faces? advanced
  • A 27
  • B 8
  • C 36
  • D 54
Show answer
Answer: 36
Why:

n = 5: 12 edges × (n − 2) = 12 × 3 = 36.

Q60 A bucket is a frustum of a cone with end radii 14 cm and 7 cm and depth 6 cm. Find its capacity (in cm³). Take π = 22/7. advanced
  • A 6468
  • B 2079
  • C 1540
  • D 2156
Show answer
Answer: 2156
Why:

(πh/3)(R² + r² + Rr) = (22/7) × (6/3) × (196 + 49 + 98) = 2156 cm³.

Q61 A frustum of a cone has end radii 14 cm and 7 cm and height 24 cm. Find its curved surface area (in cm²). Take π = 22/7. advanced
  • A 1650
  • B 550
  • C 2420
  • D 1584
Show answer
Answer: 1650
Why:

l = √(24² + 7²) = 25; π(R + r)l = (22/7) × 21 × 25 = 1650 cm².

Q62 A metal pipe is 21 cm long. Its outer radius is 4 cm and inner radius 3 cm. Find the volume of metal in it (in cm³). Take π = 22/7. intermediate
  • A 1056
  • B 462
  • C 1650
  • D 66
Show answer
Answer: 462
Why:

π × 21 × (4² − 3²) = (22/7) × 21 × 7 = 462 cm³.

Q63 A solid is a cylinder of radius 7 cm and height 10 cm with a hemisphere of the same radius on top, as shown. Find its outer surface area, leaving out the flat base (in cm²). Take π = 22/7. ![A cylinder of radius 7 and height 10 with a hemisphere on top.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-caps-1.svg) advanced
  • A 902
  • B 1056
  • C 748
  • D 440
Show answer
Answer: 748
Why:

Cylinder side 2πrh = 440 + hemisphere 2πr² = 308; total 748 cm².

Q64 A solid is a cylinder of radius 21 cm and height 20 cm with a hemisphere of the same radius on top, as shown. Find its volume (in cm³). Take π = 22/7. ![A cylinder of radius 21 and height 20 with a hemisphere on top.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-caps-2.svg) advanced
  • A 27,720
  • B 56,826
  • C 66,528
  • D 47,124
Show answer
Answer: 47,124
Why:

Cylinder πr²h = 27,720 + hemisphere ⅔πr³ = 19,404; total 47,124 cm³.

Q65 A tent is a cylinder of radius 6 m and height 9 m with a cone of height 8 m on top, as shown. How much canvas is needed, leaving out the floor (in m²)? Take π = 22/7. ![A tent: cylinder of radius 6 and height 9 with a cone of height 8 on top.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-tent-1.svg) advanced
  • A 528
  • B 490
  • C 641
  • D 717
Show answer
Answer: 528
Why:

Slant = √(6² + 8²) = 10. Canvas = 2πrh + πrl = (22/7) × 6 × (18 + 10) = 528 m².

Q66 A tent is a cylinder of radius 21 m and height 5 m with a cone of height 20 m on top, as shown. Find the volume of air inside it (in m³). Take π = 22/7. ![A tent: cylinder of radius 21 and height 5 with a cone of height 20 on top.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-tent-2.svg) advanced
  • A 6930
  • B 16,170
  • C 34,650
  • D 20,328
Show answer
Answer: 16,170
Why:

πr²h + ⅓πr²hc = 6930 + 9240 = 16,170 m³.

Q67 A tent is a cylinder of radius 21 m and height 5 m with a cone of height 20 m on top, as shown. How much canvas is needed, leaving out the floor (in m²)? Take π = 22/7. ![A tent: cylinder of radius 21 and height 5 with a cone of height 20 on top.](/syllabus/veda-skills-0-quantitative-aptitude-en/ch15/q-tent-3.svg) advanced
  • A 4488
  • B 1980
  • C 2574
  • D 3960
Show answer
Answer: 2574
Why:

Slant = √(21² + 20²) = 29. Canvas = 2πrh + πrl = (22/7) × 21 × (10 + 29) = 2574 m².

Q68 A well of diameter 7 m is dug 20 m deep. How much earth is taken out (in m³)? Take π = 22/7. intermediate
  • A 385
  • B 3080
  • C 440
  • D 770
Show answer
Answer: 770
Why:

r = 3.5; πr²h = (22/7) × 12.25 × 20 = 770 m³.

Q69 2 cm of rain falls on a flat roof 22 m × 20 m. How many litres of water fall on the roof? intermediate
  • A 8800
  • B 880
  • C 8.8
  • D 88,000
Show answer
Answer: 8800
Why:

22 × 20 × 0.02 = 8.8 m³ = 8800 litres.

Q70 A cone, a hemisphere and a cylinder have the same radius, and the cone and cylinder have height equal to that radius. Their volumes are in the ratio intermediate
  • A 3 : 2 : 1
  • B 1 : 2 : 3
  • C 1 : 1 : 1
  • D 1 : 3 : 2
Show answer
Answer: 1 : 2 : 3
Why:

⅓πr³ : ⅔πr³ : πr³ = 1 : 2 : 3.

Q71 If the radius of a sphere is doubled, its volume becomes beginner
  • A 4 times
  • B 6 times
  • C 8 times
  • D 2 times
Show answer
Answer: 8 times
Why:

Volume goes as r³ and 2³ = 8.

Q72 A cone and a cylinder have the same base and the same height. The volume of the cylinder is how many times that of the cone? beginner
  • A 4 times
  • B 2 times
  • C 1.5 times
  • D 3 times
Show answer
Answer: 3 times
Why:

V(cone) = ⅓πr²h.

Q73 Two cubes of side 5 cm are joined end to end. What is the total surface area of the cuboid formed (in cm²)? intermediate
  • A 250
  • B 200
  • C 300
  • D 275
Show answer
Answer: 250
Why:

Cuboid 10 × 5 × 5: 2(50 + 25 + 50) = 250. Two faces got hidden at the join.

Q74 When a solid is melted and recast into a different shape, which of these stays the same? beginner
  • A Its surface area
  • B Its volume
  • C Its height
  • D Its perimeter
Show answer
Answer: Its volume
Why:

The amount of metal does not change, so the volume is conserved.

Q75 The surface areas of two spheres are in the ratio 4 : 9. What is the ratio of their volumes? advanced
  • A 16 : 81
  • B 4 : 9
  • C 8 : 27
  • D 2 : 3
Show answer
Answer: 8 : 27
Why:

Radii 2 : 3, so volumes 2³ : 3³ = 8 : 27.

Q76 Assertion (A): The curved surface area of a cone of radius 7 cm and height 24 cm is 528 cm². Reason (R): The curved surface area of a cone is πrl, where l is the slant height. intermediate
  • A Both A and R are true, and R is the correct explanation of A
  • B Both A and R are true, but R is not the correct explanation of A
  • C A is true but R is false
  • D A is false but R is true
Show answer
Answer: A is false but R is true
Why:

R is true. l = 25, so πrl = 550; 528 comes from wrongly using h = 24. So A is false.

Q77 Assertion (A): When the radius of a cylinder is doubled and its height halved, its volume doubles. Reason (R): The volume of a cylinder is proportional to r²h. intermediate
  • A Both A and R are true, and R is the correct explanation of A
  • B Both A and R are true, but R is not the correct explanation of A
  • C A is true but R is false
  • D A is false but R is true
Show answer
Answer: Both A and R are true, and R is the correct explanation of A
Why:

2² × ½ = 2, so A is true and R explains it.

Q78 Consider the statements: 1. The total surface area of a solid hemisphere of radius r is 3πr². 2. A cube of side 3 cm has a longer diagonal than a cuboid 2 cm × 3 cm × 6 cm. intermediate
  • A 1 only
  • B 2 only
  • C Both 1 and 2
  • D Neither 1 nor 2
Show answer
Answer: 1 only
Why:

2πr² + πr² = 3πr², true. Cube diagonal 3√3 ≈ 5.2; cuboid diagonal 7. So 2 is false.

Q79 Which of these statements is/are true? 1. 1 m³ equals 100 litres. 2. A sphere of radius 3 cm can be recast into 27 spheres of radius 1 cm. beginner
  • A 1 only
  • B 2 only
  • C Both 1 and 2
  • D Neither 1 nor 2
Show answer
Answer: 2 only
Why:

1 m³ = 1000 L, so 1 is false. (3/1)³ = 27, so 2 is true.

Important short questions (4)

Find the volume and total surface area of a cuboid 10 cm × 8 cm × 5 cm.

V = 400 cm³; TSA = 2(80 + 40 + 50) = 340 cm².

A cone has radius 21 cm and height 20 cm. Find its slant height and curved surface area.

l = √(441 + 400) = 29; πrl = (22/7) × 21 × 29 = 1914 cm².

How many balls of radius 1 cm can be made from a solid sphere of radius 4 cm?

(4/1)³ = 64.

A cube is painted on all faces and cut into 125 equal cubes. How many have no face painted?

n = 5; (5 − 2)³ = 27.

Long answer questions (1)

A toy is a cone of radius 3.5 cm standing on a hemisphere of the same radius. The total height of the toy is 15.5 cm. (a) Find the height of the cone. (b) Find the volume of the toy. (c) Find its total surface area.

(a) h = 15.5 − 3.5 = 12 cm.

(b) Cone ⅓ × (22/7) × 12.25 × 12 = 154; hemisphere ⅔ × (22/7) × 42.875 = 89.83; total ≈ 243.83 cm³.

(c) l = √(12.25 + 144) = 12.5; cone πrl = 137.5; hemisphere 2πr² = 77; total 214.5 cm².

Flashcards (16)

Volume of a cuboid 6 × 5 × 4?
120
Total surface of a cube of side 5?
150
Diagonal of a cuboid 2 × 3 × 6?
7
Volume of a cylinder r = 7, h = 10?
1540
Curved surface of a cylinder r = 7, h = 10?
440
Slant height of a cone r = 6, h = 8?
10
Volume of a cone r = 7, h = 24?
1232
Surface of a sphere r = 7?
616
Volume of a sphere r = 21?
38,808
Total surface of a hemisphere r = 7?
462
1 m³ in litres?
1000
Balls of radius 1 from one of radius 3?
27
Cube side +10%: volume change?
+33.1%
Cone : hemisphere : cylinder (same r, height r)?
1 : 2 : 3
Cubes with 3 painted faces in a painted cube cut into 64?
8
Frustum volume formula?
(πh/3)(R² + r² + Rr)

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Veda Skills Quantitative Aptitude Unit 15 — frequently asked questions

Find the area of a triangle with sides 9 cm, 10 cm and 17 cm.

s = 18; area = √(18 × 9 × 8 × 1) = √1296 = 36 cm².

A circular field has circumference 132 m. Find its area (π = 22/7).

2πr = 132 gives r = 21; area = (22/7) × 441 = 1386 m².

Two roads 5 m wide cross at the centre of a 100 m × 80 m field, parallel to the sides. Find their area.

5 × 100 + 5 × 80 − 5 × 5 = 875 m².

Find the volume and total surface area of a cuboid 10 cm × 8 cm × 5 cm.

V = 400 cm³; TSA = 2(80 + 40 + 50) = 340 cm².

A cone has radius 21 cm and height 20 cm. Find its slant height and curved surface area.

l = √(441 + 400) = 29; πrl = (22/7) × 21 × 29 = 1914 cm².

How many balls of radius 1 cm can be made from a solid sphere of radius 4 cm?

(4/1)³ = 64.

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