Mensuration
Practice MCQsMensuration is the branch of mathematics that deals with measurement of geometric figures.
Mensuration is the branch of mathematics that deals with measurement of geometric figures. It includes the calculation of area, perimeter, surface area, lateral surface area, total surface area and volume of plane figures and solid figures. This chapter covers areas of squares, rectangles, parallelograms, triangles and circles, areas of figures that can be split into simple figures, and surface area and volume of cuboids, cylinders, cones and spheres.
What is Mensuration?
Mensuration helps us measure the size of figures. For two-dimensional figures, we calculate area and perimeter. For three-dimensional solids, we calculate surface area and volume.
In competitive exams, Mensuration questions usually test formula knowledge, unit conversion, identification of the correct figure, and the ability to split irregular figures into standard shapes such as rectangles, triangles, parallelograms and circles.
| Measurement | Meaning | Common Unit |
|---|---|---|
| Perimeter | Total boundary length of a plane figure. | cm, m, km |
| Area | Surface covered by a two-dimensional figure. | \(\text{cm}^2\), \(\text{m}^2\) |
| Lateral Surface Area | Area of curved or side surface excluding bases. | \(\text{cm}^2\), \(\text{m}^2\) |
| Total Surface Area | Total outer area of a solid including all faces or bases. | \(\text{cm}^2\), \(\text{m}^2\) |
| Volume | Space occupied by a three-dimensional solid. | \(\text{cm}^3\), \(\text{m}^3\) |
“Mensuration becomes easy when the figure is identified correctly and the right formula is applied.”
Key points
- Use area for two-dimensional figures.
- Use volume for three-dimensional solids.
- Use square units for area and surface area.
- Use cubic units for volume.
- Split irregular figures into known figures.
- Use radius, not diameter, in circle formulas.
- Use slant height for cone lateral surface area.
- Keep all units same before calculation.
Core Formula Bank
Tip: For solid figures, distinguish between curved/lateral surface area, total surface area and volume.
Mensuration Formula Selection Guide
Mensuration questions become easier when the student first identifies whether the figure is two-dimensional or three-dimensional, and whether the question asks for boundary, area, surface area or volume. The table below helps choose the correct formula quickly.
| Question Type | What to Use | Typical Clue |
|---|---|---|
| Boundary of 2D figure | Perimeter or circumference | Fence, border, boundary, round path |
| Area of square | \(A=a^2\) | All sides are equal |
| Area of rectangle | \(A=l \times b\) | Length and breadth are given |
| Area of triangle | \(A=\frac{1}{2}bh\) | Base and perpendicular height are given |
| Area of parallelogram | \(A=b \times h\) | Base and perpendicular height are given |
| Area of circle | \(A=\pi r^2\) | Radius or diameter is given |
| Composite figure | Add or subtract standard areas | Figure made of rectangle, triangle, circle or semicircle |
| Field-book type area | Split into triangles, rectangles or strips | Irregular land or field measurement |
| Cuboid or cube volume | \(V=lbh\) or \(V=a^3\) | Box, room, tank, block, cube |
| Cylinder question | \(CSA=2\pi rh\), \(TSA=2\pi r(r+h)\), \(V=\pi r^2h\) | Pipe, drum, can, tank, roller |
| Cone question | \(CSA=\pi rl\), \(TSA=\pi r(l+r)\), \(V=\frac{1}{3}\pi r^2h\) | Conical tent, cone, funnel, heap |
| Sphere question | \(SA=4\pi r^2\), \(V=\frac{4}{3}\pi r^3\) | Ball, globe, spherical object |
| Hemisphere question | \(CSA=2\pi r^2\), \(TSA=3\pi r^2\), \(V=\frac{2}{3}\pi r^3\) | Bowl, dome, half-sphere |
| Unit conversion | Convert before applying formula | Different units like cm, m, hectare, acre |
Areas of Plane Figures
Plane figures are two-dimensional figures. Their size is measured using area, and their boundary is measured using perimeter or circumference.
| Figure | Area Formula | Perimeter / Circumference | Notes |
|---|---|---|---|
| Square | \(A=a^2\) | \(P=4a\) | All sides are equal. |
| Rectangle | \(A=l \times b\) | \(P=2(l+b)\) | Opposite sides are equal. |
| Parallelogram | \(A=b \times h\) | \(P=2(a+b)\) | Use perpendicular height, not slant side. |
| Triangle | \(A=\frac{1}{2}bh\) | \(P=a+b+c\) | Use perpendicular height. |
| Circle | \(A=\pi r^2\) | \(C=2\pi r\) | Diameter \(d=2r\). |
| Semicircle | \(A=\frac{1}{2}\pi r^2\) | \(P=\pi r+2r\) | Perimeter includes curved part and diameter. |
| Quadrant | \(A=\frac{1}{4}\pi r^2\) | \(P=\frac{1}{2}\pi r+2r\) | One-fourth of a circle. |
Areas of Composite Figures and Field Book Type Questions
Some figures are not directly standard shapes. Such figures can be split into squares, rectangles, triangles, parallelograms and circles. This method is useful in field-book and land measurement type problems.
| Composite Figure Type | Method | Example Approach |
|---|---|---|
| Rectangle + Triangle | Find area of rectangle and triangle separately, then add. | Total area \(=lb+\frac{1}{2}bh\) |
| Square with Circular Part Removed | Area of square minus area of circle or semicircle. | Remaining area \(=a^2-\pi r^2\) |
| Field Split into Triangles | Divide field into triangles and add their areas. | Total area \(=\frac{1}{2}b_1h_1+\frac{1}{2}b_2h_2+\cdots\) |
| Field Split into Rectangles | Split irregular land into rectangular strips. | Total area \(=l_1b_1+l_2b_2+\cdots\) |
| Circle + Rectangle | Add circular and rectangular portions. | Total area \(=\pi r^2+lb\) |
Cuboid and Cube
A cuboid is a solid with rectangular faces. A cube is a special cuboid in which all edges are equal.
| Solid | Lateral Surface Area | Total Surface Area | Volume |
|---|---|---|---|
| Cuboid | \(LSA=2h(l+b)\) | \(TSA=2(lb+bh+lh)\) | \(V=lbh\) |
| Cube | \(LSA=4a^2\) | \(TSA=6a^2\) | \(V=a^3\) |
Right Circular Cylinder
A right circular cylinder has two equal circular bases and a uniform height. Examples include pipes, tanks, drums and cans.
| Measurement | Formula | Meaning |
|---|---|---|
| Curved / Lateral Surface Area | \(CSA=LSA=2\pi rh\) | Area of curved side only. |
| Total Surface Area | \(TSA=2\pi r(r+h)\) | Curved area plus two circular bases. |
| Volume | \(V=\pi r^2h\) | Base area multiplied by height. |
Right Circular Cone
A right circular cone has a circular base and a vertex directly above the centre of the base. It has radius \(r\), height \(h\), and slant height \(l\).
| Measurement | Formula | Meaning |
|---|---|---|
| Curved / Lateral Surface Area | \(CSA=LSA=\pi rl\) | Area of curved side only. |
| Total Surface Area | \(TSA=\pi r(l+r)\) | Curved area plus circular base. |
| Volume | \(V=\frac{1}{3}\pi r^2h\) | One-third of cylinder with same base and height. |
Sphere and Hemisphere
A sphere is a perfectly round solid in which every point on the surface is at the same distance from the centre. A hemisphere is half of a sphere.
| Solid | Curved Surface Area | Total Surface Area | Volume |
|---|---|---|---|
| Sphere | \(CSA=4\pi r^2\) | \(TSA=4\pi r^2\) | \(V=\frac{4}{3}\pi r^3\) |
| Hemisphere | \(CSA=2\pi r^2\) | \(TSA=3\pi r^2\) | \(V=\frac{2}{3}\pi r^3\) |
Unit Conversion in Mensuration
Unit conversion is very important in Mensuration. All measurements must be converted to the same unit before applying formulas.
| Conversion Type | Rule | Example |
|---|---|---|
| Length | \(1\text{ m}=100\text{ cm}\) | \(2\text{ m}=200\text{ cm}\) |
| Area | \(1\text{ m}^2=10000\text{ cm}^2\) | Because \(100 \times 100=10000\) |
| Volume | \(1\text{ m}^3=1000000\text{ cm}^3\) | Because \(100 \times 100 \times 100=1000000\) |
| Hectare | \(1\text{ hectare}=10000\text{ m}^2\) | Commonly used in land measurement. |
| Acre | \(1\text{ acre}=43560\text{ ft}^2\) | Used in field and land area questions. |
Step-by-Step Solving Method
| Step | Action | Example Focus |
|---|---|---|
| Step 1 | Identify whether the figure is 2D or 3D. | Square, rectangle, triangle, circle, cuboid, cylinder, cone, sphere. |
| Step 2 | Check what is asked. | Area, perimeter, LSA, CSA, TSA or volume. |
| Step 3 | Convert all units to one common unit. | m to cm, cm to m, hectare to square metre. |
| Step 4 | Apply the correct formula. | Use radius for circle, height for volume and slant height for cone surface area. |
| Step 5 | Write answer with correct unit. | Square units for area, cubic units for volume. |
Solved Examples
| Question | Method | Answer |
|---|---|---|
| Find the area of a square with side 9 cm. | \[ A=a^2=9^2=81 \] | 81 \(\text{cm}^2\) |
| Find the area of a rectangle with length 12 m and breadth 8 m. | \[ A=l \times b=12 \times 8=96 \] | 96 \(\text{m}^2\) |
| Find the area of a parallelogram with base 15 cm and height 6 cm. | \[ A=b \times h=15 \times 6=90 \] | 90 \(\text{cm}^2\) |
| Find the area of a triangle with base 20 cm and height 10 cm. | \[ A=\frac{1}{2}bh=\frac{1}{2}\times 20 \times 10=100 \] | 100 \(\text{cm}^2\) |
| Find the area of a circle with radius 7 cm. Use \(\pi=\frac{22}{7}\). | \[ A=\pi r^2=\frac{22}{7}\times 7 \times 7=154 \] | 154 \(\text{cm}^2\) |
| A figure is made of a rectangle \(10 \times 6\) cm and a triangle of base 10 cm and height 4 cm. Find total area. | \[ \text{Rectangle area}=10 \times 6=60 \] \[ \text{Triangle area}=\frac{1}{2}\times 10 \times 4=20 \] \[ \text{Total area}=60+20=80 \] | 80 \(\text{cm}^2\) |
| Find volume of a cuboid with \(l=8\) cm, \(b=5\) cm, \(h=3\) cm. | \[ V=lbh=8 \times 5 \times 3=120 \] | 120 \(\text{cm}^3\) |
| Find curved surface area of a cylinder with radius 7 cm and height 10 cm. Use \(\pi=\frac{22}{7}\). | \[ CSA=2\pi rh=2 \times \frac{22}{7}\times 7 \times 10=440 \] | 440 \(\text{cm}^2\) |
| Find volume of a cone with radius 7 cm and height 12 cm. Use \(\pi=\frac{22}{7}\). | \[ V=\frac{1}{3}\pi r^2h = \frac{1}{3}\times \frac{22}{7}\times 7 \times 7 \times 12 = 616 \] | 616 \(\text{cm}^3\) |
| Find surface area of a sphere with radius 7 cm. Use \(\pi=\frac{22}{7}\). | \[ SA=4\pi r^2=4 \times \frac{22}{7}\times 7 \times 7=616 \] | 616 \(\text{cm}^2\) |
Note: Always check whether the question asks for area, perimeter, lateral surface area, total surface area or volume.
Common Traps and Shortcuts
Common Traps
- Using diameter instead of radius in circle, cylinder, cone and sphere formulas.
- Forgetting \(\frac{1}{2}\) in triangle area.
- Using slant height instead of vertical height in cone volume.
- Using vertical height instead of slant height in cone surface area.
- Confusing curved surface area with total surface area.
- Writing square units for volume or cubic units for area.
- Not converting all measurements into the same unit.
- Adding areas incorrectly in composite figures.
- Forgetting base area in open or closed solid questions.
Useful Shortcuts
- If diameter is given, use \(r=\frac{d}{2}\).
- For square, area is side squared.
- For rectangle, area is length multiplied by breadth.
- For parallelogram, use perpendicular height.
- For triangle, area is half of base multiplied by height.
- For cylinder volume, use base area multiplied by height.
- Cone volume is one-third of cylinder volume with same base and height.
- Sphere surface area is \(4\pi r^2\).
- Split irregular figures into standard figures before calculation.
Practice
A) Multiple Choice Questions
-
Area of a square with side 8 cm is:
32 \(\text{cm}^2\) 48 \(\text{cm}^2\) 64 \(\text{cm}^2\) 80 \(\text{cm}^2\)
-
Area of a triangle with base 12 cm and height 5 cm is:
20 \(\text{cm}^2\) 30 \(\text{cm}^2\) 60 \(\text{cm}^2\) 120 \(\text{cm}^2\)
-
Volume of a cuboid is:
\(l+b+h\) \(2(lb+bh+lh)\) \(lbh\) \(4a^2\)
-
Curved surface area of a cylinder is:
\(\pi r^2h\) \(2\pi rh\) \(2\pi r(r+h)\) \(\frac{1}{3}\pi r^2h\)
-
Volume of a sphere is:
\(\frac{4}{3}\pi r^3\) \(4\pi r^2\) \(\pi r^2h\) \(\frac{1}{3}\pi r^2h\)
B) Solve the Higher-Order Problems
- Find the area of a rectangle whose length is 18 m and breadth is 7 m. (Hint: \(A=l \times b\).)
- Find the area of a circle with diameter 14 cm. Use \(\pi=\frac{22}{7}\). (Hint: First find radius.)
- A field is made of a rectangle \(20 \times 12\) m and a triangle of base 20 m and height 8 m. Find the total area. (Hint: Add rectangle area and triangle area.)
- Find the total surface area of a cuboid with \(l=6\) cm, \(b=4\) cm and \(h=3\) cm. (Hint: \(TSA=2(lb+bh+lh)\).)
- Find the volume of a cylinder with radius 7 cm and height 15 cm. Use \(\pi=\frac{22}{7}\). (Hint: \(V=\pi r^2h\).)
C) Match the Figure with the Correct Formula
| Figure / Solid | Correct Formula |
|---|---|
| Square Area | \(A=a^2\) |
| Rectangle Area | \(A=l \times b\) |
| Triangle Area | \(A=\frac{1}{2}bh\) |
| Circle Area | \(A=\pi r^2\) |
| Cuboid Volume | \(V=lbh\) |
| Cylinder Curved Surface Area | \(CSA=2\pi rh\) |
| Cone Volume | \(V=\frac{1}{3}\pi r^2h\) |
| Sphere Surface Area | \(SA=4\pi r^2\) |
Mensuration Reminder
Mensuration is based on choosing the correct figure and applying the correct formula. For plane figures, calculate area and perimeter. For solids, calculate lateral or curved surface area, total surface area and volume. In field-book type questions, split the given figure into simple known figures and add or subtract their areas carefully.
Task: Create five Mensuration questions using one question each from plane area, composite area, cuboid, cylinder, cone and sphere.
Show Suggested Answers
Multiple Choice
-
64 \(\text{cm}^2\)
\[ A=a^2=8^2=64 \] -
30 \(\text{cm}^2\)
\[ A=\frac{1}{2}bh=\frac{1}{2}\times 12 \times 5=30 \] -
\(lbh\)
Volume of cuboid \(=lbh\). -
\(2\pi rh\)
Curved surface area of cylinder \(=2\pi rh\). -
\(\frac{4}{3}\pi r^3\)
Volume of sphere \(=\frac{4}{3}\pi r^3\).
Higher-Order Problems
- \[ A=l \times b=18 \times 7=126 \] Answer = 126 \(\text{m}^2\).
- Diameter \(=14\) cm, so radius \(=7\) cm. \[ A=\pi r^2=\frac{22}{7}\times 7 \times 7=154 \] Answer = 154 \(\text{cm}^2\).
- Rectangle area: \[ 20 \times 12=240 \] Triangle area: \[ \frac{1}{2}\times 20 \times 8=80 \] Total area: \[ 240+80=320 \] Answer = 320 \(\text{m}^2\).
- \[ TSA=2(lb+bh+lh) \] \[ =2(6 \times 4+4 \times 3+6 \times 3) \] \[ =2(24+12+18)=108 \] Answer = 108 \(\text{cm}^2\).
- \[ V=\pi r^2h=\frac{22}{7}\times 7 \times 7 \times 15=2310 \] Answer = 2310 \(\text{cm}^3\).
Concept Matching
- Square Area → \(A=a^2\)
- Rectangle Area → \(A=l \times b\)
- Triangle Area → \(A=\frac{1}{2}bh\)
- Circle Area → \(A=\pi r^2\)
- Cuboid Volume → \(V=lbh\)
- Cylinder Curved Surface Area → \(CSA=2\pi rh\)
- Cone Volume → \(V=\frac{1}{3}\pi r^2h\)
- Sphere Surface Area → \(SA=4\pi r^2\)
Clue Explanation
Mensuration questions are solved by identifying the correct figure, selecting the correct formula, substituting values carefully and writing the final answer with the proper unit. Composite figures should be split into standard figures before calculating area.
Exam tips
- Identify whether the figure is 2D or 3D.
- Use square units for area and surface area.
- Use cubic units for volume.
- Convert diameter to radius before using circle formulas.
- Use perpendicular height for triangle and parallelogram.
- Use slant height for cone curved surface area.
- Use vertical height for cone volume.
- Split composite figures into known shapes.
- Check unit conversion before calculation.