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#22 Compulsory Coordinate Geometry, Mensuration & Trigonometry Math 22

Mensuration: 2D Areas, Perimeter & 3D Volumesक्षेत्रमिति (मेन्सुरेशन)

Mastery of 2D and 3D mensuration is critical for the UP Super TET Junior examination, accounting for high-weightage computational and formula-based questions. Aspirants must focus on unit consistency, surface area-to-volume relationships, and rapid calculation techniques for compound figures.

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Syllabus Focus & High-Yield Strategy: Mastery of 2D and 3D mensuration is critical for the UP Super TET Junior examination, accounting for high-weightage computational and formula-based questions. Aspirants must focus on unit consistency, surface area-to-volume relationships, and rapid calculation techniques for compound figures.

1. Fundamental Axioms, Definitions & Foundational Properties

Key Property 1: Perimeter and Area Axioms: Perimeter is the continuous linear boundary length measured in units (m, cm), whereas area is the two-dimensional region enclosed within this boundary measured in square units ($m^2$, $cm^2$).
Key Property 2: Volume and Capacity Axioms: Volume represents the three-dimensional space occupied by a solid object, measured in cubic units ($m^3$, $cm^3$), with the standard conversion $1 \text{ m}^3 = 1000 \text{ liters}$.
Key Property 3: Similarity and Scale Factor Theorem: If two similar 2D figures have a linear scale factor $k$, their perimeters scale by $k$, and their areas scale by $k^2$. For 3D similar solids, volumes scale by $k^3$.
Key Property 4: Additivity Principle: The area of any complex composite polygon or the volume of a compound 3D solid is strictly additive, provided the overlapping boundaries have zero measure (are non-overlapping).
Exam Pro-Tip: Always check that all linear dimensions are converted to the exact same unit (preferably SI units: meters or centimeters) before applying any area or volume formula to avoid classic unit conversion traps.
Memory Mnemonic: L-A-V — Linear is $\text{unit}^1$, Area is $\text{unit}^2$, Volume is $\text{unit}^3$!

2. Formulas, Standard Theorems & Algorithmic Procedures

Category / CaseFormula / Mathematical RuleTopper Calculation Shortcut
Right Circular Cylinder$\text{CSA} = 2\pi rh$, $\text{TSA} = 2\pi r(h+r)$, $V = \pi r^2 h$Factor out $\pi r$ immediately; use $\pi \approx \frac{22}{7}$ when dimensions are multiples of 7.
Right Circular Cone$\text{CSA} = \pi rl$, $\text{TSA} = \pi r(l+r)$, $V = \frac{1}{3}\pi r^2 h$ where $l = \sqrt{r^2+h^2}$Look for Pythagorean triplets $(3,4,5)$ or $(5,12,13)$ for radius, height, and slant height.
Theorem / Formula 1: Heron's Formula for Scalene Triangle: Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ where semi-perimeter $s = \frac{a+b+c}{2}$.
Theorem / Formula 2: Sphere and Hemisphere: Surface area of sphere $= 4\pi r^2$, Volume $= \frac{4}{3}\pi r^3$. Total surface area of solid hemisphere $= 3\pi r^2$.
Theorem / Formula 3: Frustum of a Cone: Volume $V = \frac{1}{3}\pi h (R^2 + r^2 + Rr)$, Curved Surface Area $\text{CSA} = \pi l (R + r)$ where $l = \sqrt{h^2 + (R-r)^2}$.
15-Second Speed-Math Trick: When finding the percentage change in area of a rectangle due to side modifications of $x\%$ and $y\%$, use the net percentage formula $\text{Net Change} = \left(x + y + \frac{xy}{100}\right)\%$.

3. Word Problem Modeling & Practical Applications

Problem Type 1: Melting and Recasting: When a 3D solid of volume $V_1$ is melted and recast into $n$ identical smaller solids of volume $V_2$, equate total volumes: $V_1 = n \times V_2$. Never equate surface areas during melting.
Problem Type 2: Pathways Inside/Outside Rectangular Fields: For a path of uniform width $w$ running inside a rectangular field of dimensions $L \times B$, path area $= 2w(L + B - 2w)$. If running outside, use $+2w$ instead of $-2w$.
Problem Type 3: Flow Rate Through Pipes: Volume of water discharged per second = $\text{Cross-sectional Area} \times \text{Flow Velocity}$. Equate pipe discharge volume to tank capacity to find time required.
Problem Type 4: Composite Solids (Tent/Siloin): A tent formed by a cylinder surmounted by a cone has total surface area equal to $\text{CSA}_{\text{cylinder}} + \text{CSA}_{\text{cone}}$. Do not include internal circular bases.
UP Exam High-Yield Tip: SCERT UP textbook word problems frequently test the conversion of cylindrical wells and rectangular water tanks. Always set up volume equations in terms of $\pi$ and simplify before performing long decimal multiplications.

4. Mathematical Edge Cases, Sign Conventions & Common Traps

Critical Distinction: Lateral Surface Area (LSA) vs Total Surface Area (TSA). LSA excludes the area of top and bottom bases, whereas TSA includes all bounding surfaces.
Edge Case / Exception: For a hollow cylinder with inner radius $r$ and outer radius $R$, $\text{TSA} = 2\pi(R+r)(h + R - r)$, accounting for both curved surfaces and the two flat ring bases.
Common Calculation Trap: Forgetting to add the base area of a solid hemisphere when calculating Total Surface Area, incorrectly using $2\pi r^2$ instead of $3\pi r^2$.
Trap Prevention Trick: Always underline the exact wording in questions asking for 'internal capacity' (volume) versus 'tin coating cost' (surface area) to prevent applying the wrong formula class.
Memory Mnemonic: H-S-T — Hollow cylinder: Subtract radii for thickness, Sum radii for perimeter!

Comprehensive Mathematical Matrix & Method Comparison

Parameter / DimensionMethod / Concept AMethod / Concept BExam Speed & Accuracy Takeaway
Circle vs SquareArea $= \pi r^2$, Perimeter $= 2\pi r$Area $= a^2$, Perimeter $= 4a$For equal perimeter, circle always encloses maximum area.
Cylinder vs SphereVolume $= \pi r^2 h$Volume $= \frac{4}{3}\pi r^3$Sphere gives maximum volume for minimum surface area.
Cone vs CylinderVolume $= \frac{1}{3}\pi r^2 h$Volume $= \pi r^2 h$A cone has exactly one-third the volume of a cylinder of same base and height.
Inner vs Outer PathArea $= 2w(L+B-2w)$Area $= 2w(L+B+2w)$Subtract width term for inner path; add width term for outer path.
Practice Progress: 0 of 10 Answered
Score: 0 / 10
Question 1 of 10 Super TET Junior Mathematics

1. If the radius of a right circular cylinder is decreased by $50\%$ and its height is increased by $50\%$, what is the net percentage change in its volume?

Correct Answer: Option A

Volume of cylinder $V = \pi r^2 h$. Let initial radius be $r$ and height be $h$. New radius $r' = 0.5r$ and new height $h' = 1.5h$. New volume $V' = \pi (0.5r)^2 (1.5h) = \pi (0.25r^2)(1.5h) = 0.375 \pi r^2 h$. Percentage change $= \frac{0.375 - 1}{1} \times 100 = -62.5\%$, representing a decrease of $62.5\%$.

Question 2 of 10 Super TET Junior Mathematics

2. A metallic sphere of radius $6\text{ cm}$ is melted and recast into small conical bullets of radius $2\text{ cm}$ and height $8\text{ cm}$. How many such bullets can be made?

Correct Answer: Option A

Volume of sphere $V_s = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (6)^3$. Volume of one cone $V_c = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (2)^2 (8) = \frac{1}{3}\pi (32)$. Number of bullets $n = \frac{V_s}{V_c} = \frac{\frac{4}{3}\pi (216)}{\frac{1}{3}\pi (32)} = \frac{4 \times 216}{32} = \frac{864}{32} = 27$.

Question 3 of 10 Super TET Junior Mathematics

3. The perimeter of a rectangular garden is $60\text{ m}$ and its length is twice its breadth. What is the area of the garden?

Correct Answer: Option A

Let breadth be $b$, so length $l = 2b$. Perimeter $= 2(l + b) = 2(2b + b) = 6b$. Given $6b = 60 \implies b = 10\text{ m}$ and $l = 20\text{ m}$. Area $= l \times b = 20 \times 10 = 200\text{ m}^2$.

Question 4 of 10 Super TET Junior Mathematics

4. What is the total surface area of a solid hemisphere of radius $7\text{ cm}$?

Correct Answer: Option A

Total surface area of a solid hemisphere $= 3\pi r^2$. Substituting $r = 7$ and $\pi = \frac{22}{7}$: $\text{TSA} = 3 \times \frac{22}{7} \times 7^2 = 3 \times \frac{22}{7} \times 49 = 3 \times 22 \times 7 = 66 \times 7 = 462\text{ cm}^2$.

Question 5 of 10 Super TET Junior Mathematics

5. A path $2\text{ m}$ wide is built outside a rectangular park of length $38\text{ m}$ and breadth $32\text{ m}$. What is the area of the path?

Correct Answer: Option D

The outer dimensions are 38+2(2)=42 m and 32+2(2)=36 m. The path area is 42 x 36 - 38 x 32 = 1512 - 1216 = 296 m^2. The answer is Option D.

Question 6 of 10 Super TET Junior Mathematics

6. The ratio of the radii of two cylinders is $2:3$ and the ratio of their heights is $5:4$. What is the ratio of their volumes?

Correct Answer: Option A

Volume ratio $\frac{V_1}{V_2} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} = \left(\frac{r_1}{r_2}\%2\right) \times \left(\frac{h_1}{h_2}\right) = \left(\frac{2}{3}\right)^2 \times \left(\frac{5}{4}\right) = \frac{4}{9} \times \frac{5}{4} = \frac{5}{9}$.

Question 7 of 10 Super TET Junior Mathematics

7. The diagonal of a square is $10\sqrt{2}\text{ cm}$. What is its perimeter?

Correct Answer: Option A

Diagonal of a square $d = a\sqrt{2}$. Given $a\sqrt{2} = 10\sqrt{2} \implies \text{side } a = 10\text{ cm}$. Perimeter $= 4a = 4 \times 10 = 40\text{ cm}$.

Question 8 of 10 Super TET Junior Mathematics

8. If the curved surface area of a right circular cylinder is $880\text{ cm}^2$ and its height is $10\text{ cm}$, what is its radius? (Use $\pi = \frac{22}{7}$)

Correct Answer: Option A

Curved Surface Area $\text{CSA} = 2\pi rh$. Given $2 \times \frac{22}{7} \times r \times 10 = 880 \implies \frac{440}{7} r = 880 \implies r = \frac{880 \times 7}{440} = 2 \times 7 = 14\text{ cm}$.

Question 9 of 10 Super TET Junior Mathematics

9. What is the length of the longest pole that can be placed in a room of dimensions $12\text{ m} \times 9\text{ m} \times 8\text{ m}$?

Correct Answer: Option A

The longest pole length is equal to the space diagonal of the cuboid: $d = \sqrt{l^2 + b^2 + h^2} = \sqrt{12^2 + 9^2 + 8^2} = \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m}$.

Question 10 of 10 Super TET Junior Mathematics

10. The perimeter of a semicircle is $36\text{ cm}$. What is its radius? (Use $\pi = \frac{22}{7}$)

Correct Answer: Option A

Perimeter of semicircle $= \pi r + 2r = r(\pi + 2)$. Substituting $\pi = \frac{22}{7}$: $r\left(\frac{22}{7} + 2\right) = r\left(\frac{36}{7}\right) = 36$. Therefore, $r = 7\text{ cm}$.

5-Minute Exam Simulation Mode

Experience the real pressure of Super TET Junior (Part 2C Mathematics). You will face 10 randomized speed-math questions with negative marking (-1 for incorrect answers) and a strict 5-minute countdown.

Time Limit:

5 Minutes (300s)

Marking Scheme:

+3 Marks | -1 Negative

Passing Benchmark:

21+ Marks (70%)

High-Yield Rapid Recall Formulas & Laws (Must-Remember Pointers)

#1
Perimeter of square $= 4a$, Area $= a^2$, Diagonal $= a\sqrt{2}$.
#2
Area of triangle using Heron's formula $= \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$.
#3
Area of trapezium $= \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height}$.
#4
Curved Surface Area of cylinder $= 2\pi rh$, Total Surface Area $= 2\pi r(h+r)$.
#5
Volume of cone $= \frac{1}{3}\pi r^2 h$, Slant height $l = \sqrt{r^2 + h^2}$.
#6
Total Surface Area of solid hemisphere $= 3\pi r^2$, Volume $= \frac{2}{3}\pi r^3$.
#7
Space diagonal of cuboid $= \sqrt{l^2 + b^2 + h^2}$.
#8
$1\text{ litre} = 1000\text{ cm}^3$ and $1\text{ m}^3 = 1000\text{ litres}$.

Speed Math & Memory Mnemonics Vault (Exam Retention Anchors)

Cylinder & Cone Surface Area Rule

Memory Key: R-H-L

R - Radius, H - Height, L - Slant height for cone.

Hemisphere Surface Area Anchor

Memory Key: THREE

T - Total, H - Hemisphere, R - Requires, E - Extra, E - Edge ($3\pi r^2$).

Negative Marking Traps & Common Blunders (-1 Penalty Avoidance)

Negative Marking Alert #1:

Negative Marking Trap 1: Forgetting that hemisphere total surface area includes the flat circular base ($3\pi r^2$ vs $2\pi r^2$).

Negative Marking Alert #2:

Negative Marking Trap 2: Mixing up units between centimeters and meters when calculating areas and volumes.

Negative Marking Alert #3:

Negative Marking Trap 3: Applying cylinder formulas to hollow pipes without accounting for inner and outer radii.

Negative Marking Alert #4:

Negative Marking Trap 4: Confusing perimeter and area formulas for circles ($\pi r^2$ vs $2\pi r$).

Self-Assessment Checklist

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