Chapter 11: Mensuration (پیمائش)
Chapter 11: Mensuration (پیمائش)
Comprehensive Master Study Guide & Solved Textbook Solutions: Pythagoras theorem & triples, 3D pyramid surface area & volume, sphere & hemisphere metrics, right circular cone surface area & volume, composite solids, and all textbook exercises (11.1 to 11.7 + Review 11).
🗺️ Teacher & Parent Roadmap: Chapter 11 • Mensuration (پیمائش)
This chapter develops crucial 2D to 3D spatial thinking and algebraic calculation of real-world physical capacities across 6 Student Learning Outcomes (SLOs):
- SLO 1 (Pythagoras Theorem): State theorem $c^2=a^2+b^2$, informal proofs, Pythagorean triples, acute/right/obtuse tests, and solve practical ladder/pole/wire word problems.
- SLO 2 (Pyramid Surface Area): Elements of pyramids, regular vs irregular, net unfoldings, and surface area TSA = Base Area + 1/2 Pl.
- SLO 3 (Pyramid Volume): Understand 1/3 prism relation $V = 1/3 Bh$, isosceles/hexagonal bases, and composite prism-pyramid solids.
- SLO 4 (Sphere & Hemisphere Surface Area): Archimedes cylinder equivalence, string wrapping activity, open ($2\pi r^2$) vs closed ($3\pi r^2$) hemispheres, dome cementing.
- SLO 5 (Sphere & Hemisphere Volume): Cylinder 2/3 water ratio derivation $V = 4/3 \pi r^3$, litre capacity conversions ($1\text{ dm}^3 = 1\text{ L}$), and solid metal melting/recasting.
- SLO 6 (Right Circular Cone): Slant height $l = \sqrt{r^2+h^2}$, sector dissection derivation $\text{CSA} = \pi rl$, $\text{TSA} = \pi r(l+r)$, conical tents, and volume $V = 1/3 \pi r^2 h$.
📚 11.1 Pythagoras Theorem & Pythagorean Triples
Historical Context: Thousands of years ago, the ancient Egyptians used a specialized knotted rope making a triangle with side ratios $3 : 4 : 5$ to construct perfect right angles when re-surveying agricultural land after the annual flooding of the River Nile. In the 6th century B.C., the famous Greek mathematician and philosopher Pythagoras investigated this relationship and formulated the universal theorem for all right-angled triangles.
📌 Statement of Pythagoras Theorem:
"In a right-angled triangle, the square of the hypotenuse (the length of the side opposite to the right angle) is equal to the sum of the squares of the lengths of the other two sides."
🔑 Pythagorean Triples:
A set of three positive numbers $(a, b, c)$ that satisfy $a^2 + b^2 = c^2$ is called a Pythagorean triple.
Examples: $(3, 4, 5)$, $(5, 12, 13)$, $(6, 8, 10)$, $(7, 24, 25)$, $(8, 15, 17)$, $(9, 40, 41)$.
📐 Triangle Classification Criteria:
Let $c$ be the longest side of a triangle:
• If $a^2 + b^2 = c^2 \implies$ Right-angled triangle
• If $a^2 + b^2 > c^2 \implies$ Acute-angled triangle
• If $a^2 + b^2 < c^2 \implies$ Obtuse-angled triangle
📝 Solved Exercise 11.1 — Full Step-by-Step Textbook Solutions
Question 1: Which of the following are Pythagorean triples?
(i) (6, 8, 10): $6^2 + 8^2 = 36 + 64 = 100 = 10^2$. Yes, it is a Pythagorean triple.
(ii) (5, 7, 10): $5^2 + 7^2 = 25 + 49 = 74 \neq 100$. No, not a Pythagorean triple.
(iii) (5, 12, 13): $5^2 + 12^2 = 25 + 144 = 169 = 13^2$. Yes, it is a Pythagorean triple.
(iv) (1, 1, √2): $1^2 + 1^2 = 1 + 1 = 2 = (\sqrt{2})^2$. Yes, satisfies Pythagoras theorem.
(v) (12, 16, 20): $12^2 + 16^2 = 144 + 256 = 400 = 20^2$. Yes, it is a Pythagorean triple.
(vi) (5, √5, 30): $5^2 + (\sqrt{5})^2 = 25 + 5 = 30 \neq 900$. No, not a Pythagorean triple.
(vii) (√3, √5, 2√2): $(\sqrt{3})^2 + (\sqrt{5})^2 = 3 + 5 = 8 = (2\sqrt{2})^2$. Yes, satisfies the theorem.
(viii) (8, 10, 12): $8^2 + 10^2 = 64 + 100 = 164 \neq 144$. No, not a Pythagorean triple.
Question 2: Find length of unknown side in right angled triangle ABC (∠C = 90°):
(i) a = 8 cm, b = 6 cm, c = ?
$c = \sqrt{a^2 + b^2} = \sqrt{64 + 36} = \sqrt{100} = \mathbf{10\text{ cm}}$.
(ii) a = 4 cm, c = √32 cm, b = ?
$b = \sqrt{c^2 - a^2} = \sqrt{32 - 16} = \sqrt{16} = \mathbf{4\text{ cm}}$.
(iii) b = 12 m, c = 13 m, a = ?
$a = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = \mathbf{5\text{ m}}$.
(iv) a = 1 cm, c = √2 cm, b = ?
$b = \sqrt{(\sqrt{2})^2 - 1^2} = \sqrt{2 - 1} = \mathbf{1\text{ cm}}$.
(v) b = 24 cm, c = 30 cm, a = ?
$a = \sqrt{30^2 - 24^2} = \sqrt{900 - 576} = \sqrt{324} = \mathbf{18\text{ cm}}$.
(vi) a = 4 cm, b = √3 cm, c = ?
$c = \sqrt{4^2 + (\sqrt{3})^2} = \sqrt{16 + 3} = \mathbf{\sqrt{19}\text{ cm} \approx 4.36\text{ cm}}$.
Question 3 to 11 Solutions:
Q3 (Figures):
(i) Isosceles triangle: altitude $x = \sqrt{6^2 - 3^2} = \sqrt{27} = \mathbf{3\sqrt{3}\text{ cm} \approx 5.20\text{ cm}}$.
(ii) Square side 7 cm: $x = \mathbf{7\text{ cm}}, y = \sqrt{7^2 + 7^2} = \sqrt{98} = \mathbf{7\sqrt{2}\text{ cm} \approx 9.90\text{ cm}}$.
(iii) Perpendicular $AC = 4\text{ cm}, BD = 10\text{ cm}, BC = 7\text{ cm} \implies CD = 3\text{ cm}$. In $\triangle ACD: x = \sqrt{4^2 + 3^2} = \mathbf{5\text{ cm}}$. In $\triangle ABC: y = \sqrt{4^2 + 7^2} = \sqrt{65} = \mathbf{8.06\text{ cm}}$.
(iv) $z = \sqrt{4^2 + 4^2} = \mathbf{5.66\text{ cm}}, y = \sqrt{5^2 - 4^2} = \mathbf{3\text{ cm}}, x = \sqrt{6^2 - 4^2} = \sqrt{20} = \mathbf{4.47\text{ cm}}$.
Q4 (Right Isosceles Triangle): $a = b = 10\text{ cm} \implies c = \sqrt{10^2 + 10^2} = \sqrt{200} = \mathbf{14.14\text{ cm}}$.
Q5 (Equilateral Altitude 6 cm): $s^2 = 6^2 + (s/2)^2 \implies \frac{3}{4}s^2 = 36 \implies s^2 = 48 \implies s = \sqrt{48} = \mathbf{8.49\text{ cm}}$.
Q6 (Ladder on Wall): Length $= \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = \mathbf{15\text{ m}}$.
Q7 (Pole & Wire): Distance $= \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = \mathbf{5\text{ m}}$.
Q8 (Square Diagonal 10√2 cm): (i) Side $= \mathbf{10\text{ cm}}$, (ii) Perimeter $= 4 \times 10 = \mathbf{40\text{ cm}}$, (iii) Area $= 10^2 = \mathbf{100\text{ cm}^2}$.
Q9 (Rectangle with Variable Sides): $(x+9)^2 = 6^2 + (x+7)^2 \implies 18x + 81 = 14x + 85 \implies 4x = 4 \implies \mathbf{x = 1\text{ cm}}$. (ii) Sides: $\mathbf{6\text{ cm}\text{ and }8\text{ cm}}$, (iii) Diagonal: $\mathbf{10\text{ cm}}$, (iv) Perimeter $= \mathbf{28\text{ cm}}$, Area $= \mathbf{48\text{ cm}^2}$.
Q10 (Perimeter 12 cm, Area 6 cm²): $a+b+c = 12 \implies a+b = 12-c$; $ab = 12$. $(a+b)^2 = (12-c)^2 \implies c^2 + 24 = 144 - 24c + c^2 \implies 24c = 120 \implies \mathbf{c = 5\text{ cm}}$.
Q11 (Square Leg Areas 81 & 144 cm²): $a^2 = 81 + 144 = 225 \implies \mathbf{a = 15\text{ cm}}$.
Brain Buster: Isosceles right triangle $3x = 2x + 4 \implies x = 4\text{ cm}$. Legs are $12\text{ cm}$. (i) Hypotenuse $y = \sqrt{12^2 + 12^2} = \mathbf{12\sqrt{2}\text{ cm} \approx 16.97\text{ cm}}$. (ii) Area $= \mathbf{72\text{ cm}^2}$, Perimeter $= \mathbf{24 + 12\sqrt{2}\text{ cm} \approx 40.97\text{ cm}}$.
📚 11.2 Surface Area of Pyramids (اہرام کا رقبہ)
Pyramid Definition & Elements: A pyramid is a 3D geometric solid having a polygon as its base and triangular lateral faces that converge at a single common point called the apex. Key structural elements include:
- Base: The flat bottom polygon (triangular, square, rectangular, pentagonal, hexagonal).
- Altitude (Vertical Height $h$): The perpendicular distance from the apex to the center of the base.
- Slant Height ($l$): The altitude of each triangular lateral face measured along its surface.
- Regular vs Irregular: A regular pyramid has a regular polygon as base and identical isosceles triangles as lateral faces.
📝 Solved Exercise 11.2 — Surface Area of Pyramids
Q1 (Equilateral Triangular Pyramid side 6 cm, slant height 12 cm):
(i) $\text{Base Area} = \frac{\sqrt{3}}{4}(6^2) = \mathbf{15.6\text{ cm}^2}$.
(ii) $\text{Lateral Area} = 3 \times (\frac{1}{2} \times 6 \times 12) = \mathbf{108\text{ cm}^2}$.
(iii) $\text{Total SA} = 15.6 + 108 = \mathbf{123.6\text{ cm}^2}$.
Q2 (Square Pyramid slant height 5 cm, base 4 cm):
$\text{Lateral Area} = 4 \times (\frac{1}{2} \times 4 \times 5) = \mathbf{40\text{ cm}^2}$.
Q3 (Square Pyramid Base Area 144 cm², height 8 cm):
Side $s = \sqrt{144} = 12\text{ cm}$. Slant height $l = \sqrt{8^2 + 6^2} = 10\text{ cm}$.
$\text{Lateral Area} = 4 \times (\frac{1}{2} \times 12 \times 10) = 240\text{ cm}^2 \implies \text{Total SA} = 144 + 240 = \mathbf{384\text{ cm}^2}$.
Q4 (Regular Square Pyramid base 10 cm, slant edge 10 cm):
Slant height $l = \sqrt{10^2 - 5^2} = \sqrt{75} = 8.66\text{ cm}$. Lateral Area $= 4 \times (\frac{1}{2} \times 10 \times 8.66) = 173.2\text{ cm}^2$. $\text{Total SA} = 100 + 173.2 = \mathbf{273.2\text{ cm}^2}$.
Q5 (Birdhouse Roof slant height 23 cm, square side 20 cm):
$\text{Wood required} = \text{Lateral Area} = 4 \times (\frac{1}{2} \times 20 \times 23) = \mathbf{920\text{ cm}^2}$.
Q6 (Pentagonal Pyramid side 8 cm, slant height 13 cm):
(i) $\text{Slant Area} = 5 \times (\frac{1}{2} \times 8 \times 13) = \mathbf{260\text{ cm}^2}$.
(ii) $\text{Base Area} = 500 - 260 = \mathbf{240\text{ cm}^2}$.
Q7 (Square Pyramid base 24 cm, vertical height 16 cm):
$l = \sqrt{16^2 + 12^2} = 20\text{ cm}$. (i) $\text{Slant Area} = 4 \times (\frac{1}{2} \times 24 \times 20) = \mathbf{960\text{ cm}^2}$. (ii) $\text{Whole Surface Area} = 24^2 + 960 = \mathbf{1536\text{ cm}^2}$.
Q8 (Equilateral Base 16 m, slant height 8 m):
$\text{Slant Area} = 3 \times (\frac{1}{2} \times 16 \times 8) = \mathbf{192\text{ m}^2}$.
Q9 (Hexagonal Pyramid slant edge 5 m, base edge 2 m):
$l = \sqrt{5^2 - 1^2} = \sqrt{24} \approx 4.899\text{ m}$. $\text{Lateral Area} = 6 \times (\frac{1}{2} \times 2 \times \sqrt{24}) = 6\sqrt{24} = \mathbf{29.39\text{ m}^2}$.
Q10 (Triangular Pyramid base 6 cm, slant height 10 cm):
(i) $\text{Base Area} = \mathbf{15.6\text{ cm}^2}$, (ii) $\text{Lateral Area} = \mathbf{90\text{ cm}^2}$, (iii) $\text{Total SA} = \mathbf{105.6\text{ cm}^2}$.
Brain Buster (Rectangular Pyramid 8 cm × 5 cm, height 12 cm):
$l_1 = \sqrt{12^2 + 2.5^2} = 12.26\text{ cm}$, $l_2 = \sqrt{12^2 + 4^2} = 12.65\text{ cm}$. $\text{Total SA} = 40 + (2 \times \frac{1}{2} \times 8 \times 12.26) + (2 \times \frac{1}{2} \times 5 \times 12.65) = \mathbf{201.31\text{ cm}^2}$.
📚 11.3 Volume of Pyramids & Composite Figures (اہرام کا حجم)
Core Principle: It takes exactly three hollow pyramids filled with sand/liquid to fill one prism having the identical base area and height. Therefore:
📝 Solved Exercise 11.3 — Volume of Pyramids
Q1 (Find Volumes):
(i) Square base 10 mm × 10 mm, $h = 27\text{ mm} \implies V = \frac{1}{3}(100)(27) = \mathbf{900\text{ mm}^3}$.
(ii) Square base 4 cm × 4 cm, $h = 9\text{ cm} \implies V = \frac{1}{3}(16)(9) = \mathbf{48\text{ cm}^3}$.
(iii) Rectangular base 8 cm × 5 cm, $h = 12\text{ cm} \implies V = \frac{1}{3}(40)(12) = \mathbf{160\text{ cm}^3}$.
(iv) Base Area $= 120\text{ mm}^2, h = 50\text{ mm} \implies V = \frac{1}{3}(120)(50) = \mathbf{2000\text{ mm}^3}$.
(v) Square base 24 cm × 24 cm, height 9 cm $\implies V = \frac{1}{3}(576)(9) = \mathbf{1728\text{ cm}^3}$.
Q2 (Triangular Pyramid side 4 m, height 6 m): $\text{Base Area} = 4\sqrt{3}\text{ m}^2 \implies V = \frac{1}{3}(4\sqrt{3})(6) = \mathbf{8\sqrt{3}\text{ m}^3 \approx 13.86\text{ m}^3}$.
Q3 (Square base V = 360 cm³, h = 30 cm): $\text{Base Area} = \frac{3 \times 360}{30} = \mathbf{36\text{ cm}^2}$; Side $= \mathbf{6\text{ cm}}$.
Q4 (Square base 3 m, height 7 m): (i) $V = \frac{1}{3}(9)(7) = \mathbf{21\text{ m}^3}$. (ii) Yes, doubling height doubles volume.
Q5 (Cuboid 6 × 8 × 11 cm): Pyramid volume $= \frac{1}{3}(6 \times 8 \times 11) = \mathbf{176\text{ cm}^3}$.
Q6 (V = 80 cm³, Base Area 30 cm²): Height $h = \frac{3 \times 80}{30} = \mathbf{8\text{ cm}}$.
Q7 (Equilateral base 6 cm, V = 120 cm³): $h = \frac{3 \times 120}{15.588} = \mathbf{23.1\text{ cm}}$.
Q8 (Composite Solids):
(i) Prism $(12 \times 8 \times 10 = 960) + \text{Pyramid}(\frac{1}{3} \times 96 \times 8 = 256) = \mathbf{1216\text{ cm}^3}$.
(ii) Prism $(8 \times 8 \times 13 = 832) + \text{Pyramid}(\frac{1}{3} \times 64 \times 3 = 64) = \mathbf{896\text{ cm}^3}$.
📚 11.4 Surface Area of Sphere & Hemisphere (کرہ کا رقبہ)
Archimedes Theorem: A sphere and a cylinder having the same height ($h = 2r$) and equal radii have identical total curved surface areas: $\text{Area} = 2\pi r(2r) = 4\pi r^2$.
📝 Solved Exercise 11.4 — Surface Area of Sphere & Hemisphere
Q1 (Surface Area of Spheres):
(i) $r = 14\text{ cm} \implies 4 \times \frac{22}{7} \times 14^2 = \mathbf{2464\text{ cm}^2}$.
(ii) $r = 2.1\text{ m} \implies 4 \times \frac{22}{7} \times (2.1)^2 = \mathbf{55.44\text{ m}^2}$.
(iii) $d = 35\text{ cm} (r = 17.5) \implies \mathbf{3850\text{ cm}^2}$.
(iv) $d = 2.8\text{ dm} (r = 1.4) \implies \mathbf{24.64\text{ dm}^2}$.
Q2 (Curved Area of Hemispheres):
(i) $r = 5.6\text{ cm} \implies 2 \times \frac{22}{7} \times (5.6)^2 = \mathbf{197.12\text{ cm}^2}$.
(ii) $d = 7\text{ m} (r = 3.5) \implies 2 \times \frac{22}{7} \times (3.5)^2 = \mathbf{77\text{ m}^2}$.
Q3 (Radii & Diameters from Area):
(i) $5544\text{ cm}^2 \implies r = \mathbf{21\text{ cm}}, d = \mathbf{42\text{ cm}}$.
(ii) $154\text{ cm}^2 \implies r = \mathbf{3.5\text{ cm}}, d = \mathbf{7\text{ cm}}$.
(iii) $9856\text{ m}^2 \implies r = \mathbf{28\text{ m}}, d = \mathbf{56\text{ m}}$.
Q4 (Semispherical Bowl CSA 77 m²): $2\pi r^2 = 77 \implies r = 3.5\text{ m} \implies d = \mathbf{7\text{ m}}$.
Q5 (Painting Sphere r = 7 cm @ Rs. 0.50/cm²): Area $= 616\text{ cm}^2 \implies \text{Cost} = 616 \times 0.50 = \mathbf{\text{Rs. } 308}$.
Q6 (Cementing Dome d = 8.4 m @ Rs. 100/m²): $\text{CSA} = 110.88\text{ m}^2 \implies \text{Cost} = 110.88 \times 100 = \mathbf{\text{Rs. } 11,088}$.
Q7 (Radius Doubled): Area increases by $(2)^2 = \mathbf{4\text{ times}}$.
Q8 (Masjid Dome d = 14 m @ Rs. 2000/m²): (i) Area $= 2\pi(7^2) = \mathbf{308\text{ m}^2}$. (ii) Cost $= 308 \times 2000 = \mathbf{\text{Rs. } 616,000}$.
Q9 (Hemisphere r = 14 cm): (i) Open $= 2\pi(14^2) = \mathbf{1232\text{ cm}^2}$. (ii) Closed $= 3\pi(14^2) = \mathbf{1848\text{ cm}^2}$.
Q10 (Composite Cylinder + Hemisphere Solid): Total Area $= 3\pi(3^2) + 2\pi(3)(7) = 69\pi = \mathbf{216.86\text{ cm}^2}$.
📚 11.5 Volume of Sphere & Hemisphere (کرہ کا حجم)
Capacity & Metric Equivalence: $1\text{ litre} = 1000\text{ cm}^3 = 1\text{ dm}^3$, and $1000\text{ litres} = 1\text{ m}^3$.
Volume of full sphere $= \frac{4}{3}\pi r^3$, and volume of hemisphere $= \frac{2}{3}\pi r^3$.
📝 Solved Exercise 11.5 — Volume of Sphere & Hemisphere
Q1 (Find Volumes):
(i) $r = 2.1\text{ cm} \implies V = \frac{4}{3}\pi(2.1^3) = \mathbf{38.81\text{ cm}^3}$.
(ii) $r = 4.2\text{ m} \implies V = \frac{4}{3}\pi(4.2^3) = \mathbf{310.46\text{ m}^3}$.
(iii) $r = 6.3\text{ cm} \implies V = \frac{4}{3}\pi(6.3^3) = \mathbf{1047.82\text{ cm}^3}$.
Q2 (Area to Volume): (i) $616\text{ m}^2 \implies r = \mathbf{7\text{ m}}, V = \mathbf{1437.33\text{ m}^3}$. (ii) $55.44\text{ cm}^2 \implies r = \mathbf{2.1\text{ cm}}, V = \mathbf{38.81\text{ cm}^3}$.
Q3 (Sphere & Cylinder r = 7 cm, same height 14 cm): Volume of cylinder $= \pi(7^2)(14) = \mathbf{2156\text{ cm}^3}$.
Q4 (Tank r = 10.5 dm): Volume $= \frac{4}{3}\pi(10.5^3) = 4851\text{ dm}^3 = \mathbf{4851\text{ litres}}$.
Q5 (Iron Ball r = 1.5 cm @ 42 g/cm³): Volume $= 14.143\text{ cm}^3 \implies \text{Mass} = 14.143 \times 42 = \mathbf{594\text{ g}}$.
Q6 (Two Spheres r = 6 cm & 9 cm): (a) Areas: $\mathbf{452.57\text{ cm}^2\text{ & }1018.29\text{ cm}^2}$. (b) Volumes: $\mathbf{905.14\text{ cm}^3\text{ & }3054.86\text{ cm}^3}$. (c) Area Ratio $= 6^2:9^2 = \mathbf{4:9}$; Volume Ratio $= 6^3:9^3 = \mathbf{8:27}$.
Q7 (Radius Changes): (i) Doubled $\implies \mathbf{8\text{ times}}$. (ii) Halved $\implies \mathbf{\frac{1}{8}\text{ times}}$.
Q8 (Sweets d = 14 mm): Volume of 8 sweets $= 8 \times 1437.33 = \mathbf{11498.67\text{ mm}^3}$.
Q9 (Metal Block 21 × 24 × 77 cm Melted into Sphere): (i) Volume $= \mathbf{38808\text{ cm}^3}$. (ii) $\frac{4}{3}\pi r^3 = 38808 \implies r = \mathbf{21\text{ cm}}$.
📚 11.6 Surface Area of Right Circular Cone (مخروط کا سطحی رقبہ)
📝 Solved Exercise 11.6 — Surface Area of Cone
Q1 (Find Curved & Total SA):
(i) $r = 7\text{ cm}, l = 10\text{ cm} \implies \text{CSA} = \mathbf{220\text{ cm}^2}, \text{TSA} = \mathbf{374\text{ cm}^2}$.
(ii) $r = 2.1\text{ m}, l = 5\text{ m} \implies \text{CSA} = \mathbf{33\text{ m}^2}, \text{TSA} = \mathbf{46.86\text{ m}^2}$.
(iii) $d = 15\text{ cm} (r = 7.5), l = 14\text{ cm} \implies \text{CSA} = \mathbf{330\text{ cm}^2}, \text{TSA} = \mathbf{506.79\text{ cm}^2}$.
(iv) $r = 9\text{ cm}, h = 12\text{ cm} (l = 15) \implies \text{CSA} = \mathbf{424.29\text{ cm}^2}, \text{TSA} = \mathbf{678.86\text{ cm}^2}$.
(v) $h = 24\text{ m}, l = 25\text{ m} (r = 7) \implies \text{CSA} = \mathbf{550\text{ m}^2}, \text{TSA} = \mathbf{704\text{ m}^2}$.
Q2 (CSA = 132 cm², l = 21 cm): $\pi r(21) = 132 \implies r = \mathbf{2\text{ cm}}$. Base Area $= \pi(2^2) = \mathbf{12.57\text{ cm}^2}$.
Q3 (Painting Cone r = 5 m, h = 12 m @ Rs. 20/m²): $l = 13\text{ m} \implies \text{CSA} = 204.286\text{ m}^2 \implies \text{Cost} = \mathbf{\text{Rs. } 4085.71}$.
Q4 (Tent h = 4 m, l = 5 m, r = 3 m): (i) Floor Cost $= \pi(3^2) \times 42 = \mathbf{\text{Rs. } 1188}$. (ii) Canvas Cost $= \pi(3)(5) \times 280 = \mathbf{\text{Rs. } 13,200}$.
Q5 (Hollow Cone d = 14 dm, l = 30 dm @ Rs. 10/dm²): Sheet Area $= \pi(7)(30) = 660\text{ dm}^2 \implies \text{Cost} = \mathbf{\text{Rs. } 6600}$.
Q6 (CSA = 96π cm², l = 12 cm): $r = 8\text{ cm} \implies h = \sqrt{12^2 - 8^2} = \sqrt{80} = \mathbf{8.94\text{ cm}}$.
Q7 (Base Area 154 cm², l = 10 cm): $r = 7\text{ cm} \implies \text{CSA} = \pi(7)(10) = \mathbf{220\text{ cm}^2}$.
Q8 (TSA = 198 m², l = 6r): $7\pi r^2 = 198 \implies 22r^2 = 198 \implies r = \mathbf{3\text{ m}}$.
Q9 (Sharpened Lead Pencil): $\text{Tip}(\text{CSA} = 0.397) + \text{Cylinder}(\text{CSA} = 8.80) + \text{Flat End}(0.126) = \mathbf{9.30\text{ cm}^2}$.
📚 11.7 Volume of Right Circular Cone (مخروط کا حجم)
📝 Solved Exercise 11.7 — Volume of Cone
Q1 (Missing Elements Table):
(i) $r = 6\text{ cm}, h = 8\text{ cm} \implies l = \mathbf{10\text{ cm}}, V = \mathbf{301.71\text{ cm}^3}$.
(ii) $r = 3\text{ cm}, l = 5\text{ cm} \implies h = \mathbf{4\text{ cm}}, V = \mathbf{37.71\text{ cm}^3}$.
(iii) $h = 7\text{ cm}, V = 66\text{ cm}^3 \implies r = \mathbf{3\text{ cm}}, l = \mathbf{7.62\text{ cm}}$.
(iv) $r = 6\text{ cm}, V = 528\text{ cm}^3 \implies h = \mathbf{14\text{ cm}}, l = \mathbf{15.23\text{ cm}}$.
Q2 (Conical Ice Cream Cups r = 1.5 cm, h = 2.8 cm):
(i) 1 cup $= \frac{1}{3}\pi(1.5^2)(2.8) = \mathbf{6.6\text{ cm}^3}$. (ii) 100 cups $= 100 \times 6.6 = \mathbf{660\text{ cm}^3}$.
Q3 (V = 550 dm³, r = 5 dm): (i) Height $h = \mathbf{21\text{ dm}}$. (ii) $l = 21.587\text{ dm} \implies \text{Total SA} = \mathbf{408.57\text{ dm}^2}$.
Q4 (Base Area 154 m², h = 12 m): Volume $= \frac{1}{3} \times 154 \times 12 = \mathbf{616\text{ m}^3}$.
Q5 (Sand Cone h = 9 cm, r = 3.5 cm @ 10 g/cm³): (i) Volume $= \mathbf{115.5\text{ cm}^3}$. (ii) $\text{Mass} = 115.5 \times 10 = \mathbf{1155\text{ g}}$.
Q6 (Sand Pile h = 6 m, d = 7 m): Volume $= \frac{1}{3}\pi(3.5^2)(6) = \mathbf{77\text{ m}^3}$.
Q7 (Conical Tent h = 42 dm, d = 54 dm, space/person = 2916 dm³): Tent Volume $= 32,076\text{ dm}^3 \implies \text{Capacity} = \frac{32076}{2916} = \mathbf{11\text{ persons}}$.
📝 Solved Review Exercise 11 — Complete Examination Mastery
Question 1: Textbook Multiple Choice Questions (MCQs):
- If lengths of sides of a triangle are in ratio 3 : 4 : 5, then triangle is: (c) right angled
- If c is hypotenuse of right triangle and a, b other sides: (b) c² = a² + b²
- The line segment joining vertex of cone to boundary line of base: (b) slant height
- Volume of cone of radius 3 cm and height 12 cm is: (d) 36π cm³
- Diameter of base of cone 6 cm, height 4 cm. Slant height is: (a) 5 cm
- Curved surface of right circular cone (radius 3 cm, slant height 6 cm): (c) 18π
- Radius of hemisphere is 0.7 cm. Its curved surface area is: (a) 3.08 cm²
- Hypotenuse is 13 cm, one side is 12 cm. What is other side? (d) 5 cm
- Height of apex of pyramid from base (30 × 20 cm) with V = 2000 cm³: (c) 10 cm
- If radii and heights of cone and cylinder are equal, volume of cone is: (b) one third
- All heavenly bodies look like a: (d) sphere
- Area of circle is πr². Area of sphere with same radius is: (d) 4πr²
Subjective Examination Problems (Q2 to Q11):
Q2 (Square Diagonal √50 cm): $2s^2 = 50 \implies s = 5\text{ cm} \implies \text{Area} = \mathbf{25\text{ cm}^2}, \text{Perimeter} = \mathbf{20\text{ cm}}$.
Q3 (Solid Hemisphere d = 14 cm): $\text{TSA} = 3\pi(7^2) = \mathbf{462\text{ cm}^2}$.
Q4 (Sphere Area = 154 cm²): (i) Radius $= \mathbf{3.5\text{ cm}}$. (ii) Volume $= \frac{4}{3}\pi(3.5^3) = \mathbf{179.67\text{ cm}^3}$.
Q5 (Conical Bird Cage d = 18 cm, h = 14 cm): (i) Volume $= \mathbf{1188\text{ cm}^3}$. (ii) Birds $= \frac{1188}{297} = \mathbf{4\text{ birds}}$.
Q6 (Square Pyramid h = 12 m, base 10 m): (i) Lateral Area $= \mathbf{260\text{ m}^2}$. (ii) Total SA $= \mathbf{360\text{ m}^2}$. (iii) Volume $= \mathbf{400\text{ m}^3}$.
Q7 (Square Pyramid Lateral Area 600 cm², l = 15 cm): Base side $= 20\text{ cm} \implies \text{Perimeter} = \mathbf{80\text{ cm}}, \text{Base Area} = \mathbf{400\text{ cm}^2}$.
Q8 (Equilateral Base 12√3 m, height 8 m): (i) Base Area $= 108\sqrt{3} = \mathbf{187.06\text{ m}^2}$. (ii) Volume $= \mathbf{498.83\text{ m}^3}$.
Q9 (Circle Center O, OC = 5 cm, BC = 4 cm): Diameter $AB = \mathbf{10\text{ cm}}$. Semicircle $\angle C = 90° \implies AC = \sqrt{10^2 - 4^2} = \mathbf{9.17\text{ cm}}$.
Q10 (Composite Cone + Hemisphere): $\text{Cone}(36\pi) + \text{Hemisphere}(18\pi) = 54\pi = \mathbf{169.71\text{ cm}^3}$.
Q11 (Square Diagonal 14 cm): (i) Side $= \mathbf{7\sqrt{2}\text{ cm} \approx 9.90\text{ cm}}$. (ii) Area $= \mathbf{98\text{ cm}^2}$, Perimeter $= \mathbf{28\sqrt{2}\text{ cm} \approx 39.60\text{ cm}}$.
More Chapter Notes for Class 8 (FBISE)
MathematicsTest Your Knowledge on Chapter 11: Chapter 11: Mensuration (پیمائش)
Practice textbook-aligned solved MCQs with instant answer feedback, step-by-step solutions, and timed test simulation.
Class 8 Mathematics - Ch 1: Real Numbers Chapter Mock Test
Test your complete conceptual mastery across all chapters under real board exam conditions with official timer, anti-cheat surveillance, and instant grading.