Class 6 Mathematics - Ch 9: Comprehensive Guide to Mensuration: Perimeter, Area, 3D Solids, Surface Area & Volume (FBISE)

(i) Rectangle with length $l = 15\text{ cm}$ and width $w = 8\text{ cm}$:
$\text{Perimeter} = 2(l + w) = 2(15 + 8) = 2(23) = \mathbf{46\text{ cm}}$.

(ii) Square with side $s = 6\text{ m}$:
$\text{Perimeter} = 4 \times s = 4 \times 6 = \mathbf{24\text{ m}}$.

(i) Dimensions around: $10\text{m}, 10\text{m}, 3\text{m}, 5\text{m}, 7\text{m}, 5\text{m}$.
$\text{Perimeter} = 10 + 10 + 3 + 5 + 7 + 5 = \mathbf{40\text{ m}}$.

(ii) Dimensions: $4\text{cm}, 2\text{cm}, 2\text{cm}, 1\text{cm}, 2\text{cm}, 3\text{cm}$.
$\text{Perimeter} = 4 + 2 + 2 + 1 + 2 + 3 = \mathbf{14\text{ cm}}$.

(iii) Step shape: $5\text{m}, 6\text{m}, 2\text{m}, 2\text{m}, 3\text{m}, 4\text{m}$.
$\text{Perimeter} = 5 + 6 + 2 + 2 + 3 + 4 = \mathbf{22\text{ m}}$.

(iv) Stepped polygon: $5\text{cm}, 4\text{cm}, 2\text{cm}, 1.5\text{cm}, 1.5\text{cm}, 1.5\text{cm}, 1.5\text{cm}, 1\text{cm}$.
$\text{Perimeter} = 5 + 4 + 2 + 1.5 + 1.5 + 1.5 + 1.5 + 1 = \mathbf{18\text{ cm}}$.

(v) Symmetrical cross shape with 4 protruding rectangles/squares:
Summing all outer boundary segments: $\mathbf{48\text{ m}}$.

(vi) Stepped boundary figure with labeled segments: $2\text{cm} + 3\text{cm} + 3\text{cm} + 1\text{cm} + 2\text{cm} + 1\text{cm} + 3\text{cm} + 5\text{cm} = \mathbf{20\text{ cm}}$.

(vii) L-shaped block: $5\text{m}, 12\text{m}, 2\text{m}, 8\text{m}, 3\text{m}, 4\text{m}$.
$\text{Perimeter} = 5 + 12 + 2 + 8 + 3 + 4 = \mathbf{34\text{ m}}$.

(i) Length $= 15\text{ cm}$, Width $= 12\text{ cm}$:
$\text{Perimeter} = 2(15 + 12) = 2(27) = \mathbf{54\text{ cm}}$, $\text{Area} = 15 \times 12 = \mathbf{180\text{ cm}^2}$.

(ii) Length $= 13\text{ m}$, Width $= 8\text{ m}$:
$\text{Perimeter} = 2(13 + 8) = 2(21) = \mathbf{42\text{ m}}$, $\text{Area} = 13 \times 8 = \mathbf{104\text{ m}^2}$.

(iii) Length $= 25\text{ cm}$, Width $= 10\text{ cm}$:
$\text{Perimeter} = 2(25 + 10) = 2(35) = \mathbf{70\text{ cm}}$, $\text{Area} = 25 \times 10 = \mathbf{250\text{ cm}^2}$.

(iv) Length $= 10.5\text{ m}$, Width $= 4.5\text{ m}$:
$\text{Perimeter} = 2(10.5 + 4.5) = 2(15) = \mathbf{30\text{ m}}$, $\text{Area} = 10.5 \times 4.5 = \mathbf{47.25\text{ m}^2}$.

(v) Length $= 2\frac{1}{2}\text{ cm} = 2.5\text{ cm}$, Width $= 3\frac{1}{2}\text{ cm} = 3.5\text{ cm}$:
$\text{Perimeter} = 2(2.5 + 3.5) = 2(6) = \mathbf{12\text{ cm}}$, $\text{Area} = 2.5 \times 3.5 = \mathbf{8.75\text{ cm}^2} = \mathbf{8\frac{3}{4}\text{ cm}^2}$.

(i) Composite shape partitioned into rectangles:
Dividing into Rectangle 1 ($10\text{m} \times 5\text{m} = 50\text{m}^2$) and Rectangle 2 ($7\text{m} \times 5\text{m} = 35\text{m}^2$):
$\text{Total Area} = 50\text{ m}^2 + 35\text{ m}^2 = \mathbf{85\text{ m}^2}$.

(ii) Composite L-shaped figure:
Dividing into Top Rectangle ($2\text{cm} \times 2\text{cm} = 4\text{cm}^2$) and Base Rectangle ($4\text{cm} \times 1\text{cm} = 4\text{cm}^2$):
$\text{Total Area} = 4 + 4 = \mathbf{8\text{ cm}^2}$.

(i) Outer: $16\text{m} \times 12\text{m}$, Inner: $12\text{m} \times 8\text{m}$:
$A_{\text{outer}} = 16 \times 12 = 192\text{ m}^2$, $A_{\text{inner}} = 12 \times 8 = 96\text{ m}^2$
$\text{Area of shaded path} = 192 - 96 = \mathbf{96\text{ m}^2}$.

(ii) Outer: $35\text{cm} \times 25\text{cm}$, Path width: $2.5\text{cm}$:
Inner dimensions $= (35 - 5) \times (25 - 5) = 30\text{cm} \times 20\text{cm} = 600\text{cm}^2$
$A_{\text{outer}} = 35 \times 25 = 875\text{ cm}^2 \implies \text{Path Area} = 875 - 600 = \mathbf{275\text{ cm}^2}$.

(iii) Outer square $15\text{m} \times 15\text{m}$, Inner square $11\text{m} \times 11\text{m}$:
$A_{\text{outer}} = 15^2 = 225\text{ m}^2$, $A_{\text{inner}} = 11^2 = 121\text{ m}^2$
$\text{Area of shaded path} = 225 - 121 = \mathbf{104\text{ m}^2}$.

(iv) Rectangle $15\text{m} \times 10\text{m}$ with cross paths of width $2\text{m}$:
$\text{Area of horizontal path} = 15 \times 2 = 30\text{ m}^2$, $\text{Area of vertical path} = 10 \times 2 = 20\text{ m}^2$
$\text{Intersection square} = 2 \times 2 = 4\text{ m}^2$
$\text{Total Path Area} = 30 + 20 - 4 = \mathbf{46\text{ m}^2}$.

(v) Outer: $25\text{m} \times 15\text{m}$, Inner: $20\text{m} \times 10\text{m}$:
$A_{\text{outer}} = 25 \times 15 = 375\text{ m}^2$, $A_{\text{inner}} = 20 \times 10 = 200\text{ m}^2$
$\text{Area of shaded border} = 375 - 200 = \mathbf{175\text{ m}^2}$.

(vi) Outer: $6\text{m} \times 4\text{m}$, Inner: $4\text{m} \times 2\text{m}$:
$A_{\text{outer}} = 6 \times 4 = 24\text{ m}^2$, $A_{\text{inner}} = 4 \times 2 = 8\text{ m}^2$
$\text{Area of path} = 24 - 8 = \mathbf{16\text{ m}^2}$.

(vii) Outer square $10\text{m} \times 10\text{m}$, Inner square $6\text{m} \times 6\text{m}$:
$A_{\text{outer}} = 100\text{ m}^2$, $A_{\text{inner}} = 36\text{ m}^2 \implies \text{Area of path} = 100 - 36 = \mathbf{64\text{ m}^2}$.

(viii) Outer rectangle $50\text{m} \times 30\text{m}$ with cross paths of width $3\text{m}$:
$\text{Area} = (50 \times 3) + (30 \times 3) - (3 \times 3) = 150 + 90 - 9 = \mathbf{231\text{ m}^2}$.

Question 2: A painting $30\text{cm}$ long and $20\text{cm}$ wide is pasted on cardboard such that there is a margin of $1.5\text{cm}$ along each of its sides. Find the total area of the margin.
Inner area (painting) $= 30 \times 20 = 600\text{ cm}^2$.
Outer length $= 30 + 2(1.5) = 33\text{ cm}$, Outer width $= 20 + 2(1.5) = 23\text{ cm}$.
Outer area $= 33 \times 23 = 759\text{ cm}^2$.
$\text{Area of margin} = 759 - 600 = \mathbf{159\text{ cm}^2}$.

Question 3: A rectangular lawn is $60\text{m}$ long and $40\text{m}$ wide. A path $2.5\text{m}$ wide is constructed outside around it. Find the area of the path.
Inner area $= 60 \times 40 = 2400\text{ m}^2$.
Outer length $= 60 + 2(2.5) = 65\text{ m}$, Outer width $= 40 + 2(2.5) = 45\text{ m}$.
Outer area $= 65 \times 45 = 2925\text{ m}^2$.
$\text{Area of path} = 2925 - 2400 = \mathbf{525\text{ m}^2}$.

Question 4: A sheet of steel is $2\text{m}$ long and $1.5\text{m}$ wide. A strip of $0.25\text{m}$ is cut all around it. Find the area of the remaining sheet and the area of the cut strip.
Total initial area $= 2 \times 1.5 = 3\text{ m}^2$.
Remaining length $= 2 - 2(0.25) = 1.5\text{ m}$, Remaining width $= 1.5 - 2(0.25) = 1.0\text{ m}$.
$\text{Area of remaining sheet} = 1.5 \times 1.0 = \mathbf{1.5\text{ m}^2}$.
$\text{Area of cut strip} = 3 - 1.5 = \mathbf{1.5\text{ m}^2}$.

Question 1: Find the cost of fencing a square field of side $18\text{m}$ at the rate of Rs. 40 per metre.
$\text{Perimeter} = 4 \times 18 = 72\text{ m}$.
$\text{Total Cost} = 72 \times 40 = \mathbf{\text{Rs. } 2880}$.

Question 2: A rectangular playground is $75\text{m}$ long and $35\text{m}$ wide. How much wire is required to fence it 3 times? Also find the cost of wire at Rs. 15 per metre.
$\text{One round perimeter} = 2(75 + 35) = 2(110) = 220\text{ m}$.
$\text{Wire needed for 3 rounds} = 3 \times 220 = \mathbf{660\text{ m}}$.
$\text{Cost} = 660 \times 15 = \mathbf{\text{Rs. } 9900}$.

Question 3: A rectangular room is $6\text{m}$ long and $4.5\text{m}$ wide. Find the cost of carpeting its floor at the rate of Rs. 120 per $\text{m}^2$.
$\text{Floor Area} = 6 \times 4.5 = 27\text{ m}^2$.
$\text{Cost} = 27 \times 120 = \mathbf{\text{Rs. } 3240}$.

Question 4: A square lawn of side $25\text{m}$ has a flower bed of size $5\text{m} \times 4\text{m}$ in its centre. Find the remaining area of the lawn.
$\text{Total lawn area} = 25 \times 25 = 625\text{ m}^2$.
$\text{Flower bed area} = 5 \times 4 = 20\text{ m}^2$.
$\text{Remaining area} = 625 - 20 = \mathbf{605\text{ m}^2}$.

Question 5: A floor measuring $5\text{m} \times 4\text{m}$ is to be paved with square tiles of side $20\text{cm}$. How many tiles are required?
$\text{Floor Area} = 500\text{ cm} \times 400\text{ cm} = 200{,}000\text{ cm}^2$.
$\text{Area of 1 tile} = 20 \times 20 = 400\text{ cm}^2$.
$\text{Number of tiles} = \frac{200{,}000}{400} = \mathbf{500\text{ tiles}}$.

Question 6: The perimeter of a square field is $160\text{m}$. Find its area.
$\text{Side } s = \frac{160}{4} = 40\text{ m}$.
$\text{Area} = s^2 = 40 \times 40 = \mathbf{1600\text{ m}^2}$.

Question 7: The area of a rectangular hall is $180\text{m}^2$. If its breadth is $12\text{m}$, find its length and perimeter.
$\text{Length} = \frac{\text{Area}}{\text{breadth}} = \frac{180}{12} = \mathbf{15\text{ m}}$.
$\text{Perimeter} = 2(15 + 12) = 2(27) = \mathbf{54\text{ m}}$.

Question 8: A wire of length $36\text{cm}$ is bent to form a square. Find the side and area of the square.
$\text{Side} = \frac{36}{4} = \mathbf{9\text{ cm}}$, $\text{Area} = 9 \times 9 = \mathbf{81\text{ cm}^2}$.

Question 9: If the same wire of length $36\text{cm}$ is now bent into a rectangle of length $10\text{cm}$, find its width and area. Compare which shape encloses more area.
$2(10 + w) = 36 \implies 10 + w = 18 \implies w = \mathbf{8\text{ cm}}$.
$\text{Area of rectangle} = 10 \times 8 = \mathbf{80\text{ cm}^2}$.
$\text{Comparison: The square encloses more area } (81\text{ cm}^2 > 80\text{ cm}^2)$.

Question 10: Find the cost of levelling a rectangular garden $40\text{m}$ long and $25\text{m}$ broad at Rs. 15 per $\text{m}^2$.
$\text{Area} = 40 \times 25 = 1000\text{ m}^2$.
$\text{Cost} = 1000 \times 15 = \mathbf{\text{Rs. } 15{,}000}$.

Question 11: A path of width $2\text{m}$ runs around the inside of a square field of side $30\text{m}$. Find the area of the path and the cost of gravelling it at Rs. 25 per $\text{m}^2$.
Outer area $= 30 \times 30 = 900\text{ m}^2$.
Inner side $= 30 - 2(2) = 26\text{ m} \implies \text{Inner area} = 26^2 = 676\text{ m}^2$.
$\text{Area of path} = 900 - 676 = \mathbf{224\text{ m}^2}$.
$\text{Cost} = 224 \times 25 = \mathbf{\text{Rs. } 5600}$.

(i) Base $= 12\text{ cm}$, Altitude $= 8\text{ cm}$:
$\text{Area} = 12 \times 8 = \mathbf{96\text{ cm}^2}$.

(ii) Base $= 18\text{ m}$, Altitude $= 15\text{ m}$:
$\text{Area} = 18 \times 15 = \mathbf{270\text{ m}^2}$.

(iii) Base $= 25\text{ cm}$, Area $= 375\text{ cm}^2$:
$\text{Altitude} = \frac{375}{25} = \mathbf{15\text{ cm}}$.

(iv) Altitude $= 6.5\text{ m}$, Area $= 130\text{ m}^2$:
$\text{Base} = \frac{130}{6.5} = \mathbf{20\text{ m}}$.

(i) Parallelogram: Base $= 9\text{ cm}$, Altitude $= 4\text{ cm}$:
$\text{Area} = 9 \times 4 = \mathbf{36\text{ cm}^2}$.

(ii) Triangle: Base $= 10\text{ cm}$, Altitude $= 6\text{ cm}$:
$\text{Area} = \frac{1}{2} \times 10 \times 6 = \mathbf{30\text{ cm}^2}$.

(iii) Triangle: Base $= 14\text{ m}$, Altitude $= 9\text{ m}$:
$\text{Area} = \frac{1}{2} \times 14 \times 9 = \mathbf{63\text{ m}^2}$.

(iv) Parallelogram: Base $= 16\text{ m}$, Altitude $= 10\text{ m}$:
$\text{Area} = 16 \times 10 = \mathbf{160\text{ m}^2}$.

(v) Right-angled Triangle: Legs $= 8\text{ cm}$ and $6\text{ cm}$:
$\text{Area} = \frac{1}{2} \times 8 \times 6 = \mathbf{24\text{ cm}^2}$.

(vi) Triangle: Base $= 12.4\text{ cm}$, Altitude $= 5\text{ cm}$:
$\text{Area} = \frac{1}{2} \times 12.4 \times 5 = \mathbf{31\text{ cm}^2}$.

(vii) Obtuse-angled Triangle: Base $= 7\text{ cm}$, Altitude $= 4\text{ cm}$:
$\text{Area} = \frac{1}{2} \times 7 \times 4 = \mathbf{14\text{ cm}^2}$.

Question 3: The base of a parallelogram is $15\text{cm}$ and its altitude is $8\text{cm}$. Find its area.
$\text{Area} = 15 \times 8 = \mathbf{120\text{ cm}^2}$.

Question 4: The area of a parallelogram is $144\text{m}^2$. If its altitude is $9\text{m}$, find the length of its corresponding base.
$\text{Base} = \frac{\text{Area}}{\text{altitude}} = \frac{144}{9} = \mathbf{16\text{ m}}$.

Question 5: A triangular field has a base of $80\text{m}$ and height $45\text{m}$. Find the cost of cultivating it at Rs. 5 per $\text{m}^2$.
$\text{Area} = \frac{1}{2} \times 80 \times 45 = 1800\text{ m}^2$.
$\text{Cost} = 1800 \times 5 = \mathbf{\text{Rs. } 9000}$.

Question 6: The altitude of a triangle is $12\text{cm}$ and its area is $72\text{cm}^2$. Find its base.
$\text{Base} = \frac{2 \times \text{Area}}{h} = \frac{2 \times 72}{12} = \mathbf{12\text{ cm}}$.

Question 7: A parallelogram-shaped parking lot has a base of $50\text{m}$ and height $24\text{m}$. Find the cost of paving it at Rs. 150 per $\text{m}^2$.
$\text{Area} = 50 \times 24 = 1200\text{ m}^2$.
$\text{Cost} = 1200 \times 150 = \mathbf{\text{Rs. } 180{,}000}$.

(i) Parallel sides $a = 4\text{cm}, b = 3\text{cm}$, Height $h = 2\text{cm}$:
$\text{Area} = \frac{1}{2} \times 2 \times (4 + 3) = 1 \times 7 = \mathbf{7\text{ cm}^2}$.

(ii) Parallel sides $a = 6\text{cm}, b = 4\text{cm}$, Height $h = 3\text{cm}$:
$\text{Area} = \frac{1}{2} \times 3 \times (6 + 4) = \frac{3 \times 10}{2} = \mathbf{15\text{ cm}^2}$.

(iii) Parallel sides $a = 9\text{m}, b = 6\text{m}$, Height $h = 7\text{m}$:
$\text{Area} = \frac{1}{2} \times 7 \times (9 + 6) = \frac{7 \times 15}{2} = \mathbf{52.5\text{ m}^2}$.

(i) Trapezium with parallel sides $120\text{m}, 60\text{m}$, non-parallel sides $75\text{m}, 75\text{m}$, height $50\text{m}$:
$\text{Area} = \frac{1}{2} \times 50 \times (120 + 60) = 25 \times 180 = \mathbf{4500\text{ m}^2}$.
$\text{Perimeter} = 120 + 60 + 75 + 75 = \mathbf{330\text{ m}}$.

(ii) Trapezium with parallel sides $50\text{m}, 30\text{m}$, non-parallel sides $33\text{m}, 33\text{m}$, height $30\text{m}$:
$\text{Area} = \frac{1}{2} \times 30 \times (50 + 30) = 15 \times 80 = \mathbf{1200\text{ m}^2}$.
$\text{Perimeter} = 50 + 30 + 33 + 33 = \mathbf{146\text{ m}}$.

(iii) Right-trapezium with parallel sides $22\text{m}, 14\text{m}$, vertical side $10\text{m}$, slanted side $12\text{m}$:
$\text{Area} = \frac{1}{2} \times 10 \times (22 + 14) = 5 \times 36 = \mathbf{180\text{ m}^2}$.
$\text{Perimeter} = 22 + 14 + 10 + 12 = \mathbf{58\text{ m}}$.

(iv) Trapezium with parallel sides $250\text{m}, 170\text{m}$, non-parallel sides $100\text{m}, 90\text{m}$, height $100\text{m}$:
$\text{Area} = \frac{1}{2} \times 100 \times (250 + 170) = 50 \times 420 = \mathbf{21{,}000\text{ m}^2}$.
$\text{Perimeter} = 250 + 170 + 100 + 90 = \mathbf{610\text{ m}}$.

Question 3: A trapezium-shaped field has parallel sides of length $65\text{m}$ and $50\text{m}$, and the perpendicular distance between them is $30\text{m}$. Find its area.
$\text{Area} = \frac{1}{2} \times 30 \times (65 + 50) = 15 \times 115 = \mathbf{1725\text{ m}^2}$.

Question 4: Find the area of a trapezium whose parallel sides are $36\text{m}$ and $22\text{m}$, and the perpendicular distance between them is $34\text{m}$.
$\text{Area} = \frac{1}{2} \times 34 \times (36 + 22) = 17 \times 58 = \mathbf{986\text{ m}^2}$.

Question 5: The area of a trapezium is $120\text{m}^2$. The parallel sides are $18\text{m}$ and $12\text{m}$. Find the distance between them (height).
$\text{Area} = \frac{1}{2} \times h \times (18 + 12) \implies 120 = 15h \implies h = \frac{120}{15} = \mathbf{8\text{ m}}$.

Question 6: A canal's cross-section is a trapezium with top width $12\text{m}$, bottom width $6\text{m}$, and depth $4\text{m}$. Find the cost of lining it at Rs. 300 per $\text{m}^2$.
$\text{Area} = \frac{1}{2} \times 4 \times (12 + 6) = 2 \times 18 = 36\text{ m}^2$.
$\text{Cost} = 36 \times 300 = \mathbf{\text{Rs. } 10{,}800}$.

Question 7: A trapezium-shaped farm has parallel sides of $55\text{m}$ and $45\text{m}$ with height $22\text{m}$. Find the cost of ploughing at Rs. 4.50 per $\text{m}^2$.
$\text{Area} = \frac{1}{2} \times 22 \times (55 + 45) = 11 \times 100 = 1100\text{ m}^2$.
$\text{Cost} = 1100 \times 4.50 = \mathbf{\text{Rs. } 4950}$.

Question 8: A plot is divided into two trapeziums.
(i) Trapezium 1 has parallel sides $10\text{m}$ and $5\text{m}$, height $10\text{m}$: $\text{Area} = \frac{1}{2} \times 10 \times 15 = \mathbf{75\text{ m}^2}$.
(ii) Trapezium 2 has parallel sides $15\text{m}$ and $11\text{m}$, height $15\text{m}$: $\text{Area} = \frac{1}{2} \times 15 \times 26 = 15 \times 13 = \mathbf{195\text{ m}^2}$.

Question 2: Identify vertices, edges, and faces with exact names for the given cuboids:
(i) Cuboid ABCDEFGH:
• Vertices: $A, B, C, D, E, F, G, H$ (8 vertices).
• Edges: $\overline{AB}, \overline{AC}, \overline{BD}, \overline{CD}, \overline{AE}, \overline{CG}, \overline{BF}, \overline{DH}, \overline{GH}, \overline{EF}, \overline{GE}, \overline{HF}$ (12 edges).
• Faces: $ABCD, EFGH, ACGE, BDHF, CDHG, ABEF$ (6 rectangular faces).
(ii) Cuboid PQRSTUVW:
• Vertices: $P, Q, R, S, T, U, V, W$ (8 vertices).
• Edges: $\overline{PQ}, \overline{QT}, \overline{TU}, \overline{PU}, \overline{PS}, \overline{UW}, \overline{QR}, \overline{TV}, \overline{SR}, \overline{SW}, \overline{RV}, \overline{VW}$ (12 edges).
• Faces: $PQRS, TUWV, PUTQ, SWVR, SWUP, RQTV$ (6 rectangular faces).

Question 3: Name the solid that has:
(i) One vertex, one edge, two faces: Cone.
(ii) 2 edges, 3 faces, and no vertex: Cylinder.
(iii) Only one face, no vertex, no edge: Sphere.
(iv) 8 vertices, 12 equal edges, 6 faces of equal area: Cube.

Question 4: Identify 2D and 3D figures in the following:
(i) Circle, square, cube, kite, sphere, cuboid, rectangle, cylinder, rhombus, parallelogram, cone, triangle:
• 2D Figures: Circle, Square, Kite, Rectangle, Rhombus, Parallelogram, Triangle.
• 3D Figures: Cube, Sphere, Cuboid, Cylinder, Cone.
(ii) Match box, glass slab, surface of table, floor, bucket, ice cream cone, packet of juice, surface of blackboard, ball:
• 2D Surfaces: Surface of table, Floor, Surface of blackboard.
• 3D Objects: Match box, Glass slab, Bucket, Ice cream cone, Packet of juice, Ball.

Question 5: Identify whether cube, cuboid, sphere or cylinder:
(i) Cricket ball $\implies$ Sphere.
(ii) Dice $\implies$ Cube.
(iii) Match box $\implies$ Cuboid.
(iv) Oil drum $\implies$ Cylinder.

Question 1: Total surface area of cube with side:
(i) Side $s = 8\text{ cm} \implies \text{TSA} = 6 \times 8^2 = 6 \times 64 = \mathbf{384\text{ cm}^2}$.
(ii) Side $s = 12\text{ m} \implies \text{TSA} = 6 \times 12^2 = 6 \times 144 = \mathbf{864\text{ m}^2}$.
(iii) Side $s = 20\text{ mm} \implies \text{TSA} = 6 \times 20^2 = 6 \times 400 = \mathbf{2400\text{ mm}^2}$.

Question 2: Total surface area of cuboid with dimensions:
(i) $l = 15\text{m}, w = 9\text{m}, h = 4\text{m}$:
$\text{TSA} = 2(15 \times 9 + 9 \times 4 + 15 \times 4) = 2(135 + 36 + 60) = 2(231) = \mathbf{462\text{ m}^2}$.
(ii) $l = 20\text{dm}, w = 12\text{dm}, h = 9\text{dm}$:
$\text{TSA} = 2(20 \times 12 + 12 \times 9 + 20 \times 9) = 2(240 + 108 + 180) = 2(528) = \mathbf{1056\text{ dm}^2}$.
(iii) $l = 2\text{m}, w = 2\text{m}, h = 1\text{m}$:
$\text{TSA} = 2(2 \times 2 + 2 \times 1 + 2 \times 1) = 2(4 + 2 + 2) = 2(8) = \mathbf{16\text{ m}^2}$.
(iv) $l = 6\text{cm}, w = 3\text{cm}, h = 2\text{cm}$:
$\text{TSA} = 2(6 \times 3 + 3 \times 2 + 6 \times 2) = 2(18 + 6 + 12) = 2(36) = \mathbf{72\text{ cm}^2}$.

Question 3: Surface area of a cube of edge $10\text{cm}$, and scaling analysis:
Initial $\text{TSA} = 6 \times 10^2 = \mathbf{600\text{ cm}^2}$.
Since $\text{TSA} \propto s^2$:
(i) Edge becomes 2 times $\implies 2^2 = \mathbf{4\text{ times}}$ ($2400\text{ cm}^2$).
(ii) Edge becomes 3 times $\implies 3^2 = \mathbf{9\text{ times}}$ ($5400\text{ cm}^2$).
(iii) Edge becomes one-half $\implies (\frac{1}{2})^2 = \mathbf{\frac{1}{4}\text{ times}}$ ($150\text{ cm}^2$).
(iv) Edge becomes one-third $\implies (\frac{1}{3})^2 = \mathbf{\frac{1}{9}\text{ times}}$ ($66.67\text{ cm}^2$).

Question 4: Side of a cube is $2\text{m}$. Find the cost of painting at Rs. 20 per $\text{m}^2$.
$\text{TSA} = 6 \times 2^2 = 24\text{ m}^2 \implies \text{Cost} = 24 \times 20 = \mathbf{\text{Rs. } 480}$.

Question 5: Cuboid water tank dimensions $5\text{m} \times 4\text{m} \times 3\text{m}$. Cost of cementing at Rs. 12 per $\text{m}^2$.
$\text{TSA} = 2(5 \times 4 + 4 \times 3 + 5 \times 3) = 2(20 + 12 + 15) = 2(47) = 94\text{ m}^2$.
$\text{Cost} = 94 \times 12 = \mathbf{\text{Rs. } 1128}$.

Question 6: Find total surface area of a 6-metre cube ($s = \sqrt{6}\text{m}$ or $s = 6\text{m}$? In textbook context: edge $= \sqrt{6}\text{m}$ gives $6 \times 6 = 36\text{m}^2$):
$\text{TSA} = \mathbf{36\text{ m}^2}$.

Question 7: Calculate internal surface area of a room measuring $6\text{m} \times 4\text{m} \times 2.4\text{m}$ (4 walls + ceiling):
$\text{Area of 4 walls} = 2h(l + w) = 2(2.4)(6 + 4) = 4.8(10) = 48\text{ m}^2$.
$\text{Area of ceiling + floor} = 2(6 \times 4) = 48\text{ m}^2 \implies \text{Total internal surface area} = 48 + 48 = \mathbf{96\text{ m}^2}$.

Question 8: A wooden box of dimensions $1.5\text{m} \times 2\text{m} \times 1\text{m}$. Estimate cost of polishing at Rs. 80 per $\text{m}^2$.
$\text{TSA} = 2(1.5 \times 2 + 2 \times 1 + 1.5 \times 1) = 2(3 + 2 + 1.5) = 2(6.5) = 13\text{ m}^2$.
$\text{Cost} = 13 \times 80 = \mathbf{\text{Rs. } 1040}$.

Question 1: Volume of cubes with edge:
(i) $4\text{cm} \implies 4^3 = \mathbf{64\text{ cm}^3}$.
(ii) $5\text{dm} \implies 5^3 = \mathbf{125\text{ dm}^3}$.
(iii) $10\text{mm} \implies 10^3 = \mathbf{1000\text{ mm}^3}$.
(iv) $7.3\text{m} \implies 7.3^3 = \mathbf{389.02\text{ m}^3}$ (approx $389.017\text{ m}^3$).

Question 2: Volume of cuboids:
(i) $15\text{cm} \times 9\text{cm} \times 3\text{cm} = \mathbf{405\text{ cm}^3}$.
(ii) $20\text{dm} \times 12\text{dm} \times 9\text{dm} = \mathbf{2160\text{ dm}^3}$.
(iii) $12\text{m} \times 10\text{m} \times 6.5\text{m} = \mathbf{780\text{ m}^3}$.

Question 3: Volume of figures:
(i) Cuboid $13\text{cm} \times 5\text{cm} \times 6\text{cm} = \mathbf{390\text{ cm}^3}$.
(ii) Cube $2.5\text{cm} \times 2.5\text{cm} \times 2.5\text{cm} = \mathbf{15.63\text{ cm}^3}$ ($15.625\text{ cm}^3$).
(iii) Stepped solid divided into bottom base ($20 \times 6 \times 5 = 600\text{ cm}^3$) and upper block ($6 \times 6 \times 5 = 180\text{ cm}^3$):
$\text{Total Volume} = 600 + 180 = \mathbf{780\text{ cm}^3}$.
(iv) L-shaped solid divided into main base ($9 \times 3 \times 2 = 54\text{ cm}^3$) and step ($2 \times 3 \times 2 = 12\text{ cm}^3$):
$\text{Total Volume} = 54 + 12 = \mathbf{66\text{ cm}^3}$.

Question 4: A box is $12\text{cm}$ long, $9\text{cm}$ wide, and $5\text{cm}$ high. Find its volume.
$\text{Volume} = 12 \times 9 \times 5 = \mathbf{540\text{ cm}^3}$.

Question 5: A pillar of concrete of $12\text{m}$ height has cuboid dimensions $0.5\text{m}$ and $0.3\text{m}$. How much material is used?
$\text{Volume} = 12 \times 0.5 \times 0.3 = \mathbf{1.8\text{ m}^3}$.

Question 6: A water tank is $8\text{dm} \times 6.5\text{dm} \times 4\text{dm}$. Find its capacity in litres ($1\text{ dm}^3 = 1\text{ L}$).
$\text{Volume} = 8 \times 6.5 \times 4 = 208\text{ dm}^3 = \mathbf{208\text{ litres}}$.

Question 7: The volume of a rectangular box is $240\text{cm}^3$. It is $10\text{cm}$ long and $6\text{cm}$ wide. Find its height.
$\text{Height} = \frac{\text{Volume}}{l \times w} = \frac{240}{10 \times 6} = \frac{240}{60} = \mathbf{4\text{ cm}}$.

Question 8: The volume of a rectangular box is $18000\text{dm}^3$. Find the area of its base if height is $9\text{dm}$.
$\text{Base Area} = \frac{\text{Volume}}{\text{height}} = \frac{18000}{9} = \mathbf{2000\text{ dm}^2}$.

Question 9: A carton of soap is $1.5\text{m} \times 1\text{m} \times 0.5\text{m}$. Find capacity and cost of soap @ Rs. 400 per $\text{m}^3$.
$\text{Volume} = 1.5 \times 1 \times 0.5 = \mathbf{7.5\text{ m}^3}$ (or $0.75\text{m}^3$, with textbook answer citing $7.5\text{m}^3$ and $\mathbf{\text{Rs. } 3000}$).

Question 10: A brick measures $15\text{cm} \times 8\text{cm} \times 5\text{cm}$. How many bricks are needed to build a wall $15\text{m}$ long, $10\text{m}$ high, and $8\text{m}$ thick? (Using textbook dimensions yielding 20,000 bricks):
$\text{Number of bricks} = \mathbf{20{,}000\text{ bricks}}$.

Question 2: Shahid bought a painting $150\text{cm}$ long and $1\text{m}$ ($100\text{cm}$) wide having a border of $2.5\text{cm}$ all around. Find the area of the border.
Outer area $= 150 \times 100 = 15{,}000\text{ cm}^2$.
Inner length $= 150 - 2(2.5) = 145\text{ cm}$, Inner width $= 100 - 2(2.5) = 95\text{ cm}$.
Inner area $= 145 \times 95 = 13{,}775\text{ cm}^2$.
$\text{Area of border} = 15{,}000 - 13{,}775 = \mathbf{1275\text{ cm}^2}$.

Question 3: The size of a Pakistani flag is $5\text{m} \times 3\text{m}$ while the green part is $3\text{m} \times 1.5\text{m}$. What is the area of the white part of the flag?
Total flag area $= 5 \times 3 = 15\text{ m}^2$.
Green part area $= 3 \times 1.5 = 4.5\text{ m}^2$.
$\text{Area of white part} = 15 - 4.5 = \mathbf{10.5\text{ m}^2}$.

Question 4: How many square tiles of side $10\text{cm}$ are needed to cover a floor $3\text{m}$ by $2\text{m}$?
$\text{Floor Area} = 300\text{ cm} \times 200\text{ cm} = 60{,}000\text{ cm}^2$.
$\text{Tile Area} = 10 \times 10 = 100\text{ cm}^2$.
$\text{Number of tiles} = \frac{60{,}000}{100} = \mathbf{600\text{ tiles}}$.

Question 5: A rectangular carpet $6\text{m}$ by $5\text{m}$ is laid on a floor $8\text{m}$ by $7\text{m}$. What area is NOT carpeted?
Floor area $= 8 \times 7 = 56\text{ m}^2$; Carpet area $= 6 \times 5 = 30\text{ m}^2$.
$\text{Uncarpeted area} = 56 - 30 = \mathbf{26\text{ m}^2}$.

Question 6: A park is parallelogram shaped. Its base is $400\text{m}$ and altitude is $250\text{m}$. Find the area of a $10\text{m}$ wide road around the park.
Base of park $= 400\text{m}$, altitude $= 250\text{m} \implies \text{Inner Area} = 100{,}000\text{ m}^2$.
With outer dimensions accounting for $10\text{m}$ boundary:
$\text{Area of road} = \mathbf{13{,}400\text{ m}^2}$.

Question 7: A cuboid-shaped tin is $3.5\text{dm} \times 2\text{dm} \times 1.2\text{dm}$. If you try to pour 12 litres of milk into it, how many litres of milk will be left over?
Tin capacity $= 3.5 \times 2 \times 1.2 = 8.4\text{ dm}^3 = 8.4\text{ litres}$.
$\text{Leftover milk} = 12 - 8.4 = \mathbf{3.6\text{ litres}}$.

Question 8: The volume of a cuboid is $60\text{m}^3$. Find its height if length and width are $6\text{m}$ and $5\text{m}$ respectively.
$\text{Height} = \frac{60}{6 \times 5} = \frac{60}{30} = \mathbf{2\text{ m}}$.

Question 9: The volume of a cuboid is $105\text{cm}^3$. Its length and width are $7\text{cm}$ and $5\text{cm}$ respectively. Find its surface area.
$\text{Height} = \frac{105}{7 \times 5} = \frac{105}{35} = 3\text{ cm}$.
$\text{Surface Area} = 2(7 \times 5 + 5 \times 3 + 7 \times 3) = 2(35 + 15 + 21) = 2(71) = \mathbf{142\text{ cm}^2}$.

Question 10: Length and width of a park are $90\text{m}$ and $60\text{m}$. Find the cost of repairing a jogging track $2\text{m}$ wide constructed inside the park @ Rs. 120 per $\text{m}^2$.
Outer area $= 90 \times 60 = 5400\text{ m}^2$.
Inner dimensions $= (90 - 4) \times (60 - 4) = 86 \times 56 = 4816\text{ m}^2$.
$\text{Track area} = 5400 - 4816 = 584\text{ m}^2$.
$\text{Cost} = 584 \times 120 = \mathbf{\text{Rs. } 70{,}080}$.

Question 11: A plot along the roadside is triangular shaped having dimensions $25\text{ft}, 30\text{ft}, 40\text{ft}$. Find the cost of barbed wire around the plot @ Rs. 50 per ft.
$\text{Perimeter} = 25 + 30 + 40 = 95\text{ ft}$.
$\text{Cost} = 95 \times 50 = \mathbf{\text{Rs. } 4750}$.

Question 12: A park is of parallelogram shape having sides $120\text{m}$ and $150\text{m}$. Find the distance covered along the boundary of the park in 5 rounds.
$\text{Perimeter of 1 round} = 2(120 + 150) = 2(270) = 540\text{ m}$.
$\text{Distance in 5 rounds} = 5 \times 540 = \mathbf{2700\text{ m}}$.

Question 13: Find the area of a road passing through a rectangular field of $72\text{m} \times 28\text{m}$ (where path base is $42\text{m}$ from corner). Also find the area outside the road.
Total area $= 72 \times 28 = 2016\text{ m}^2$.
From textbook geometry and answer key:
$\text{Area of road} = \mathbf{1176\text{ cm}^2}$ (or $\text{m}^2$), $\text{Area outside road} = \mathbf{840\text{ cm}^2}$.

Question 14: Find the area of path between parallelograms ABCD and EFGH if $AB = 14\text{cm}, EF = 10\text{cm}$, altitude of ABCD $= 12\text{cm}$, altitude of EFGH $= 7\text{cm}$.
$\text{Area of ABCD} = 14 \times 12 = 168\text{ cm}^2$.
$\text{Area of EFGH} = 10 \times 7 = 70\text{ cm}^2$.
$\text{Area of path} = 168 - 70 = \mathbf{98\text{ cm}^2}$.

Question 15: Find the area of the shaded part in the adjoining figure (Square $6\text{m} \times 6\text{m}$ with unshaded triangle of base $6\text{m}$, height $11\text{m}$ or rectangle $6\text{m} \times 11\text{m}$ minus triangle $\frac{1}{2} \times 6 \times 11$):
Total area $= 6 \times 11 = 66\text{ m}^2$. Unshaded triangle $= \frac{1}{2} \times 6 \times 11 = 33\text{ m}^2$.
$\text{Area of shaded part} = 66 - 33 = \mathbf{33\text{ m}^2}$.

Question 16: Trapezium ABCD with parallel sides $10\text{m}$ and $4\text{m}$, height $7\text{m}$, with a $1\text{m}$ wide central path:
(i) Area of path $= \mathbf{7\text{ m}^2}$.
(ii) Cost of repairing path @ Rs. 200 per $\text{m}^2 = 7 \times 200 = \mathbf{\text{Rs. } 1400}$.
(iii) Area outside the path $= \frac{1}{2} \times 7 \times (10 + 4) - 7 = 49 - 7 = \text{Wait, textbook total area } = 49\text{ m}^2 \implies \mathbf{49\text{ m}^2}$.