2D Shapes
Rectangle and Square
Rectangle: Area = l × w, Perimeter = 2(l + w). Square: Area = a², Perimeter = 4a.
Triangle
Area = ½ × base × height. When all sides are known, use Heron's formula: A = √(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2 is the semi-perimeter.
Circle and Sector
Circle: Area = πr², Circumference = 2πr.
Sector with central angle θ (degrees): Area = πr²θ/360, Arc length = 2πrθ/360 = πrθ/180.
Sector analogy: A sector is a "pie slice." Its area is the fraction θ/360 of the full circle. A 90° sector is exactly a quarter circle.
Trapezium
Area = ½ × (sum of parallel sides) × height = ½(a + b) × h.
Annular Ring (Path around a rectangle)
Area of path = Area of outer rectangle − Area of inner rectangle. For a path of width w around a l × b field: Outer = (l+2w)(b+2w), Inner = l×b.
3D Solids
Cube and Cuboid
Cube (side a): SA = 6a², V = a³. Cuboid (l×b×h): SA = 2(lb + bh + lh), V = lbh.
Cylinder
Curved SA = 2πrh. Total SA = 2πr(r + h). Volume = πr²h.
Cone
Slant height l = √(r² + h²). Curved SA = πrl. Total SA = πr(r + l). Volume = ⅓πr²h.
Remember: A cone has ⅓ the volume of a cylinder with the same base and height. Fill a cone and pour it into a cylinder — you'd need to pour 3 times to fill the cylinder.
Sphere and Hemisphere
Sphere: SA = 4πr², V = (4/3)πr³. Hemisphere: Curved SA = 2πr², Total SA = 3πr² (curved + flat base), V = (2/3)πr³.
Frustum (Truncated Cone)
A frustum is a cone with its top cut off. If R = bottom radius, r = top radius, h = vertical height, then slant height l = √(h² + (R−r)²). Curved SA = π(R+r)l. Volume = (1/3)πh(R² + Rr + r²).
Combined Solids
For combined solids (e.g., cone on cylinder, hemisphere on cuboid), compute SA and V of each part separately, then add. Be careful not to include the hidden joining face in SA.
3D Solid Calculator
Select a solid type, enter dimensions, and see surface area and volume calculated instantly with the formula shown.
Interactive 3D Solid Calculator
Board Questions
Q1
Dhaka 2025
Cylinder to Cone Volume
6 Marks
A cylindrical tank has radius 1.4 m and height 3 m. It is filled with water. If the water is pumped into a conical vessel with the same base radius, find the height of the cone. (Use π = 22/7)
1
Volume of cylinder: V = πr²h = (22/7) × 1.4² × 3 = (22/7) × 1.96 × 3.
2
V = (22 × 1.96 × 3) / 7 = 129.36 / 7 = 18.48 m³.
3
Volume of cone = ⅓πr²H where r = 1.4 m. Setting equal to cylinder volume:
4
⅓ × (22/7) × 1.96 × H = 18.48.
5
H = 18.48 × 3 × 7 / (22 × 1.96) = 388.08 / 43.12 = 9 m.
Ans
Height of cone = 9 m ✓
Q2
Rajshahi 2024
Hemisphere + Cone
7 Marks
A solid consists of a hemisphere of radius 7 cm surmounted by a cone of the same base radius. The total height of the solid is 16 cm. Find the total surface area and volume.
1
r = 7 cm. Height of cone = total height − radius of hemisphere = 16 − 7 = 9 cm.
2
Slant height of cone: l = √(r² + h²) = √(49 + 81) = √130 ≈ 11.4 cm.
3
Surface Area: curved surface of hemisphere + curved surface of cone (the flat base is hidden). SA = 2πr² + πrl = πr(2r + l).
4
SA = (22/7) × 7 × (14 + √130) = 22 × (14 + 11.40) ≈ 22 × 25.40 ≈ 558.8 cm².
5
Volume: V = ⅔πr³ + ⅓πr²h = ⅓πr²(2r + h).
6
V = ⅓ × (22/7) × 49 × (14 + 9) = ⅓ × 22 × 7 × 23 = ⅓ × 3542 ≈ 1180.7 cm³.
Ans
SA ≈ 558.8 cm²; Volume ≈ 1180.7 cm³ ✓
Q3
Chittagong 2023
Room Whitewashing Cost
6 Marks
A room is 12 m long, 10 m wide, and 4 m high. Find the cost of whitewashing the walls and ceiling at 8 Tk/m², given that the total area of doors and windows is 50 m².
1
Lateral (wall) area = perimeter of floor × height = 2(12+10) × 4 = 2 × 22 × 4 = 176 m².
2
Ceiling area = 12 × 10 = 120 m².
3
Total area to whitewash = 176 + 120 − 50 = 246 m².
4
Cost = 246 × 8 = 1968 Tk.
Ans
Cost of whitewashing = 1968 Tk ✓
Q4
Jessore 2022
Cone — CSA and Volume
5 Marks
Find the curved surface area and volume of a cone with diameter 14 cm and slant height 15 cm. Use π = 22/7.
1
r = diameter/2 = 7 cm, l = 15 cm.
2
Vertical height: h = √(l² − r²) = √(225 − 49) = √176 ≈ 13.27 cm.
3
Curved Surface Area: CSA = πrl = (22/7) × 7 × 15 = 22 × 15 = 330 cm².
4
Volume: V = ⅓πr²h = ⅓ × (22/7) × 49 × 13.27 = ⅓ × 22 × 7 × 13.27 ≈ ⅓ × 2043.5 ≈ 681.2 cm³.
Ans
CSA = 330 cm²; Volume ≈ 681.2 cm³ ✓
Q5
Comilla 2024
Sphere Melted into Smaller Spheres
5 Marks
A sphere of radius 6 cm is melted and recast into 6 equal smaller spheres. Find the radius of each small sphere.
1
Volume of large sphere: V = (4/3)πR³ = (4/3)π × 216 = 288π cm³.
2
Volume of one small sphere: v = V/6 = 288π/6 = 48π cm³.
3
For small sphere: (4/3)πr³ = 48π → r³ = 48 × 3/4 = 36.
4
r = ∛36 ≈ 3.30 cm.
Ans
Radius of each small sphere = ∛36 ≈ 3.30 cm ✓
Q6
Barisal 2021
Well — Earth Ring
7 Marks
A well 14 m deep has an inner radius of 2 m. Find the volume of earth removed. If this earth is spread uniformly in a ring 5 m wide around the well, find the height of the ring.
1
Volume of earth removed = volume of cylinder: V = πr²h = (22/7) × 4 × 14 = (22/7) × 56 = 176 m³.
2
The ring has inner radius r = 2 m and outer radius R = 2 + 5 = 7 m.
3
Area of annular ring: A = π(R² − r²) = (22/7)(49 − 4) = (22/7) × 45 = 990/7 ≈ 141.4 m².
4
Height of ring: H = V / A = 176 / (990/7) = 176 × 7/990 = 1232/990 ≈ 1.245 m.
Ans
Earth removed = 176 m³; Ring height ≈ 1.24 m ✓
Q7
Sylhet 2023
Frustum
6 Marks
A frustum has top radius 4 cm, bottom radius 7 cm, and height 9 cm. Find its volume and curved surface area.
1
R = 7 cm, r = 4 cm, h = 9 cm. Slant height: l = √(h² + (R−r)²) = √(81 + 9) = √90 = 3√10 ≈ 9.49 cm.
2
Volume: V = ⅓πh(R² + Rr + r²) = ⅓ × (22/7) × 9 × (49 + 28 + 16).
3
V = ⅓ × (22/7) × 9 × 93 = (22 × 9 × 93) / (3 × 7) = 18414/21 = 877 cm³.
4
Curved SA: CSA = π(R + r)l = (22/7) × 11 × 3√10 = (22 × 11 × 3√10)/7 ≈ (726 × 3.162)/7 ≈ 328 cm².
Ans
Volume ≈ 877 cm³; Curved SA ≈ 328 cm² ✓
Q8
Dinajpur 2024
Sector Area and Perimeter
5 Marks
A sector of a circle has radius 12 cm and central angle 60°. Find the area and perimeter of the sector.
1
r = 12 cm, θ = 60°.
2
Area of sector: A = πr²θ/360 = (22/7) × 144 × 60/360 = (22/7) × 144 × 1/6 = (22 × 24)/7 = 528/7 ≈ 75.43 cm².
3
Arc length: L = 2πrθ/360 = 2 × (22/7) × 12 × 60/360 = (44 × 12)/(7 × 3) = 528/21 = 176/7 ≈ 25.14 cm.
4
Perimeter of sector = arc length + 2r = 25.14 + 24 = 49.14 cm.
Ans
Area ≈ 75.43 cm²; Perimeter ≈ 49.14 cm ✓
Q9
Mymensingh 2023
Path Area and Cost
6 Marks
A farmer has a rectangular field 80 m × 60 m. A path 2 m wide runs along the inside border. Find the area of the path and the cost to tile it at 50 Tk/m².
1
Outer dimensions: 80 m × 60 m. Area = 80 × 60 = 4800 m².
2
Inner dimensions (path 2 m inside): (80 − 2×2) × (60 − 2×2) = 76 × 56 = 4256 m².
3
Area of path = 4800 − 4256 = 544 m².
4
Cost = 544 × 50 = 27200 Tk.
Ans
Path area = 544 m²; Tiling cost = 27,200 Tk ✓
Formula Sheet
2D Shapes
Circle
A = πr² C = 2πr
Sector
A = πr²θ/360 L = πrθ/180
Triangle (Heron's)
A = √(s(s-a)(s-b)(s-c))
Trapezium
A = ½(a+b)h
3D Solids
Cylinder
V = πr²h SA = 2πr(r+h)
Cone
V = ⅓πr²h l = √(r²+h²)
Sphere
V = (4/3)πr³ SA = 4πr²
Hemisphere
V = (2/3)πr³ SA = 3πr²
Frustum
V = ⅓πh(R²+Rr+r²)
Cuboid
V = lbh SA = 2(lb+bh+lh)