CBSE Class 10 Mathematics Surface Area and Volume Assignment Set 03

Exam Tip: Always keep the value of \(\pi\) in fractional form until the final calculation step to avoid rounding errors.

Question 2. A bucket of height 8 cm and made up of copper sheet is in the form of frustum of right circular cone with radii of its lower and upper ends as 3 cm and 9 cm respectively. Calculate i) the height of the cone of which the bucket is a part ii) the volume of water which can be filled in the bucket iii) the area of copper sheet required to make the bucket (Leave the answer in terms of \(\pi\))
Answer: Let the total height of the original cone be \(h'\) cm and the height of the small upper cone be \(x\) cm.
Given height of the frustum (\(h\)) = 8 cm, lower radius (\(r_1\)) = 3 cm, upper radius (\(r_2\)) = 9 cm.
(i) Using similar triangles properties of cones:
\(\frac{x}{x + 8} = \frac{3}{9}\)
\(\frac{x}{x + 8} = \frac{1}{3}\)
\(3x = x + 8\)
\(2x = 8 \implies x = 4 \text{ cm}\)
Total height of the original cone (\(h'\)) = \(4 + 8 = 12 \text{ cm}\)
(ii) Volume of water in the bucket (Volume of frustum) = \(\frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1 r_2)\)
\(= \frac{1}{3}\pi \times 8 \times (3^2 + 9^2 + 3 \times 9)\)
\(= \frac{8\pi}{3} \times (9 + 81 + 27) = \frac{8\pi}{3} \times 117 = 312\pi \text{ cm}^3\)
(iii) Slant height of the frustum (\(l\)) = \(\sqrt{h^2 + (r_2 - r_1)^2}\)
\(l = \sqrt{8^2 + (9 - 3)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ cm}\)
Area of copper sheet required (Total Surface Area of Frustum) = \(\pi (r_1 + r_2)l + \pi r_1^2\)
\(= \pi (3 + 9) \times 10 + \pi (3)^2\)
\(= 120\pi + 9\pi = 129\pi \text{ cm}^2\)
In simple words: First find the height of the missing top cone using similar triangles, then apply standard frustum formulas for volume and surface area.

Exam Tip: Remember that the area of the bottom circular base must be added when calculating the total sheet metal required for an open bucket with a closed bottom.

Question 3. A sphere and a cube have equal surface areas. Show that the ratio of the volume of the sphere to that of the cube is \(\sqrt{6} : \sqrt{\pi}\).
Answer: Let the radius of the sphere be \(r\) and the side length of the cube be \(a\).
Surface area of sphere = Surface area of cube
\(4\pi r^2 = 6a^2\)
\(a^2 = \frac{4\pi r^2}{6} = \frac{2\pi r^2}{3}\)
\(a = \sqrt{\frac{2\pi}{3}} r\)
Volume of sphere (\(V_1\)) = \(\frac{4}{3}\pi r^3\)
Volume of cube (\(V_2\)) = \(a^3 = \left( \sqrt{\frac{2\pi}{3}} r \right)^3\)
Ratio of volumes = \(\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r^3}{\left(\frac{2\pi}{3}\right)^{\frac{3}{2}} r^3}\)
\(= \frac{\frac{4}{3}\pi}{\left(\frac{2\pi}{3}\right)\sqrt{\frac{2\pi}{3}}} = \frac{2}{\sqrt{\frac{2\pi}{3}}} = \frac{2\sqrt{3}}{\sqrt{2\pi}} = \sqrt{\frac{6}{\pi}} = \sqrt{6} : \sqrt{\pi}\)
In simple words: Express the cube's side length in terms of the sphere's radius using the given surface area equality, then substitute into the volume formulas to find the ratio.

Exam Tip: Rationalize radical expressions carefully when simplifying volume ratios involving \(\pi\).

Question 4. A right triangle whose sides are 15 cm and 20 cm is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.
Answer: Let the right-angled triangle have perpendicular sides \(AB = 15 \text{ cm}\) and \(AC = 20 \text{ cm}\).
Hypotenuse \(BC = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \text{ cm}\)
When revolved about the hypotenuse \(BC\), a double cone is formed meeting at the common radius \(OA\).
Area of right triangle \(\triangle ABC = \frac{1}{2} \times BC \times OA = \frac{1}{2} \times AB \times AC\)
\(\frac{1}{2} \times 25 \times OA = \frac{1}{2} \times 15 \times 20\)
\(25 \times OA = 300 \implies OA = 12 \text{ cm}\) (common radius \(r = 12 \text{ cm}\))
Let \(BO = x\), then \(OC = 25 - x\).
In right \(\triangle OAB\): \(AB^2 = BO^2 + OA^2 \implies 15^2 = x^2 + 12^2 \implies x = 9 \text{ cm}\)
Height of first cone = 9 cm, Height of second cone = \(25 - 9 = 16 \text{ cm}\)
Volume of double cone = Volume of cone 1 + Volume of cone 2
\(= \frac{1}{3}\pi r^2 h_1 + \frac{1}{3}\pi r^2 h_2 = \frac{1}{3}\pi r^2 (h_1 + h_2)\)
\(= \frac{1}{3} \times \pi \times 12^2 \times 25 = \frac{1}{3} \times 3.1416 \times 144 \times 25 = 3768 \text{ cm}^3\)
Surface area of double cone = Curved surface area of cone 1 + Curved surface area of cone 2
\(= \pi r l_1 + \pi r l_2 = \pi \times 12 \times 15 + \pi \times 12 \times 20\)
\(= 180\pi + 240\pi = 420\pi = 420 \times 3.14 = 1318.8 \text{ cm}^2\)
In simple words: Revolving a triangle about its hypotenuse creates two cones joined base-to-base, where the triangle's altitude to the hypotenuse serves as the common radius.

Exam Tip: Use the triangle area property to easily determine the radius without solving quadratic equations.

Question 5. Water in a canal 30 dm wide and 12 dm deep is flowing with a velocity of 10 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is required for irrigation?
Answer: Width of canal = \(30 \text{ dm} = 3 \text{ m}\)
Depth of canal = \(12 \text{ dm} = 1.2 \text{ m}\)
Speed of water = \(10 \text{ km/h} = 10,000 \text{ m/h}\)
Length of water column flowing in 30 minutes (\(0.5 \text{ hours}\)) = \(10,000 \times 0.5 = 5,000 \text{ m}\)
Volume of water flowing out in 30 minutes = Width \(\times\) Depth \(\times\) Length
\(= 3 \times 1.2 \times 5,000 = 18,000 \text{ m}^3\)
Required standing water height for irrigation = \(8 \text{ cm} = 0.08 \text{ m}\)
Let the irrigated area be \(A \text{ m}^2\).
Volume of water in irrigated field = Area \(\times\) Height
\(A \times 0.08 = 18,000\)
\(A = \frac{18,000}{0.08} = 225,000 \text{ m}^2\)
In simple words: Equate the total volume of water discharged from the canal in the given time to the volume of water spread over the field.

Exam Tip: Ensure all units are converted consistently to meters before multiplying volumes.

Question 6. A cylindrical vessel of diameter 14 cm and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16 cm and height 42 cm. The total space between two vessels is filled with cork dust for heat insulation purposes. How many cubic centimetres of cork dust will be required?
Answer: Outer diameter = 16 cm, Outer radius (\(R\)) = 8 cm
Inner diameter = 14 cm, Inner radius (\(r\)) = 7 cm
Height of vessels (\(h\)) = 42 cm
Volume of cork dust required = Volume of outer cylinder - Volume of inner cylinder
\(= \pi R^2 h - \pi r^2 h = \pi h (R^2 - r^2)\)
\(= \frac{22}{7} \times 42 \times (8^2 - 7^2)\)
\(= 22 \times 6 \times (64 - 49)\)
\(= 132 \times 15 = 1980 \text{ cm}^3\)
In simple words: The volume of insulation is the difference between the outer and inner cylinders, which share the same height.

Exam Tip: Use the algebraic identity \(R^2 - r^2 = (R - r)(R + r)\) for quick mental calculation.

Question 7. An ice-cream cone has a hemispherical top. If the height of the cone is 9 cm and base radius is 2.5 cm, find the volume of ice cream cone.
Answer: Radius of cone and hemisphere (\(r\)) = 2.5 cm = \(\frac{5}{2} \text{ cm}\)
Height of the cone (\(h\)) = 9 cm
Total volume of ice-cream cone = Volume of cone + Volume of hemispherical top
\(= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3\)
\(= \frac{1}{3}\pi r^2 (h + 2r)\)
\(= \frac{1}{3} \times \frac{22}{7} \times \left(\frac{5}{2}\right)^2 \times \left(9 + 2 \times \frac{5}{2}\right)\)
\(= \frac{1}{3} \times \frac{22}{7} \times \frac{25}{4} \times (9 + 5)\)
\(= \frac{1}{3} \times \frac{22}{7} \times \frac{25}{4} \times 14\)
\(= \frac{1}{3} \times \frac{22}{7} \times \frac{175}{2} = \frac{275}{3} = 91\frac{2}{3} \text{ cm}^3\)
In simple words: Add the volume of the conical base and the hemispherical scoop on top.

Exam Tip: Factor out common terms like \(\frac{1}{3}\pi r^2\) to simplify calculations.

Question 8. A building is in the form of a cylinder surrounded by a hemispherical vaulted dome and contains \(41\frac{19}{21} \text{ m}^3\) of air. If the internal diameter of the building is equal to its total height above the floor, find the height of the building.
Answer: Let the radius of the cylindrical part be \(r\).
Height of the cylindrical part = \(h\).
Internal diameter = \(2r\). Total height = \(h + r\).
Given that diameter equals total height: \(2r = h + r \implies h = r\).
Volume of building = Volume of cylinder + Volume of hemisphere
\(= \pi r^2 h + \frac{2}{3}\pi r^3 = \pi r^2 (r) + \frac{2}{3}\pi r^3 = \frac{5}{3}\pi r^3\)
Given volume = \(41\frac{19}{21} = \frac{880}{21} \text{ m}^3\)
\(\frac{5}{3}\pi r^3 = \frac{880}{21}\)
\(\frac{5}{3} \times \frac{22}{7} \times r^3 = \frac{880}{21}\)
\(\frac{110}{21} r^3 = \frac{880}{21}\)
\(r^3 = \frac{880}{21} \times \frac{21}{110} = 8\)
\(r = 2 \text{ m}\)
Height of the building = Total height = \(2r = 2 \times 2 = 4 \text{ m}\)
In simple words: Set the total volume equation in terms of radius, solve for \(r\), and then use the geometric condition to find the total height.

Exam Tip: Pay close attention when compound shapes have height conditions linked to the diameter.

Question 9. The height of the Cone is 30 cm A small cone is cut off at the top by a plane parallel to its base if its volume be \(\frac{1}{27}\) of the volume of the given cone at what height above the base is the section cut
Answer: Let the height of the large cone be \(H = 30 \text{ cm}\) and radius be \(R\).
Let the height of the small upper cone cut off be \(h\) and its radius be \(r\).
By similar triangles (\(\triangle VO'B' \sim \triangle VOB\)):
\(\frac{h}{H} = \frac{r}{R} \implies \frac{h}{30} = \frac{r}{R} \quad \text{--- (1)}\)
Given volume of small cone = \(\frac{1}{27} \times\) Volume of large cone
\(\frac{1}{3}\pi r^2 h = \frac{1}{27} \left( \frac{1}{3}\pi R^2 H \right)\)
\(\frac{r^2}{R^2} \times \frac{h}{H} = \frac{1}{27}\)
Using equation (1), \(\left(\frac{h}{30}\right)^2 \times \frac{h}{30} = \frac{1}{27}\)
\(\frac{h^3}{27,000} = \frac{1}{27}\)
\(h^3 = 1,000 \implies h = 10 \text{ cm}\)
Height of the section cut above the base = \(H - h = 30 - 10 = 20 \text{ cm}\)
In simple words: Use the ratio of volumes to find the height of the small top cone, then subtract it from the total height to find the distance from the base.

Exam Tip: The ratio of volumes of similar cones is the cube of the ratio of their heights.

Question 10. A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface of the remainder is \(\frac{8}{9}\) th of the curved surface of the whole cone, find the ratio of the line segments into which the cone’s altitude is divided by the plane.
Answer: Let the total height of the cone be \(H\) and radius \(R\). Let the height of the removed top cone be \(h\) and radius \(r\).
By similar triangles: \(\frac{h}{H} = \frac{r}{R} = \frac{l}{L}\), where \(L\) and \(l\) are the slant heights.
Curved surface area of the remainder (frustum) = \(\frac{8}{9} \times\) CSA of whole cone
\(\text{CSA of frustum} = \text{CSA of whole cone} - \text{CSA of top cone}\)
\(\pi RL - \pi rl = \frac{8}{9}\pi RL\)
\(\pi rl = \pi RL - \frac{8}{9}\pi RL = \frac{1}{9}\pi RL\)
\(\frac{rl}{RL} = \frac{1}{9}\)
Since \(\frac{r}{R} = \frac{l}{L} = \frac{h}{H}\), we have:
\(\left(\frac{h}{H}\right)^2 = \frac{1}{9} \implies \frac{h}{H} = \frac{1}{3} \implies H = 3h\)
The altitude is divided into segments of height \(h\) and \(H - h = 3h - h = 2h\).
Required ratio of line segments = \(\frac{h}{H - h} = \frac{h}{2h} = 1:2\)
In simple words: Relate the surface area of the top cone to the total cone, find the height ratio, and then determine the ratio of the upper segment to the lower segment.

Exam Tip: The ratio of surface areas of similar cones is the square of the ratio of their linear dimensions.

Question 11. Two right circular cones X and Y are made X having 3 times the radius of Y and Y having half the Volume of X. Calculate the ratio of heights of X and Y.
Answer: Let the radius of cone Y be \(r_Y = r\), then radius of cone X is \(r_X = 3r\).
Let height of cone X be \(h_X\) and height of cone Y be \(h_Y\).
Volume of cone X (\(V_X\)) = \(\frac{1}{3}\pi (3r)^2 h_X = \frac{1}{3}\pi (9r^2) h_X\)
Volume of cone Y (\(V_Y\)) = \(\frac{1}{3}\pi r^2 h_Y\)
Given \(V_Y = \frac{1}{2} V_X\):
\(\frac{1}{3}\pi r^2 h_Y = \frac{1}{2} \left( \frac{1}{3}\pi (9r^2) h_X \right)\)
\(r^2 h_Y = \frac{9}{2} r^2 h_X\)
\(h_Y = \frac{9}{2} h_X\)
\(\frac{h_X}{h_Y} = \frac{2}{9} \implies 9:2 \quad (\text{Ratio of heights of X to Y is } 9:2)\)
In simple words: Set up the volume formula for both cones using the given radius relationship and solve for the ratio of their heights.

Exam Tip: Be careful with the order of variables when asked for the ratio of X to Y versus Y to X.

Question 12. If the areas of three adjacent faces of cuboid are x, y, z respectively, Find the volume of the cuboids.
Answer: Let the length, breadth, and height of the cuboid be \(l, b, h\).
Area of the three adjacent faces are:
\(lb = x\)
\(bh = y\)
\(hl = z\)
Volume of the cuboid (\(V\)) = \(lbh\)
Multiply the three face areas:
\(x \cdot y \cdot z = (lb) \cdot (bh) \cdot (hl)\)
\(xyz = l^2 b^2 h^2\)
\(xyz = (lbh)^2 = V^2\)
\(V = \sqrt{xyz}\)
In simple words: The volume of a cuboid is the square root of the product of the areas of its three adjacent faces.

Exam Tip: This is a standard geometric result that is very useful as a direct formula in competitive exams.

Question 13. A shuttlecock used for playing badminton has the shape of a frustum of a Cone mounted on a hemisphere. The external diameters of the frustum are 5 cm and 2 cm, and the height of the entire shuttlecock is 7cm. Find the external surface area.
Answer: Lower radius of frustum (\(r_1\)) = \(\frac{2}{2} = 1 \text{ cm}\) (matches hemisphere radius)
Upper radius of frustum (\(r_2\)) = \(\frac{5}{2} = 2.5 \text{ cm}\)
Radius of hemisphere = 1 cm
Total height of shuttlecock = 7 cm
Height of hemisphere = 1 cm
Height of the frustum (\(h\)) = \(7 - 1 = 6 \text{ cm}\)
Slant height of the frustum (\(l\)) = \(\sqrt{h^2 + (r_2 - r_1)^2}\)
\(l = \sqrt{6^2 + (2.5 - 1)^2} = \sqrt{36 + 1.5^2} = \sqrt{36 + 2.25} = \sqrt{38.25} \approx 6.18 \text{ cm}\)
External surface area = CSA of frustum + Surface area of hemisphere
\(= \pi (r_1 + r_2)l + 2\pi r_1^2\)
\(= \frac{22}{7} (1 + 2.5) \times 6.18 + 2 \times \frac{22}{7} \times 1^2\)
\(= \frac{22}{7} \times 3.5 \times 6.18 + \frac{44}{7}\)
\(= 11 \times 6.18 + 6.285 = 67.98 + 6.285 = 74.265 \approx 74.26 \text{ cm}^2\)
In simple words: The total surface area is the sum of the curved surface area of the conical frustum and the curved surface area of the hemisphere.

Exam Tip: Subtract the hemisphere's radius from the total height to find the correct height of the frustum component.

Question 14. A Solid toy in the form of a hemisphere surmounted by the right circular cone of height 2cm and diameter of the base 4 cm. If a right circular cylinder circumscribes the toy, find how much more space than the toy it will cover.
Answer: Radius of hemisphere and cone base (\(r\)) = \(\frac{4}{2} = 2 \text{ cm}\)
Height of cone (\(h_c\)) = 2 cm
Radius of hemisphere = 2 cm
Total height of the toy = Height of cone + Height of hemisphere = \(2 + 2 = 4 \text{ cm}\)
The circumscribing cylinder will have:
Radius = Radius of toy = 2 cm
Height = Total height of toy = 4 cm
Volume of cylinder = \(\pi R^2 H = \pi (2)^2 (4) = 16\pi \text{ cm}^3\)
Volume of toy = Volume of cone + Volume of hemisphere
\(= \frac{1}{3}\pi r^2 h_c + \frac{2}{3}\pi r^3\)
\(= \frac{1}{3}\pi (2)^2 (2) + \frac{2}{3}\pi (2)^3 = \frac{8\pi}{3} + \frac{16\pi}{3} = \frac{24\pi}{3} = 8\pi \text{ cm}^3\)
Extra space covered by cylinder = Volume of cylinder - Volume of toy
\(= 16\pi - 8\pi = 8\pi \text{ cm}^3\)
In simple words: Find the volume of the circumscribing cylinder and subtract the volume of the toy inside it.

Exam Tip: A circumscribing cylinder takes the maximum radius and total height of the enclosed solid figure.

Question 15. A conical vessel of radius 6cm and height 8cm is completely filled with water. A sphere is lowered into the water and its size is such that when it touches the sides, it is just immersed as shown in the figure. What fraction of water flows out.
Answer: Radius of cone (\(R\)) = 6 cm, Height of cone (\(H\)) = 8 cm.
Let \(r\) be the radius of the sphere.
In the right-angled triangle formed by the cone's axis and slant height, let vertex angle half be \(\theta\).
\(\tan\theta = \frac{R}{H} = \frac{6}{8} = \frac{3}{4} \implies \sin\theta = \frac{3}{5}\)
In the triangle formed by the sphere's center, cone vertex, and touch point:
\(\sin\theta = \frac{r}{VO} = \frac{r}{H - r}\)
\(\frac{3}{5} = \frac{r}{8 - r}\)
\(24 - 3r = 5r \implies 8r = 24 \implies r = 3 \text{ cm}\)
Volume of sphere = \(\frac{4}{3}\pi (3)^3 = 36\pi \text{ cm}^3\)
Volume of cone = \(\frac{1}{3}\pi R^2 H = \frac{1}{3}\pi (6)^2 (8) = 96\pi \text{ cm}^3\)
Fraction of water that overflows = \(\frac{\text{Volume of sphere}}{\text{Volume of cone}} = \frac{36\pi}{96\pi} = \frac{3}{8}\)
In simple words: The fraction of water overflowing equals the ratio of the volume of the fully immersed sphere to the total volume of the conical vessel.

Exam Tip: Use trigonometric ratios of the semi-vertical angle of the cone to easily find the radius of the inscribed touching sphere.

Question 16. A golf ball has a diameter equal to 4.1 cm. Its surface has 150 dimples each of radius 2 mm. Calculate the total surface area which is exposed to the surroundings assuming that the dimples are hemispherical.
Answer: Diameter of golf ball = 4.1 cm, Radius of ball (\(R\)) = \(\frac{4.1}{2} = 2.05 \text{ cm}\)
Surface area of spherical ball = \(4\pi R^2 = 4\pi (2.05)^2 = 16.8\pi \text{ cm}^2\)
Radius of each dimple (\(r\)) = \(2 \text{ mm} = 0.2 \text{ cm}\)
Each hemispherical dimple of radius \(r\) removes a flat circular area of \(\pi r^2\) from the sphere's surface, but adds a hemispherical curved surface area of \(2\pi r^2\).
Net increase in area per dimple = \(2\pi r^2 - \pi r^2 = \pi r^2\)
Total surface area exposed = Surface area of ball - Area of flat circles removed + CSA of dimples added
\(= 4\pi R^2 - 150\pi r^2 + 150(2\pi r^2)\)
\(= 4\pi R^2 + 150\pi r^2\)
\(= 16.8\pi + 150 \times \pi \times (0.2)^2\)
\(= 16.8\pi + 150 \times 0.04\pi = 16.8\pi + 6\pi = 22.8\pi\)
\(= 22.8 \times 3.1416 = 71.63 \approx 71.68 \text{ cm}^2\)
In simple words: The total exposed area is the original ball surface minus the flat tops of the dimples plus the curved inner surfaces of the dimples.

Exam Tip: Always convert all units to the same system (centimetres) before calculating surface areas.

Question 17. A solid metallic circular cone 20cm height with vertical angle 60 is cut into two parts at the middle point of its height by a plane parallel to the base. If the frustum, so obtained be drawn into a wire of diameter \(\frac{1}{16}\) cm Find the length of the wire.
Answer: Height of cone (\(H\)) = 20 cm, vertical angle = \(60^\circ\), so semi-vertical angle = \(30^\circ\).
The cone is cut at the midpoint of its height, so height of frustum (\(h\)) = 10 cm, and height of top removed cone = 10 cm.
Lower radius of frustum (\(r_2\)) = \(H \tan 30^\circ = 20 \times \frac{1}{\sqrt{3}} = \frac{20}{\sqrt{3}} \text{ cm}\)
Upper radius of frustum (\(r_1\)) = \(10 \tan 30^\circ = 10 \times \frac{1}{\sqrt{3}} = \frac{10}{\sqrt{3}} \text{ cm}\)
Volume of frustum = \(\frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1 r_2)\)
\(= \frac{1}{3}\pi \times 10 \left( \left(\frac{10}{\sqrt{3}}\right)^2 + \left(\frac{20}{\sqrt{3}}\right)^2 + \frac{10}{\sqrt{3}} \times \frac{20}{\sqrt{3}} \right)\)
\(= \frac{10\pi}{3} \left( \frac{100}{3} + \frac{400}{3} + \frac{200}{3} \right) = \frac{10\pi}{3} \times \frac{700}{3} = \frac{7000\pi}{9} \text{ cm}^3\)
Wire diameter = \(\frac{1}{16} \text{ cm} \implies \text{Radius } (r_w) = \frac{1}{32} \text{ cm}\)
Let length of wire be \(x \text{ cm}\). Volume of wire (cylinder) = \(\pi r_w^2 x\)
\(\pi \left(\frac{1}{32}\right)^2 x = \frac{7000\pi}{9}\)
\(\frac{1}{1024} x = \frac{7000}{9}\)
\(x = \frac{7000 \times 1024}{9} = \frac{7,168,000}{9} \text{ cm} = \frac{71,680}{9} \text{ m} = 7964.44 \text{ m}\)
In simple words: Equate the volume of the frustum to the volume of the cylindrical wire to find its length.

Exam Tip: Keep units straight by converting centimeters to meters at the final step when required.

Question 18. If the areas of the circular bases of a frustum of a cone are \(4\text{ cm}^2\) and \(9\text{ cm}^2\) respectively and the height of the frustum is 12cm. What is the volume of the frustum.
Answer: Area of top base (\(A_1\)) = \(\pi r_1^2 = 4 \text{ cm}^2\)
Area of bottom base (\(A_2\)) = \(\pi r_2^2 = 9 \text{ cm}^2\)
Height of the frustum (\(h\)) = 12 cm
Volume of frustum = \(\frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1 r_2)\)
\(= \frac{1}{3} h (\pi r_1^2 + \pi r_2^2 + \sqrt{\pi r_1^2 \cdot \pi r_2^2})\)
\(= \frac{1}{3} \times 12 \times (4 + 9 + \sqrt{4 \times 9})\)
\(= 4 \times (13 + 6) = 4 \times 19 = 76 \text{ cm}^3\)
In simple words: Substitute the given base areas directly into the modified frustum volume formula.

Exam Tip: When base areas are given directly, avoid finding individual radii to save time.

Question 19. The lower portion of a hay stack is an inverted cone frustum and the upper part is a cone find the total volume of the hay stack.
Answer: From the accompanying diagram dimensions:
Lower frustum part: Height = 7 cm, lower diameter = 4 cm (radius \(r_1 = 2 \text{ cm}\)), upper diameter = 6 cm (radius \(r_2 = 3 \text{ cm}\)).
Upper cone part: Base radius = 3 cm, height = \(10.5 - 7 = 3.5 \text{ cm}\).
Volume of lower frustum = \(\frac{1}{3}\pi h_1 (r_1^2 + r_2^2 + r_1 r_2)\)
\(= \frac{1}{3} \times \frac{22}{7} \times 7 \times (2^2 + 3^2 + 2 \times 3)\)
\(= \frac{22}{3} \times (4 + 9 + 6) = \frac{22}{3} \times 19 = \frac{418}{3} = 139.33 \text{ cm}^3\)
Volume of upper cone = \(\frac{1}{3}\pi r_2^2 h_2 = \frac{1}{3} \times \frac{22}{7} \times 3^2 \times 3.5\)
\(= \frac{1}{3} \times \frac{22}{7} \times 9 \times 3.5 = 33 \text{ cm}^3\)
Total volume of hay stack = Volume of frustum + Volume of upper cone
\(= 139.33 + 33 = 172.33 \text{ cm}^3\)
In simple words: Calculate the volume of the frustum and the cone separately and add them together to get the total volume.

Exam Tip: Double-check the height subtraction for the upper conical section from the total height.

Question 20. A vessel in shape of an inverted cone is surmounted by a cylinder has a common radius of 7cm this was filled with liquid till it covered one third the height of the cylinder. If the height of each part is 9cm and the vessel is turned upside down. Find the volume of the liquid and to what height will it reach in the cylindrical part.
Answer: Common radius (\(r\)) = 7 cm
Height of each part (cone and cylinder) = 9 cm
The vessel is initially upright and filled with liquid up to one-third the height of the cylinder.
Height of liquid in cylinder = \(\frac{1}{3} \times 9 = 3 \text{ cm}\).
Volume of liquid = Volume of inverted cone + Volume of cylinder filled with liquid
\(= \frac{1}{3}\pi r^2 h_{\text{cone}} + \pi r^2 h_{\text{liquid in cyl}}\)
\(= \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 9 + \frac{22}{7} \times 7^2 \times 3\)
\(= \frac{22}{7} \times 49 \times (3 + 3) = 154 \times 6 = 924 \text{ cm}^3\)
When the vessel is turned upside down, the cone is at the bottom and the cylinder is at the top.
The total volume of liquid is 924 \(\text{ cm}^3\).
Volume of the inverted cone at the bottom = \(\frac{1}{3}\pi r^2 h_{\text{cone}} = \frac{1}{3} \times \frac{22}{7} \times 49 \times 9 = 462 \text{ cm}^3\)
Remaining volume of liquid in the cylindrical part = \(924 - 462 = 462 \text{ cm}^3\)
Height of liquid in the cylindrical part (\(h'\)) = \(\frac{\text{Remaining volume}}{\pi r^2}\)
\(h' = \frac{462}{\frac{22}{7} \times 49} = \frac{462}{154} = 3 \text{ cm}\)
In simple words: Calculate the total liquid volume in the upright position, then subtract the cone's capacity when inverted to find the height the liquid reaches in the cylinder.

Exam Tip: Turning a vessel upside down swaps the positions of the cone and cylinder, altering how liquid distributes by volume levels.