Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 9 Set 9.2 Surface Area and Volume Solutions

Question 1. Perpendicular height of a cone is 12 cm and its slant height is 13 cm. Find the radius of the base of the cone.
Given: Height (h) = 12 cm, length (l) = 13 cm
To find: Radius of the base of the cone (r)
Answer:
Solution:
We know the relation between slant height (l), radius (r), and height (h) of a cone:
\[l^2 = r^2 + h^2\]
Substituting the given values:
\[13^2 = r^2 + 12^2\]
\[169 = r^2 + 144\]
\[169 - 144 = r^2\]
\[r^2 = 25\]
\[r = \sqrt{25}\] ... [Taking square root on both sides]
\[r = 5 \text{ cm}\]
The radius of base of the cone is 5 cm.
In simple words: We used the Pythagorean theorem, which relates the slant height, radius, and perpendicular height of a cone, to find the unknown radius.

🎯 Exam Tip: Remember the relationship \(l^2 = r^2 + h^2\) for a cone, as it's fundamental for solving problems involving these three dimensions.

Question 2. Find the volume of a cone, if its total surface area is 7128 sq.cm and radius of base is 28 cm. \( \left( \pi = \frac{22}{7} \right) \)
Given: Radius (r) = 28 cm,
Total surface area of cone = 7128 sq.cm
To find: Volume of the cone
Answer:
Solution:
i. Total surface area of cone = \( \pi r (l + r) \)
Therefore, \(7128 = \pi \times 28 \times (l + 28)\)
\(7128 = \frac{22}{7} \times 28 \times (l + 28)\)
\(7128 = 22 \times 4 \times (l + 28)\)
Therefore, \(l + 28 = \frac{7128}{22 \times 4}\)
\(l + 28 = 81\)
Therefore, \(l = 81 - 28\)
\(l = 53 \text{cm}\)
ii. Now, \(l^2 = r^2 + h^2\)
Therefore, \(53^2 = 28^2 + h^2\)
\(2809 = 784 + h^2\)
Therefore, \(2809 - 784 = h^2\)
Therefore, \(h^2 = 2025\)
Therefore, \(h = \sqrt{2025}\) ...... [Taking square root on both sides]
\(h = 45 \text{ cm}\)
Therefore, Volume of cone = \( \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times 28^2 \times 45 \)
\( = \frac{1}{3} \times \frac{22}{7} \times 28 \times 28 \times 45 \)
\( = 22 \times 4 \times 28 \times 15 \)
\( = 36960 \text{ cubic.cm}\)
Therefore, The volume of the cone is 36960 cubic.cm.
In simple words: First, we calculated the slant height using the total surface area and radius. Then, we used the slant height and radius to find the perpendicular height. Finally, we calculated the volume of the cone using the radius and the perpendicular height.

🎯 Exam Tip: When multiple unknown dimensions (like slant height, height, radius) are involved, break down the problem into steps. Use the given formulas for surface area or volume to find intermediate values, often using the Pythagorean theorem for height or slant height.

Question 3. Curved surface area of a cone is 251.2 cm² and radius of its base is 8 cm. Find its slant height and perpendicular height, \( (\pi = 3.14) \)
Given: Radius (r) = 8 cm, curved surface area of cone = 251.2 cm²
To find: Slant height (l) and the perpendicular height (h) of the cone
Answer:
Solution:
i. Curved surface area of cone = \( \pi r l \)
Therefore, \(251.2 = 3.14 \times 8 \times l\)
\[l = \frac{251.2}{3.14 \times 8}\]
\[ = \frac{25120}{314 \times 8}\]
\[ = \frac{3140}{314}\]
Therefore, \(l = 10 \text{ cm}\)
ii. Now, \(l^2 = r^2 + h^2\)
Therefore, \(10^2 = 8^2 + h^2\)
Therefore, \(100 = 64 + h^2\)
Therefore, \(100 - 64 = h^2\)
Therefore, \(h^2 = 36\)
Therefore, \(h = \sqrt{36}\) ... [Taking square root on both sides]
\(h = 6 \text{ cm}\)
Therefore, The slant height and the perpendicular height of the cone are 10 cm and 6 cm respectively.
In simple words: We first found the slant height using the given curved surface area and radius, then used the Pythagorean theorem with the slant height and radius to calculate the perpendicular height.

🎯 Exam Tip: Always identify which dimensions are given and which need to be found. Use the appropriate surface area or volume formula to find one unknown, and then the Pythagorean relation \(l^2 = r^2 + h^2\) to find the other, if needed.

Question 4. What will be the cost of making a closed cone of tin sheet having radius of base 6 m and slant height 8 m if the rate of making is Rs. 10 per sq.m?
Given: Radius (r) = 6 m, length (l) = 8 m
To find: Total cost of making the cone
Answer:
Solution:
i. To find the total cost of making the cone of tin sheet, first we need to find the total surface area of the cone.
Total surface area of the cone = \( \pi r (l + r) \)
\( = \frac{22}{7} \times 6 \times (8 + 6) \)
\( = \frac{22}{7} \times 6 \times 14 \)
\( = 22 \times 6 \times 2 = 264 \text{ sq.m}\)
ii. Rate of making the cone = Rs. 10 per sq.m
Therefore, Total cost = Total surface area x Rate of making the cone
\( = 264 \times 10 \)
\( = \text{Rs. } 2640 \)
Therefore, The total cost of making the cone of tin sheet is Rs. 2640.
In simple words: First, we calculated the total surface area of the cone, which represents the amount of tin sheet needed. Then, we multiplied this area by the given rate per square meter to find the total cost.

🎯 Exam Tip: For cost-related problems, always ensure you calculate the correct area (curved surface area for covering, total surface area for making a closed object) before multiplying by the rate.

Question 5. Volume of a cone is 6280 cubic cm and base radius of the cone is 20 cm. Find its perpendicular height, \( (\pi = 3.14) \)
Given: Radius (r) = 20 cm,
Volume of cone = 6280 cubic cm
To find: Perpendicular height (h) of the cone
Answer:
Solution:
Volume of cone = \( \frac{1}{3} \pi r^2 h \)
Therefore, \(6280 = \frac{1}{3} \times 3.14 \times 20^2 \times h\)
Therefore, \(h = \frac{6280 \times 3}{3.14 \times 400}\)
\[h = \frac{6280 \times 3}{1256}\]
\[h = \frac{20 \times 3}{4}\]
\[h = 5 \times 3\]
\(h = 15 \text{ cm}\)
Therefore, The perpendicular height of the cone is 15 cm.
In simple words: We used the formula for the volume of a cone and rearranged it to solve for the perpendicular height, substituting the given volume, radius, and value of pi.

🎯 Exam Tip: When a specific value for \(\pi\) is provided (e.g., 3.14 or 22/7), use that value consistently in your calculations to ensure accuracy. Make sure to isolate the unknown variable (height 'h' in this case) correctly.

Question 6. Surface area of a cone is 188.4 sq.cm and its slant height is 10 cm. Find its perpendicular height \( (\pi = 3.14) \).
Given: Length (l) = 10 cm, curved surface area of the cone = 188.4 sq.cm
To find: Perpendicular height (h) of the cone
Answer:
Solution:
i. Curved surface area of the cone = \( \pi r l \)
Therefore, \(188.4 = 3.14 \times r \times 10\)
\[r = \frac{188.4}{3.14 \times 10}\]
\[r = \frac{188.4}{31.4}\]
\[r = \frac{1884}{314}\]
\(r = 6 \text{ cm}\)
ii. Now, \(l^2 = r^2 + h^2\)
Therefore, \(10^2 = 6^2 + h^2\)
Therefore, \(100 = 36 + h^2\)
Therefore, \(100 - 36 = h^2\)
Therefore, \(h^2 = 64\)
Therefore, \(h = \sqrt{64}\) ... [Taking square root on both sides]
\(h = 8 \text{ cm}\)
Therefore, The perpendicular height of the cone is 8 cm.
In simple words: We first found the radius using the given curved surface area and slant height, and then applied the Pythagorean theorem to calculate the perpendicular height of the cone.

🎯 Exam Tip: This problem requires two main steps: first, finding the radius using the curved surface area formula, and second, using the derived radius and given slant height in the Pythagorean relation to find the perpendicular height. Ensure accurate calculations at each stage.

Question 7. Volume of a cone is 1232 cm³ and its height is 24 cm. Find the surface area of the cone. \( \left( \pi = \frac{22}{7} \right) \)
Given: Height (h) = 24 cm,
Volume of cone = 1232 cm³
To find: Surface area of the cone
Answer:
Solution:
i. Volume of cone = \( \frac{1}{3} \pi r^2 h \)
Therefore, \(1232 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 24\)
Therefore, \(r^2 = \frac{1232 \times 3 \times 7}{22 \times 24}\)
\[r^2 = \frac{1232 \times 21}{528}\]
\[r^2 = \frac{25872}{528}\]
\[r^2 = \frac{56 \times 1 \times 7}{1 \times 8}\]
\(r^2 = 49\)
Therefore, \(r = \sqrt{49}\) ... [Taking square root on both sides]
\(r = 7 \text{ cm}\)
ii. Now, \(l^2 = r^2 + h^2\)
Therefore, \(l^2 = 7^2 + 24^2\)
\( = 49 + 576 = 625 \)
Therefore, \(l = \sqrt{625}\) ... [Taking square root on both sides]
\(l = 25 \text{ cm}\)
iii. Curved surface area of cone = \( \pi r l \)
\( = \frac{22}{7} \times 7 \times 25 \)
\( = 22 \times 25 \)
\( = 550 \text{ sq.cm}\)
Therefore, The surface area of the cone is 550 sq.cm.
In simple words: We first used the given volume and height to find the radius of the cone. Then, we used the calculated radius and given height to find the slant height. Finally, we calculated the curved surface area using the radius and slant height.

🎯 Exam Tip: "Surface area of the cone" usually refers to the curved surface area unless "total surface area" is specified. Be careful with this distinction in questions. This problem involves finding radius, then slant height, and finally the curved surface area.

Question 8. The curved surface area of a cone is 2200 sq.cm and its slant height is 50 cm. Find the total surface area of cone. \( \left( \pi = \frac{22}{7} \right) \)
Given: Length (l) = 50 cm, curved surface area of cone = 2200 sq.cm
To find: Total surface area of the cone
Answer:
Solution:
i. Curved surface area of cone = \( \pi r l \)
Therefore, \(2200 = \frac{22}{7} \times r \times 50\)
Therefore, \(r = \frac{2200 \times 7}{22 \times 50}\)
\[r = \frac{100 \times 7}{50}\]
\[r = 2 \times 7\]
\(r = 14 \text{ cm}\)
ii. Total surface area of cone = \( \pi r (l + r) \)
\( = \frac{22}{7} \times 14 \times (50 + 14) \)
\( = \frac{22}{7} \times 14 \times 64 \)
\( = 22 \times 2 \times 64 \)
\( = 2816 \text{ sq.cm}\)
Therefore, The total surface area of the cone is 2816 sq.cm.
In simple words: First, we used the given curved surface area and slant height to find the radius of the cone. Then, we used the calculated radius and given slant height to find the total surface area.

🎯 Exam Tip: For total surface area, you need both the radius and the slant height. If one is missing but other surface area values are given, you can often derive the missing dimension first.

Question 9. There are 25 persons in a tent which is conical in shape. Every person needs an area of 4 sq.m, of the ground inside the tent. If height of the tent is 18 m, find the volume of the tent.
Given: For the tent,
height (h) = 18m,
number of people in the tent = 25,
area required for each person = 4 sq.m
To find: Volume of the tent
Answer:
Solution:
i. Every person needs an area of 4 sq.m, of the ground inside the tent.
Surface area of the base of the tent = number of people in the tent \( \times \) area required for each person
\( = 25 \times 4 \)
\( = 100 \text{ sq.m}\)
ii. Surface area of the base of the tent = \( \pi r^2 \)
Therefore, \(100 = \pi r^2\)
Therefore, \( \pi r^2 = 100 \)
iii. Volume of the tent = \( \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times 100 \times 18 \) ...... [Since \( \pi r^2 = 100 \)]
\( = 100 \times 6 \)
\( = 600 \text{ cubic metre}\)
Therefore, The volume of the tent is 600 cubic metre.
In simple words: We first calculated the total base area required by all persons. Since the tent is conical, this base area is \( \pi r^2 \). We then substituted this value into the volume formula for a cone along with the given height to find the tent's volume.

🎯 Exam Tip: Pay attention to how seemingly unrelated information (like people and area per person) translates into a geometrical dimension (base area or radius). Look for opportunities to substitute known compound values (like \( \pi r^2 \)) directly into formulas to simplify calculations.

Question 10. In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is 2.1 m and diameter of base is 7.2 m. Find the volume of the heap of the fodder. If it is to be covered by polythene in rainy season then how much minimum polythene sheet is needed? \( \left( \pi = \frac{22}{7} \text{ and } \sqrt{17.37} = 4.17 \right) \)
Given: Height of the heap (h) = 2.1 m.
diameter of the base (d) = 7.2 m
Therefore, Radius of the base (r) = \( \frac{d}{2} = \frac{7.2}{2} = 3.6 \text{ m} \)
To find: Volume of the heap of the fodder and polythene sheet required
Answer:
Solution:
i. Volume of the heap of fodder = \( \frac{1}{3} \pi r^2 h \)
\( = \frac{1}{3} \times \frac{22}{7} \times (3.6)^2 \times 2.1 \)
\( = \frac{1}{3} \times \frac{22}{7} \times 3.6 \times 3.6 \times 2.1 \)
\( = 1 \times 22 \times 1.2 \times 3.6 \times 0.3 \)
\( = 28.51 \text{ cubic metre}\)
ii. Now, \(l^2 = r^2 + h^2\)
\( = (3.6)^2 + (2.1)^2\)
\( = 12.96 + 4.41 \)
Therefore, \(l^2 = 17.37\)
Therefore, \(l = \sqrt{17.37}\) ... [Taking square root on both sides]
\(l = 4.17 \text{ m}\)
iii. Area of the polythene sheet needed to cover the heap of the fodder = Curved surface area of the conical heap
\( = \pi r l \)
\( = \frac{22}{7} \times 3.6 \times 4.17 \)
\( = 47.18 \text{ sq.m}\)
Therefore, The volume of the heap of the fodder is 28.51 cubic metre and a polythene sheet of 47.18 sq.m will be required to cover it.
In simple words: First, we calculated the volume of the conical fodder heap using its radius (derived from diameter) and height. Then, to find the amount of polythene sheet needed, we calculated the slant height using the Pythagorean theorem, and finally computed the curved surface area of the cone.

🎯 Exam Tip: This question requires calculating both volume and curved surface area. Remember that the polythene sheet covers only the curved surface, not the base. Also, be sure to use the correct value of \(\sqrt{17.37}\) if provided, to maintain accuracy.