Surface Areas & Volumes
NCERT Solutions • Class 9 Maths • Chapter 11Exercise 11.1
1. Diameter of the base of a cone is $10.5 \text{ cm}$ and its slant height is $10 \text{ cm}$. Find its curved surface area.
Given: Diameter $d = 10.5 \text{ cm} \Rightarrow \text{Radius } r = \frac{10.5}{2} = 5.25 \text{ cm}$.
Slant height $l = 10 \text{ cm}$.
Formula: CSA of cone = $\pi r l$
$= \frac{22}{7} \times 5.25 \times 10$
$= 22 \times 0.75 \times 10 = 16.5 \times 10 =$ $165 \text{ cm}^2$.
Slant height $l = 10 \text{ cm}$.
Formula: CSA of cone = $\pi r l$
$= \frac{22}{7} \times 5.25 \times 10$
$= 22 \times 0.75 \times 10 = 16.5 \times 10 =$ $165 \text{ cm}^2$.
2. Find the total surface area of a cone, if its slant height is $21 \text{ m}$ and diameter of its base is $24 \text{ m}$.
Given: $l = 21 \text{ m}$, $d = 24 \text{ m} \Rightarrow r = 12 \text{ m}$.
Formula: TSA of cone = $\pi r(l + r)$
$= \frac{22}{7} \times 12 \times (21 + 12)$
$= \frac{22}{7} \times 12 \times 33 = \frac{8712}{7}$
$= 1244.571… \approx$ $1244.57 \text{ m}^2$.
Formula: TSA of cone = $\pi r(l + r)$
$= \frac{22}{7} \times 12 \times (21 + 12)$
$= \frac{22}{7} \times 12 \times 33 = \frac{8712}{7}$
$= 1244.571… \approx$ $1244.57 \text{ m}^2$.
3. Curved surface area of a cone is $308 \text{ cm}^2$ and its slant height is $14 \text{ cm}$. Find (i) radius of the base and (ii) total surface area of the cone.
(i) Find Radius:
CSA $= \pi r l = 308$
$\frac{22}{7} \times r \times 14 = 308 \Rightarrow 44r = 308$
$r = \frac{308}{44} =$ $7 \text{ cm}$.
CSA $= \pi r l = 308$
$\frac{22}{7} \times r \times 14 = 308 \Rightarrow 44r = 308$
$r = \frac{308}{44} =$ $7 \text{ cm}$.
(ii) Find Total Surface Area:
TSA $= \text{CSA} + \text{Base Area} = 308 + \pi r^2$
$= 308 + \frac{22}{7} \times 7 \times 7$
$= 308 + 154 =$ $462 \text{ cm}^2$.
TSA $= \text{CSA} + \text{Base Area} = 308 + \pi r^2$
$= 308 + \frac{22}{7} \times 7 \times 7$
$= 308 + 154 =$ $462 \text{ cm}^2$.
4. A conical tent is $10 \text{ m}$ high and the radius of its base is $24 \text{ m}$. Find (i) slant height of the tent. (ii) cost of the canvas required to make the tent, if the cost of $1 \text{ m}^2$ canvas is ₹ 70.
(i) Slant height ($l$):
$l = \sqrt{r^2 + h^2} = \sqrt{24^2 + 10^2}$
$= \sqrt{576 + 100} = \sqrt{676} =$ $26 \text{ m}$.
$l = \sqrt{r^2 + h^2} = \sqrt{24^2 + 10^2}$
$= \sqrt{576 + 100} = \sqrt{676} =$ $26 \text{ m}$.
(ii) Cost of canvas:
Canvas Area = CSA $= \frac{22}{7} \times 24 \times 26 = \frac{13728}{7} \text{ m}^2$.
Cost $= \text{Area} \times \text{Rate} = \frac{13728}{7} \times 70$
$= 13728 \times 10 =$ ₹ $1,37,280$.
Canvas Area = CSA $= \frac{22}{7} \times 24 \times 26 = \frac{13728}{7} \text{ m}^2$.
Cost $= \text{Area} \times \text{Rate} = \frac{13728}{7} \times 70$
$= 13728 \times 10 =$ ₹ $1,37,280$.
5. What length of tarpaulin $3 \text{ m}$ wide will be required to make conical tent of height $8 \text{ m}$ and base radius $6 \text{ m}$? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately $20 \text{ cm}$ (Use $\pi = 3.14$).
Step 1: Find Slant Height ($l$)
$l = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m}$.
$l = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m}$.
Step 2: Find Area of Tarpaulin (CSA)
CSA $= \pi r l = 3.14 \times 6 \times 10 = 188.4 \text{ m}^2$.
CSA $= \pi r l = 3.14 \times 6 \times 10 = 188.4 \text{ m}^2$.
Step 3: Find Length
Area of Rectangular Tarpaulin $= L \times \text{Width}$
$188.4 = L \times 3 \Rightarrow L = 62.8 \text{ m}$.
Total Length = $L + \text{Extra} = 62.8 \text{ m} + 0.2 \text{ m}$ ($20 \text{ cm} = 0.2 \text{ m}$)
Total Length $= 63 \text{ m}$.
Area of Rectangular Tarpaulin $= L \times \text{Width}$
$188.4 = L \times 3 \Rightarrow L = 62.8 \text{ m}$.
Total Length = $L + \text{Extra} = 62.8 \text{ m} + 0.2 \text{ m}$ ($20 \text{ cm} = 0.2 \text{ m}$)
Total Length $= 63 \text{ m}$.
6. The slant height and base diameter of a conical tomb are $25 \text{ m}$ and $14 \text{ m}$ respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per $100 \text{ m}^2$.
Step 1: Area to be white-washed (CSA)
$l = 25 \text{ m}, r = 7 \text{ m}$.
CSA $= \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550 \text{ m}^2$.
$l = 25 \text{ m}, r = 7 \text{ m}$.
CSA $= \frac{22}{7} \times 7 \times 25 = 22 \times 25 = 550 \text{ m}^2$.
Step 2: Calculate Cost
Rate = ₹ 210 per $100 \text{ m}^2$.
Cost $= \frac{550}{100} \times 210 = 5.5 \times 210 =$ ₹ $1155$.
Rate = ₹ 210 per $100 \text{ m}^2$.
Cost $= \frac{550}{100} \times 210 = 5.5 \times 210 =$ ₹ $1155$.
7. A joker’s cap is in the form of a right circular cone of base radius $7 \text{ cm}$ and height $24 \text{ cm}$. Find the area of the sheet required to make 10 such caps.
Step 1: Slant Height ($l$)
$l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ cm}$.
$l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ cm}$.
Step 2: Area for 1 Cap (CSA)
CSA $= \frac{22}{7} \times 7 \times 25 = 550 \text{ cm}^2$.
CSA $= \frac{22}{7} \times 7 \times 25 = 550 \text{ cm}^2$.
Step 3: Area for 10 Caps
Total Area $= 10 \times 550 =$ $5500 \text{ cm}^2$.
Total Area $= 10 \times 550 =$ $5500 \text{ cm}^2$.
8. A bus stop is barricaded… using 50 hollow cones. Base diameter $40 \text{ cm}$ and height $1 \text{ m}$. Cost of painting is ₹ 12 per $\text{m}^2$. What will be the cost of painting all these cones? (Use $\pi = 3.14$ and $\sqrt{1.04} = 1.02$)
Step 1: Convert units and find $l$
$r = \frac{40}{2} = 20 \text{ cm} = 0.2 \text{ m}$. Height $h = 1 \text{ m}$.
$l = \sqrt{r^2 + h^2} = \sqrt{(0.2)^2 + 1^2} = \sqrt{0.04 + 1} = \sqrt{1.04}$.
Given $\sqrt{1.04} = 1.02$, so $l = 1.02 \text{ m}$.
$r = \frac{40}{2} = 20 \text{ cm} = 0.2 \text{ m}$. Height $h = 1 \text{ m}$.
$l = \sqrt{r^2 + h^2} = \sqrt{(0.2)^2 + 1^2} = \sqrt{0.04 + 1} = \sqrt{1.04}$.
Given $\sqrt{1.04} = 1.02$, so $l = 1.02 \text{ m}$.
Step 2: CSA of 1 Cone
CSA $= \pi r l = 3.14 \times 0.2 \times 1.02 = 0.64056 \text{ m}^2$.
CSA $= \pi r l = 3.14 \times 0.2 \times 1.02 = 0.64056 \text{ m}^2$.
Step 3: Total Cost
Area of 50 cones $= 50 \times 0.64056 = 32.028 \text{ m}^2$.
Cost $= 32.028 \times 12 = 384.336$.
Approx Cost = ₹ $384.34$.
Area of 50 cones $= 50 \times 0.64056 = 32.028 \text{ m}^2$.
Cost $= 32.028 \times 12 = 384.336$.
Approx Cost = ₹ $384.34$.