Exercise-11.1
Maths - Exercise
Q1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Solution:
The diameter of the base of the cone is given as \(10.5~\text{cm}\). To find the radius, we divide by 2: $$ r = \frac{10.5}{2} = 5.25~\text{cm} $$
The slant height of the cone is $$l = 10~\text{cm}$$
The formula for the curved surface area of a cone is $$\pi r l$$
Substituting the given values:
Curved Surface Area
$$
\small\begin{aligned}
&= \pi \times r \times l \\
&= \frac{22}{7} \times 5.25 \times 10
\end{aligned}
$$
Calculating step by step: $$\begin{aligned} &=\frac{22}{7} \times 5.25 \\&= \frac{22 \times 5.25}{7} \\&= \frac{115.5}{7} \\&= 16.5 \end{aligned}$$ Then, $$ 16.5 \times 10 = 165~\text{cm}^2 $$
Therefore, the curved surface area of the cone is $$165~\text{cm}^2$$
Q2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
The slant height of the cone is given as $$l = 21~\text{m}$$ and the diameter of the base is $$24~\text{m}$$ To find the radius, divide the diameter by 2: $$ r = \frac{24}{2} = 12~\text{m} $$
The formula for the total surface area (TSA) of a cone is $$ \pi r (r + l) $$
Substituting the values into the formula: $$ \small\begin{aligned} \text{TSA} &= \pi r (r + l) \\\\ &= \frac{22}{7} \times 12 \times (12 + 21) \\\\ &= \frac{22}{7} \times 12 \times 33 \\\\ &= \frac{22 \times 12 \times 33}{7} \\\\ &= \frac{8712}{7} \\\\ &\approx 1244.57~\text{m}^2 \end{aligned} $$
Therefore, the total surface area of the cone is $$1244.57~\text{m}^2$$
Q3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
(i) radius of the base and
(ii) total surface area of the cone.
Solution:
The curved surface area of the cone is given as $$308~\text{cm}^2$$ and its slant height is $$14~\text{cm}$$
First, to find the radius \(r\) of the base: $$ \begin{aligned} \text{CSA} &= \pi r l \\ 308 &= \frac{22}{7} \times r \times 14 \\ 308 &= 2 \times 22 \times r \\ 308 &= 44r \\ r &= \frac{308}{44} \\ r &= 7~\text{cm} \end{aligned} $$
Next, to find the total surface area \(\text{TSA}\) of the cone: $$ \begin{aligned} \text{TSA} &= \pi r (r + l) \\ &= \frac{22}{7} \times 7 \times (7 + 14) \\ &= 22 \times 21 \\ &= 462~\text{cm}^2 \end{aligned} $$
Therefore, the radius of the base is $$7~\text{cm}$$ and the total surface area of the cone is $$462~\text{cm}^2$$
Q4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of \(1\ m^2\) canvas is ₹70.
Solution:
The height of the conical tent is $$h = 10~\text{m}$$ and the radius of its base is $$r = 24~\text{m}$$
Slant height \(l\), $$ \begin{aligned} l^2 &= r^2 + h^2 \\ &= 24^2 + 10^2 \\ &= 576 + 100 \\ &= 676 \\ l &= \sqrt{676} \\&= 26~\text{m} \end{aligned} $$
Next, calculate the curved surface area (CSA) of the tent: $$ \begin{aligned} \text{CSA} &= \pi r l \\ &= \frac{22}{7} \times 24 \times 26 \\ &= \frac{22}{7} \times 624 \\ &= \frac{13728}{7} \\ &\approx 1961.14~\text{m}^2 \end{aligned} $$
If the cost of canvas per $$1~\text{m}^2$$ is ₹70, then the total cost is: $$ \begin{aligned} \text{Total Cost} &= 70 \times \text{CSA} \\ &= 70 \times 1961.14 \\ &\approx ₹137,280 \end{aligned} $$
Therefore, the slant height of the tent is $$26~\text{m}$$ and the estimated cost of the canvas required to make the tent is ₹137,280.
Q5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).
Solution:
Given that the height of the conical tent is $$8~\text{m}$$ and the radius of the base is $$6~\text{m}$$ let's first find the slant height \(l\): $$ \begin{aligned} l^2 &= h^2 + r^2 \\ &= 8^2 + 6^2 \\ &= 64 + 36 \\ &= 100 \\ l &= \sqrt{100} \\&= 10~\text{m} \end{aligned} $$
The curved surface area (CSA) of the tent is: $$ \begin{aligned} \text{CSA} &= \pi r l \\ &= 3.14 \times 6 \times 10 \\ &= 188.4~\text{m}^2 \end{aligned} $$
The tarpaulin is $$3~\text{m}$$ wide, so the required length of tarpaulin (without extra margin) is: $$\begin{aligned} \text{Length} &= \frac{\text{CSA}}{\text{width}} \\\\&= \frac{188.4}{3} \\\\&= 62.8~\text{m} \end{aligned}$$
Since an extra length of $$20~\text{cm} = 0.2~\text{m}$$ is needed for stitching margins and wastage, the total length required is: $$\begin{aligned} \text{Total length}& = 62.8~\text{m} + 0.2~\text{m} \\&= 63~\text{m} \end{aligned}$$
Therefore, a tarpaulin of width $$3~\text{m}$$ and total length $$63~\text{m}$$ will be required to make the conical tent.
Q6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹210 per \(100\ m^2\).
Solution:
The slant height of the conical tomb is $$25~\text{m}$$ and its base diameter is $$14~\text{m}$$ The radius is half the diameter: $$ r = \frac{14}{2} = 7~\text{m} $$
To find the curved surface area (CSA) of the tomb: $$ \begin{aligned} \text{CSA} &= \pi r l \\ &= \frac{22}{7} \times 7 \times 25 \\ &= 22 \times 25 \\ &= 550~\text{m}^2 \end{aligned} $$
The cost to whitewash \(100~\text{m}^2\) is ₹210.
To find the total cost for \(550~\text{m}^2\)
$$
\begin{aligned}
\text{Total cost} &= \frac{210}{100} \times 550 \\
&= 2.1 \times 550 \\
&= ₹1155
\end{aligned}
$$
Therefore, the cost of white-washing the curved surface of the conical tomb will be ₹1155.
Q7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Solution:
The base radius of the joker’s cap is $$7~\text{cm}$$ and its height is $$24~\text{cm}$$
Slant height \(l\): $$ \begin{aligned} l^2 &= r^2 + h^2 \\ &= 7^2 + 24^2 \\ &= 49 + 576 \\ &= 625 \\ l &= \sqrt{625} \\&= 25~\text{cm} \end{aligned} $$
The curved surface area (CSA) for one cap is: $$ \begin{aligned} \text{CSA} &= \pi r l \\ &= \frac{22}{7} \times 7 \times 25 \\ &= 22 \times 25 \\ &= 550~\text{cm}^2 \end{aligned} $$
For 10 caps, the area of the sheet required is: $$ \begin{aligned} \text{Total area} &= 550 \times 10 \\ &= 5500~\text{cm}^2 \end{aligned} $$
Therefore, to make 10 such caps, the area of the sheet required is $$5500~\text{cm}^2$$
Q8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is \(₹\,12\, per\, m^2\), what will be the cost of painting all these cones? (Use π = 3.14 and take \(\sqrt{1.04}= 1.02\)
Solution:
The base diameter of each barricade cone is $$40~\text{cm}$$ so the radius is: $$ r = \frac{40}{2} = 20~\text{cm} = 0.2~\text{m} $$
The height of each cone is $$h = 1~\text{m}$$ Now, calculate the slant height \(l\): $$ \begin{aligned} l^2 &= r^2 + h^2 \\ &= (0.2)^2 + (1)^2 \\ &= 0.04 + 1 \\ &= 1.04 \\ l &= \sqrt{1.04} \\&= 1.02~\text{m} \end{aligned} $$
The curved surface area (CSA) of one cone is: $$ \begin{aligned} \text{CSA} &= \pi r l \\ &= 3.14 \times 0.2 \times 1.02 \\ &= 0.64~\text{m}^2 \end{aligned} $$
The cost of painting \(1~\text{m}^2\) is ₹12.
So, the cost to paint one cone:
$$\begin{aligned}
&= 0.64 \times 12 \\&= ₹7.68
\end{aligned}$$
For 50 cones, the total cost will be: $$ \begin{aligned} \text{Total cost} &= 0.64 \times 12 \times 50 \\ &= 7.68 \times 50 \\ &= ₹384 \end{aligned} $$
Therefore, the total cost of painting all the cones will be ₹384.
