Exercise 11.1 Practice
Surface Area of a Right Circular Cone
Q1: Curved Surface Area
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Diameter of the base of a cone is $14 \text{ cm}$ and its slant height is $12 \text{ cm}$. Find its curved surface area.
Radius $r = \frac{14}{2} = 7 \text{ cm}$.
Slant height $l = 12 \text{ cm}$.
Slant height $l = 12 \text{ cm}$.
Curved Surface Area (CSA) $= \pi r l$
$= \frac{22}{7} \times 7 \times 12$
$= 22 \times 12 = 264 \text{ cm}^2$.
$= \frac{22}{7} \times 7 \times 12$
$= 22 \times 12 = 264 \text{ cm}^2$.
CSA $= 264 \text{ cm}^2$
Q2: Total Surface Area
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Find the total surface area of a cone, if its slant height is $25 \text{ m}$ and diameter of its base is $14 \text{ m}$.
Radius $r = \frac{14}{2} = 7 \text{ m}$.
Slant height $l = 25 \text{ m}$.
Slant height $l = 25 \text{ m}$.
Total Surface Area (TSA) $= \pi r (l + r)$
$= \frac{22}{7} \times 7 \times (25 + 7)$
$= 22 \times 32 = 704 \text{ m}^2$.
$= \frac{22}{7} \times 7 \times (25 + 7)$
$= 22 \times 32 = 704 \text{ m}^2$.
TSA $= 704 \text{ m}^2$
Q3: Finding Radius
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Curved surface area of a cone is $550 \text{ cm}^2$ and its slant height is $25 \text{ cm}$. Find (i) radius of the base and (ii) total surface area of the cone.
(i) Radius of the base:
CSA $= \pi r l = 550$.
$\frac{22}{7} \times r \times 25 = 550$.
$r = \frac{550 \times 7}{22 \times 25} = 7 \text{ cm}$.
CSA $= \pi r l = 550$.
$\frac{22}{7} \times r \times 25 = 550$.
$r = \frac{550 \times 7}{22 \times 25} = 7 \text{ cm}$.
(ii) Total Surface Area:
TSA $= \pi r (l + r)$
$= \frac{22}{7} \times 7 \times (25 + 7)$
$= 22 \times 32 = 704 \text{ cm}^2$.
TSA $= \pi r (l + r)$
$= \frac{22}{7} \times 7 \times (25 + 7)$
$= 22 \times 32 = 704 \text{ cm}^2$.
Radius $= 7 \text{ cm}$, TSA $= 704 \text{ cm}^2$
Q4: Conical Tent
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A conical tent is $24 \text{ m}$ high and the radius of its base is $7 \text{ m}$. 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 ₹$100$.
(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 ₹$100$.
(i) Slant height ($l$):
$l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2}$
$= \sqrt{49 + 576} = \sqrt{625} = 25 \text{ m}$.
$l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2}$
$= \sqrt{49 + 576} = \sqrt{625} = 25 \text{ m}$.
(ii) Cost of canvas:
Canvas area = CSA $= \pi r l = \frac{22}{7} \times 7 \times 25$.
$= 550 \text{ m}^2$.
Cost $= \text{Area} \times \text{Rate} = 550 \times 100$
$=$ ₹$55000$.
Canvas area = CSA $= \pi r l = \frac{22}{7} \times 7 \times 25$.
$= 550 \text{ m}^2$.
Cost $= \text{Area} \times \text{Rate} = 550 \times 100$
$=$ ₹$55000$.
Slant height $= 25 \text{ m}$, Cost = ₹$55000$
Q5: Tarpaulin Length
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What length of tarpaulin $2 \text{ m}$ wide will be required to make conical tent of height $4 \text{ m}$ and base radius $3 \text{ m}$? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately $25 \text{ cm}$. (Use $\pi = 3.14$)
Height $h = 4 \text{ m}$, Radius $r = 3 \text{ m}$.
Slant height $l = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m}$.
Slant height $l = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m}$.
Area of tarpaulin required = CSA of tent $= \pi r l$
$= 3.14 \times 3 \times 5 = 47.1 \text{ m}^2$.
$= 3.14 \times 3 \times 5 = 47.1 \text{ m}^2$.
Width of tarpaulin $= 2 \text{ m}$.
Length required for area $= \frac{\text{Area}}{\text{Width}} = \frac{47.1}{2} = 23.55 \text{ m}$.
Length required for area $= \frac{\text{Area}}{\text{Width}} = \frac{47.1}{2} = 23.55 \text{ m}$.
Extra length for margins $= 25 \text{ cm} = 0.25 \text{ m}$.
Total length $= 23.55 + 0.25 = 23.8 \text{ m}$.
Total length $= 23.55 + 0.25 = 23.8 \text{ m}$.
Length $= 23.8 \text{ m}$
Q6: White Washing
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The slant height and base diameter of a conical tomb are $50 \text{ m}$ and $28 \text{ m}$ respectively. Find the cost of white-washing its curved surface at the rate of ₹$150$ per $100 \text{ m}^2$.
Slant height $l = 50 \text{ m}$.
Radius $r = \frac{28}{2} = 14 \text{ m}$.
Radius $r = \frac{28}{2} = 14 \text{ m}$.
CSA $= \pi r l = \frac{22}{7} \times 14 \times 50 = 44 \times 50 = 2200 \text{ m}^2$.
Rate = ₹$150$ per $100 \text{ m}^2$.
Cost $= \frac{2200}{100} \times 150 = 22 \times 150 = $ ₹$3300$.
Cost $= \frac{2200}{100} \times 150 = 22 \times 150 = $ ₹$3300$.
Cost = ₹$3300$
Q7: Joker's Cap
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A joker's cap is in the form of a right circular cone of base radius $3.5 \text{ cm}$ and height $12 \text{ cm}$. Find the area of the sheet required to make $20$ such caps.
Radius $r = 3.5 \text{ cm}$, Height $h = 12 \text{ cm}$.
Slant height $l = \sqrt{r^2 + h^2} = \sqrt{3.5^2 + 12^2}$
$= \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \text{ cm}$.
Slant height $l = \sqrt{r^2 + h^2} = \sqrt{3.5^2 + 12^2}$
$= \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \text{ cm}$.
Area of sheet for 1 cap = CSA $= \pi r l$
$= \frac{22}{7} \times 3.5 \times 12.5 = 11 \times 12.5 = 137.5 \text{ cm}^2$.
$= \frac{22}{7} \times 3.5 \times 12.5 = 11 \times 12.5 = 137.5 \text{ cm}^2$.
Area for 20 caps $= 137.5 \times 20 = 2750 \text{ cm}^2$.
Area $= 2750 \text{ cm}^2$
Q8: Bus Stop Barricade
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A bus stop is barricaded from the remaining part of the road, by using $40$ hollow cones made of recycled cardboard. Each cone has a base diameter of $60 \text{ cm}$ and height $40 \text{ cm}$. If the outer side of each of the cones is to be painted and the cost of painting is ₹$25$ per $\text{m}^2$, what will be the cost of painting all these cones? (Use $\pi = 3.14$)
Diameter $= 60 \text{ cm} \Rightarrow r = 30 \text{ cm} = 0.3 \text{ m}$.
Height $h = 40 \text{ cm} = 0.4 \text{ m}$.
Height $h = 40 \text{ cm} = 0.4 \text{ m}$.
Slant height $l = \sqrt{r^2 + h^2} = \sqrt{(0.3)^2 + (0.4)^2}$
$= \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 \text{ m}$.
$= \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 \text{ m}$.
CSA of 1 cone $= \pi r l = 3.14 \times 0.3 \times 0.5$
$= 0.471 \text{ m}^2$.
$= 0.471 \text{ m}^2$.
CSA of 40 cones $= 40 \times 0.471 = 18.84 \text{ m}^2$.
Cost $= \text{Area} \times \text{Rate} = 18.84 \times 25$
$=$ ₹$471$.
$=$ ₹$471$.
Cost = ₹$471$