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Q7:
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1$ m and $4$ m respectively, and the slant height of the top is $2.8$ m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of $₹ 500$ per m$^2$. (Note that the base of the tent will not be covered with canvas.)
Solution :
Given:
- Shape of the tent: A cylinder surmounted by a cone.
- Height of the cylindrical part ($h_c$) = $2.1$ m.
- Diameter of the cylindrical part ($d$) = $4$ m.
- Slant height of the conical part ($l$) = $2.8$ m.
- Rate of canvas = $₹ 500$ per m$^2$.
To Find:
- Total area of the canvas used for the tent.
- Total cost of the canvas.
Step 1: Determine the radius of the base.
Since the diameter of the cylindrical part is $4$ m, the radius ($r$) is half of the diameter.
$r = \frac{d}{2} = \frac{4}{2} = 2$ m.
[Since the cone is mounted on the cylinder, the radius of the cone is also $r = 2$ m.]
Step 2: Calculate the surface area of the canvas.
The canvas covers the curved surface area of the cylinder and the curved surface area of the cone.
Formula for Curved Surface Area (CSA) of a cylinder = $2\pi rh_c$.
Formula for Curved Surface Area (CSA) of a cone = $\pi rl$.
Total Area ($A$) = $CSA_{cylinder} + CSA_{cone} = 2\pi rh_c + \pi rl = \pi r(2h_c + l)$.
Substituting the values ($r=2, h_c=2.1, l=2.8, \pi = \frac{22}{7}$):
$A = \frac{22}{7} \times 2 \times (2 \times 2.1 + 2.8)$
$A = \frac{44}{7} \times (4.2 + 2.8)$
$A = \frac{44}{7} \times 7$
$A = 44$ m$^2$.
Step 3: Calculate the total cost of the canvas.
Cost = Total Area $\times$ Rate per m$^2$.
Cost = $44 \times 500$.
Cost = $22,000$.
Final Answer: The total area of the canvas used is $44$ m$^2$ and the total cost of the canvas is $₹ 22,000$.
More Questions from Class 10 Mathematics Surface Areas and Volumes EXERCISE 12.1
- Q1: 2 cubes each of volume $64$ cm$^3$ are joined end to end. Find the surface area of the resulting cuboid.
- Q2: A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14$ cm and the total height of the vessel is $13$ cm. Find the inner surface area of the vessel.
- Q3: A toy is in the form of a cone of radius $3.5$ cm mounted on a hemisphere of same radius. The total height of the toy is $15.5$ cm. Find the total surface area of the toy.
- Q4: A cubical block of side $7$ cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
- Q5: A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $l$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
- Q6: A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is $14$ mm and the diameter of the capsule is $5$ mm. Find its surface area.
- Q8: From a solid cylinder whose height is $2.4$ cm and diameter $1.4$ cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm$^2$.
- Q9: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is $10$ cm, and its base is of radius $3.5$ cm, find the total surface area of the article.
CBSE Solutions for Class 10 Mathematics Surface Areas and Volumes
Chapters in CBSE - Class 10 Mathematics
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