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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.
Solution :
Given:
1. The toy consists of a cone mounted on a hemisphere.
2. The radius of the cone ($r$) = $3.5$ cm.
3. The radius of the hemisphere ($r$) = $3.5$ cm.
4. The total height of the toy ($H$) = $15.5$ cm.
To Find:
The total surface area (TSA) of the toy.
Step 1: Determine the height of the conical part ($h$)
The total height of the toy is the sum of the height of the cone ($h$) and the radius of the hemisphere ($r$).
$H = h + r$
$15.5 = h + 3.5$
$h = 15.5 - 3.5$
$h = 12$ cm
Step 2: Calculate the slant height of the cone ($l$)
The formula for the slant height of a cone is $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{(3.5)^2 + (12)^2}$
$l = \sqrt{12.25 + 144}$
$l = \sqrt{156.25}$
$l = 12.5$ cm
Step 3: Formulate the Total Surface Area (TSA) of the toy
The total surface area of the toy is the sum of the curved surface area (CSA) of the cone and the curved surface area (CSA) of the hemisphere.
$TSA = \text{CSA of cone} + \text{CSA of hemisphere}$
$TSA = \pi rl + 2\pi r^2$
$TSA = \pi r (l + 2r)$
Step 4: Perform the calculation
Substitute the values $r = 3.5$ cm, $l = 12.5$ cm, and $\pi = \frac{22}{7}$:
$TSA = \frac{22}{7} \times 3.5 \times (12.5 + 2(3.5))$
$TSA = \frac{22}{7} \times 3.5 \times (12.5 + 7)$
$TSA = \frac{22}{7} \times 3.5 \times 19.5$
Since $\frac{3.5}{7} = 0.5$:
$TSA = 22 \times 0.5 \times 19.5$
$TSA = 11 \times 19.5$
$TSA = 214.5$ cm$^2$
Final Answer: The total surface area of the toy is 214.5 cm$^2$.
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.
- 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.
- 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.)
- 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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