Surface Areas and Volumes | Exercise 12.1

Question 3

A toy is in the form of a cone of radius 3.5 cm3.5\text{ cm} mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm15.5\text{ cm}. Find the total surface area of the toy.

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Solution

The toy is made of two parts joined together — a cone on top and a hemisphere (half of a sphere) at the bottom.

Total surface area = curved surface area (CSA) of the cone + curved surface area (CSA) of the hemisphere. (The flat circular faces are hidden inside where the two parts join, so we do not count them.)

Step 1 — Find the height of the cone

Let's first note down what is given:

  • Radius of the cone, r=3.5 cmr = 3.5\text{ cm}
  • Radius of the hemisphere =3.5 cm= 3.5\text{ cm} (same as the cone)
  • Total height of the toy =15.5 cm= 15.5\text{ cm}

For a hemisphere, its height is equal to its radius. So the hemisphere is 3.5 cm3.5\text{ cm} tall.

Looking at the figure, the total height is the cone's height plus the hemisphere's height. So to find the cone's height hh, we subtract:

h=Total heightHeight of hemisphereh = \text{Total height} - \text{Height of hemisphere}

h=15.53.5h = 15.5 - 3.5

h=12 cm\boxed{h = 12 \text{ cm}}

Diagram 1

Step 2 — Find the slant height of the cone

To find the CSA of a cone, we first need its slant height ll (the distance along the sloping side).

The radius, the height, and the slant height form a right-angled triangle, where the slant height is the hypotenuse. So by the Pythagoras theorem:

(hypotenuse)2=(height)2+(base)2(\text{hypotenuse})^2 = (\text{height})^2 + (\text{base})^2

l2=h2+r2l^2 = h^2 + r^2

Taking the square root on both sides:

l=h2+r2l = \sqrt{h^2 + r^2}

l=(12)2+(3.5)2l = \sqrt{(12)^2 + (3.5)^2}

l=144+12.25l = \sqrt{144 + 12.25}

l=156.25l = \sqrt{156.25}

l=12.5 cm\boxed{l = 12.5 \text{ cm}}

Step 3 — Find the curved surface areas

Now we find the CSA of each part. We will use π=227\pi = \dfrac{22}{7}.

Cone — the formula is CSA =πrl= \pi r l:

CSA of cone=227×3.5×12.5\text{CSA of cone} = \frac{22}{7} \times 3.5 \times 12.5

Here 3.5÷7=0.53.5 \div 7 = 0.5, which makes the multiplication simpler:

CSA of cone=22×0.5×12.5\text{CSA of cone} = 22 \times 0.5 \times 12.5

CSA of cone=137.5 cm2\boxed{\text{CSA of cone} = 137.5 \text{ cm}^2}

Hemisphere — the formula is CSA =2πr2= 2\pi r^2:

CSA of hemisphere=2×227×(3.5)2\text{CSA of hemisphere} = 2 \times \frac{22}{7} \times (3.5)^2

CSA of hemisphere=2×227×12.25\text{CSA of hemisphere} = 2 \times \frac{22}{7} \times 12.25

Again 12.25÷7=1.7512.25 \div 7 = 1.75:

CSA of hemisphere=2×22×1.75\text{CSA of hemisphere} = 2 \times 22 \times 1.75

CSA of hemisphere=77 cm2\boxed{\text{CSA of hemisphere} = 77 \text{ cm}^2}

Step 4 — Find the total surface area

Finally, we add the two curved surface areas to get the total surface area of the toy:

Total Surface Area=CSA of cone+CSA of hemisphere\text{Total Surface Area} = \text{CSA of cone} + \text{CSA of hemisphere}

Total Surface Area=137.5+77\text{Total Surface Area} = 137.5 + 77

Total Surface Area=214.5 cm2\boxed{\text{Total Surface Area} = 214.5 \text{ cm}^2}

Answer

The total surface area of the toy is 214.5 cm².

More questions in Exercise 12.1

Q1

Unless stated otherwise, take π=227\pi = \frac{22}{7}.

2 cubes each of volume 64 cm364\text{ 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 cm14\text{ cm} and the total height of the vessel is 13 cm13\text{ cm}. Find the inner surface area of the vessel.

Q3

A toy is in the form of a cone of radius 3.5 cm3.5\text{ cm} mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm15.5\text{ cm}. Find the total surface area of the toy.

Q4

A cubical block of side 7 cm7\text{ 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 ll 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 mm14\text{ mm} and the diameter of the capsule is 5 mm5\text{ 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 m². (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 cm².

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.

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