GSEB Class 10 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.1

Question 1. 2 cubes of volume 64 \( \text{cm}^3 \) are joined end to end. Find the surface area of the resulting cuboid.
Answer: Let the side length of each cube be 'a' \( \text{cm} \). The volume of a cube is given as 64 \( \text{cm}^3 \).
\( a^3 = 64 \)
\( \implies a = 4 \text{ cm} \)
When both cubes are put together, they form a new cuboid. The cuboid's length will be \( l = 4 + 4 = 8 \text{ cm} \). The breadth \( b \) and height \( h \) will remain 4 \( \text{cm} \) each.
The surface area of the cuboid is calculated using the formula:
\( = 2[lb + bh + lh] \)
\( = 2[8 \times 4 + 4 \times 4 + 8 \times 4] \)
\( = 2[32 + 16 + 32] \)
\( = 2[80] \)
\( = 160 \text{ cm}^2 \)
In simple words: Two cubes are joined together. First, find the side length of one cube. Then, calculate the dimensions of the new cuboid and use the formula to find its total surface area.

Exam Tip: When cubes are joined end to end, only the length changes; the breadth and height stay the same. Always remember the formula for the surface area of a cuboid: \( 2(lb+bh+lh) \).

Question 2. 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. (CBSE 2013)
Answer: The hemisphere's diameter is 14 \( \text{cm} \). So, its radius \( r = 7 \text{ cm} \).
The total height of the vessel is 13 \( \text{cm} \). Since the hemisphere is mounted on the cylinder, its radius is also the height of the hemispherical part.
The cylinder's height will be \( h = 13 - 7 = 6 \text{ cm} \).
The inner surface area of the entire vessel is the sum of the curved surface area (CSA) of the cylinder and the CSA of the hemisphere.
Inner surface area of vessel \( = 2\pi rh + 2\pi r^2 \)
Factoring out \( 2\pi r \), we get \( = 2\pi r(h + r) \)
\( = 2 \times \frac{22}{7} \times 7 (6 + 7) \)
\( = 2 \times 22 \times (13) \)
\( = 44 \times 13 \)
\( = 572 \text{ cm}^2 \)
In simple words: The vessel looks like a cylinder with a hemisphere on top. First, find the radius and the height of the cylindrical part. Then, add the curved surface areas of both the cylinder and the hemisphere to get the total inner surface area.

Exam Tip: When a hemisphere is mounted on a cylinder, its radius is also its height. Remember that the base of the cylinder is not included in the inner surface area in this context.

Question 3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Answer: The radius of both the cone and hemisphere is \( r = 3.5 \text{ cm} \).
The overall height of the toy is 15.5 \( \text{cm} \). Since the cone is mounted on the hemisphere, the height of the cone is the total height minus the radius of the hemisphere.
Height of cone \( h = 15.5 - 3.5 = 12 \text{ cm} \).
We need to determine the slant height \( l \) of the cone using the Pythagorean theorem:
\( l = \sqrt{r^2+h^2} \)
\( = \sqrt{(3.5)^2+(12)^2} \)
\( = \sqrt{12.25+144} \)
\( = \sqrt{156.25} \)
\( = 12.5 \text{ cm} \).
The total surface area of the toy is the sum of the curved surface area (CSA) of the cone and the CSA of the hemisphere.
Total surface area of toy \( = \text{CSA of cone} + \text{CSA of hemisphere} \)
\( = \pi rl + 2\pi r^2 \)
Factoring out \( \pi r \), we get \( = \pi r (l + 2r) \)
\( = \frac{22}{7} \times 3.5 (12.5 + 2 \times 3.5) \)
\( = \frac{22}{7} \times 3.5 (12.5 + 7) \)
\( = \frac{22}{7} \times 3.5 \times 19.5 \)
\( = 22 \times 0.5 \times 19.5 \)
\( = 214.5 \text{ cm}^2 \)
In simple words: The toy is a cone sitting on a hemisphere. Calculate the cone's slant height. Then, find the curved surface areas of both the cone and the hemisphere and add them to get the total surface area of the toy.

Exam Tip: When a cone is mounted on a hemisphere, the flat circular bases are joined, so they are not part of the total surface area. Only the curved surfaces are included.

Question 4. 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.
Answer: The cubical block has a side length \( a = 7 \text{ cm} \).
For the hemisphere to fit perfectly on top, the maximum diameter it can have is equal to the side of the cube, which is 7 \( \text{cm} \).
So, the diameter of the hemisphere \( = 7 \text{ cm} \).
Its radius \( r = \frac{7}{2} \text{ cm} \).
The surface area of the combined solid is calculated as the area of the 6 faces of the cube, minus the area of the base of the hemisphere (which covers part of the top face), plus the curved surface area of the hemisphere.
Surface area of solid \( = 6a^2 - \pi r^2 + 2\pi r^2 \)
\( = 6a^2 + \pi r^2 \)
\( = 6 \times (7)^2 + \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \)
\( = 6 \times 49 + \frac{11 \times 7}{2} \)
\( = 294 + \frac{77}{2} \)
\( = 294 + 38.5 \)
\( = 332.5 \text{ cm}^2 \)
In simple words: A hemisphere sits on top of a cube. The hemisphere's biggest diameter will be the same as the cube's side. Calculate the area of the cube's exposed faces, plus the curved surface area of the hemisphere.

Exam Tip: When one solid sits on another, be careful to identify which areas are covered (and thus not part of the external surface area) and which new surfaces (like the curved surface of the top solid) are exposed.

Question 5. 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.
Answer: The diameter of the hemispherical depression is given as \( l \), which is equal to the edge of the cubical block. So, the side length of the cube is \( a = l \).
Therefore, the radius of the hemispherical depression is \( r = \frac{l}{2} \).
The surface area of the remaining solid is calculated by taking the total surface area of the cube, subtracting the flat circular area where the hemisphere is cut out, and then adding the curved surface area of the hemispherical depression.
Surface area of solid \( = 6a^2 - \pi r^2 + 2\pi r^2 \)
\( = 6a^2 + \pi r^2 \)
Substitute \( a = l \) and \( r = \frac{l}{2} \):
\( = 6l^2 + \pi \left(\frac{l}{2}\right)^2 \)
\( = 6l^2 + \frac{\pi l^2}{4} \)
Taking \( l^2 \) common:
\( = l^2 \left( 6 + \frac{\pi}{4} \right) \)
\( = l^2 \left( \frac{24+\pi}{4} \right) \)
\( = \frac{1}{4} l^2 (24 + \pi) \)
In simple words: A hemisphere is carved out of a cube. The hemisphere's diameter is the same as the cube's side. To find the surface area, calculate the cube's area, subtract the circular hole, and add the curved area of the carved-out hemisphere.

Exam Tip: When a depression is cut out, the surface area increases by the curved surface area of the depression, while the flat area where it was cut is subtracted from the original solid's face.

Question 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Answer: The capsule's diameter is 5 \( \text{mm} \). So, its radius \( r = \frac{5}{2} = 2.5 \text{ mm} \).
The total length of the capsule is 14 \( \text{mm} \). Since there are two hemispheres at each end, the total height of the hemispherical parts is \( 2 \times r = 2 \times 2.5 = 5 \text{ mm} \).
The height of the cylindrical portion \( h \) is the total length minus the height of the two hemispheres:
\( h = 14 - (2.5 + 2.5) = 14 - 5 = 9 \text{ mm} \).
The surface area of the capsule is the sum of the curved surface area (CSA) of the cylinder and the CSA of the two hemispheres.
Surface area of capsule \( = \text{CSA of cylinder} + 2 \times \text{CSA of hemisphere} \)
\( = 2\pi rh + 2(2\pi r^2) \)
\( = 2\pi rh + 4\pi r^2 \)
Factoring out \( 2\pi r \), we get \( = 2\pi r(h + 2r) \)
\( = 2 \times \frac{22}{7} \times \frac{5}{2} \left(9 + 2 \times \frac{5}{2}\right) \)
\( = \frac{22}{7} \times 5 (9 + 5) \)
\( = \frac{22}{7} \times 5 \times 14 \)
\( = 22 \times 5 \times 2 \)
\( = 220 \text{ mm}^2 \)
In simple words: The capsule looks like a cylinder with a hemisphere on each end. Find the radius and the height of the cylinder part. Then, add the curved surface areas of the cylinder and both hemispheres to find the total surface area of the capsule.

Exam Tip: For capsules, remember that the height of each hemisphere is its radius. This means you subtract two radii from the total length to find the cylinder's height.

Question 7. 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 4m 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).
Answer: The height of the cylindrical part is \( h = 2.1 \text{ m} \).
The diameter of the cylindrical part is 4 \( \text{m} \), so its radius \( r = \frac{4}{2} = 2 \text{ m} \).
The slant height of the conical top is \( l = 2.8 \text{ m} \).
The area of canvas needed for making the tent is the sum of the curved surface area (CSA) of the cylinder and the CSA of the cone. The base of the tent is not covered.
Area of canvas \( = \text{CSA of cylinder} + \text{CSA of cone} \)
\( = 2\pi rh + \pi rl \)
Factoring out \( \pi r \), we get \( = \pi r(2h + l) \)
\( = \frac{22}{7} \times 2 (2 \times 2.1 + 2.8) \)
\( = \frac{44}{7} (4.2 + 2.8) \)
\( = \frac{44}{7} \times 7 \)
\( = 44 \text{ m}^2 \)
The cost of the canvas is Rs. 500 for each square meter.
Total cost of canvas used \( = \text{Rs. } 500 \times 44 \)
\( = \text{Rs. } 22000 \)
In simple words: The tent is shaped like a cylinder with a cone on top. Calculate the curved surface area for the cylinder and the cone separately. Add these areas to find the total canvas needed. Then, multiply this total area by the cost per square meter to get the final cost.

Exam Tip: Carefully read what areas need to be included in the calculation. For a tent, the base is usually not covered, so exclude its area from the total surface area calculation.

Question 8. 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 \( \text{cm}^2 \).
Answer: The height of the solid cylinder is \( h = 2.4 \text{ cm} \).
Its diameter is 1.4 \( \text{cm} \), so the radius \( r = \frac{1.4}{2} = 0.7 \text{ cm} \).
A conical cavity of the same height and diameter is carved out. So, for the cone, height \( h = 2.4 \text{ cm} \) and radius \( r = 0.7 \text{ cm} \).
First, we calculate the slant height \( l \) of the cone:
\( l = \sqrt{r^2+h^2} \)
\( = \sqrt{(0.7)^2+(2.4)^2} \)
\( = \sqrt{0.49+5.76} \)
\( = \sqrt{6.25} \)
\( = 2.5 \text{ cm} \).
The total surface area of the remaining solid includes the curved surface area (CSA) of the cylinder, the area of its base (since the cavity is cut from the top, the bottom base remains), and the CSA of the hollowed-out conical cavity.
Total surface area of remaining solid \( = \text{CSA of cylinder} + \text{Area of cylindrical base} + \text{CSA of cone} \)
\( = 2\pi rh + \pi r^2 + \pi rl \)
\( = 2 \times \frac{22}{7} \times 0.7 \times 2.4 + \frac{22}{7} \times (0.7)^2 + \frac{22}{7} \times 0.7 \times 2.5 \)
\( = (44 \times 0.1 \times 2.4) + (22 \times 0.1 \times 0.7) + (22 \times 0.1 \times 2.5) \)
\( = 10.56 + 1.54 + 5.5 \)
\( = 17.6 \text{ cm}^2 \)
Rounded to the nearest square centimeter, the surface area is \( 18 \text{ cm}^2 \).
In simple words: A cone-shaped hole is carved out of a solid cylinder. The surface area of the remaining solid includes the cylinder's curved side, its bottom base, and the curved inside surface of the carved-out cone. Add these three areas together and then round the answer to the closest whole number.

Exam Tip: When a shape is hollowed out, its curved surface area is added to the total surface area of the remaining solid. Make sure to include all exposed surfaces, including the base that was not affected by the hollowing.

Question 9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10cm and its base is of radius 3.5 cm, find the total surface area of the article.
Answer: The cylinder's height is \( h = 10 \text{ cm} \).
The base radius of the cylinder is \( r = 3.5 \text{ cm} \).
Two hemispheres are carved out from each end of the cylinder. The total surface area of the article will be the curved surface area (CSA) of the cylinder plus the CSA of the two scooped-out hemispheres.
Total surface area of article \( = \text{CSA of cylinder} + 2 \times \text{CSA of hemisphere} \)
\( = 2\pi rh + 2(2\pi r^2) \)
\( = 2\pi rh + 4\pi r^2 \)
Factoring out \( 2\pi r \), we get \( = 2\pi r(h + 2r) \)
\( = 2 \times \frac{22}{7} \times 3.5 (10 + 2 \times 3.5) \)
\( = 2 \times 22 \times 0.5 (10 + 7) \)
\( = 22 (17) \)
\( = 374 \text{ cm}^2 \)
In simple words: The article is a cylinder with a hemisphere carved out of each end. To find the total surface area, add the curved surface area of the cylinder and the curved surface areas of the two carved-out hemispheres.

Exam Tip: When parts are scooped out from a solid, their curved surface areas contribute to the total surface area of the modified solid, replacing the flat areas they were carved from.