NCERT Solutions for Class 10 Maths Chapter 13 - Surface Areas and Volumes

Important Points

1. Solid Figures : The objects which occupy space (i.e., they have three dimensions) are called solids. The solid figures can be derived from the plane figures.

e.g., In Fig. (a), we have a paper cut in the form as shown. It is a plane figure but when we fold the paper along the dotted lines, a box can be made as shown in Fig. (b).

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2. Surface Area : The sum of the areas of the plane figures making up the boundary of a solid figure is called its surface area e.g., the area of paper in fig. (a) is the surface area of box.

3. Volume : The measure of part of space occupied by a solid is called its volume.

4. Parallelopiped (cuboid) : A cuboid or parallelopiped is a region bounded by its six rectangular faces. A parallelopiped whose faces are rectangles is called a rectangular parallelopiped or a rectangular solid or a cuboid.

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(i) A cuboid has six rectangular plane surfaces called faces.

(ii) A cuboid has 8 corners called the vertices.

(iii) Volume of cuboid = l × b × h

where, l = length, b = breadth and h = height

(iv) Whole surface of cuboid = 2(lb + bh + lh)

$$\text{(v) Diagonal of the cuboid =}\sqrt{l^{2}+b^{2}+h^{2}}$$

(vi) Area of 4 walls = 2(l + b).h

5. Cube : If the faces of a rectangular parallelopiped be squares, then it is called a cube.

(i) Edge of cube = Length = Breadth = Height

(ii) Volume of a cube = (Edge)3

(iii) Total Surface Area (TSA) of a cube = 6 × (Edge)2

$$\text{(iv) Diagonal of a cube =}\space\sqrt{3}×\text{Edge}$$

6. Right Circular Cylinder : A right circular cylinder is a solid generated by the revolution of a rectangle about one of its sides which remains fixed. A cylinder shown in figure is described by revolving the rectangle ABCD about the side BC. The examples of right circular cylinder are many such as water pipes, powder box, laboratory beakers etc.

(i) Volume of the cylinder

= (Area of base) × Height

= πr2h cu units

where, r is the radius of the base and h is the height of the cylinder.

(ii) Curved surface area
= Circumference of the base × Height
= 2 πrh sq. units

(iii) Total surface area = Curved surface area + Area of two ends

= 2πry + 2πr2 = 2πr(h + r)

$$\text{Also, note h}=\frac{\text{V}}{\pi r^{2}}\\\Rarr r=\sqrt{\frac{\text{V}}{\pi h}}$$

7. Hollow Cylinder : The volume of material in a hollow cylinder is the differences between the volume of a cylinder having the external dimensions and the volume of a cylinder having the internal dimensions.

Let R and r be the external and internal radii of the hollow cylinder and h be its height. Then,

(i) Volume of material = π(R2 – r2)h

(ii) Total surface area = 2π(R + r)(h + R – r)

(iii) Curved surface area

= 2πRh + 2πrh = 2π(R + r)h

(iv) Total outer surface area

= 2πrh + πR2 + π(R2 – r2)

8. Right Circular Cone : A right circular cone is a solid generated by the revolution of a right angled triangle about one of its sides containing the right angle is axis. It is a 3-D shape that has one circular base and narrows smoothly from base to a point, called vertex. In the adjoining figure a cone of height h and radius r is generated by revolving the right ΔAOB along AO.

$$\text{The slant height of the cone is\space}\text{l = AC}\\=\sqrt{r^{2}+h^{2}}\\\text{(i) Volume of cone =}\frac{1}{3}\pi r^{2}h\space\text{cu units}$$

(ii) Curved surface area of cone = πrl sq units.

(iii) Total surface area of a cone = πr(l + r) sq. units.

9. Frustum of a Cone : If a cone is cut by a plane parallel to the base of the cone, then the portion between the plane and base is called the frustum of a cone.

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Let R and r be the radii of base and top of the frustum of a cone. Let h be the height, then

(i) Volume of frustum of right circular cone

$$=\frac{\pi h}{3}[\text{R}^{2}+\text{r}^{2}+\text{Rr}]\space\text{cm}^{3}$$

(ii) Lateral surface area of frustum of right circular cone

= π(R+r) l sq.units

$$\text{where, slant height, l =}\sqrt{h^{2}+(R^{2}-r^{2})}$$

(iii) Total surface area of frustum of right circular cone = Area of base + Area of top + Lateral surface area

=  π [R2 + r2 + l(R + r)]

(iv) Total surface area of bucket

= π [(R2 + r) l + r2]

(∵ It is open at the bigger end)

10. Sphere : A sphere is a solid that is round in shape and the points on its surface are at equidistant from the centre.

(i) A sphere is the locus of a point which moves in space such that its distance from a fixed point in space remains constant. The fixed point is called the centre of sphere and the constant distance is called the radius of sphere.

(ii) If figure, O is centre and r is radius, all radii are equal.

(iii) The section of a sphere cut by any plane is a circle.

(iv) If the cutting plane passes through the centre of the sphere the section is called a great circle.

$$\text{(v)\space}\text{Volume of sphere =}\frac{4}{3}\pi r^{3}\space\text{cu units}$$

(vi) Surface area of sphere = 4πr2 sq. units.

(vii) Volume of a hollow sphere =

$$\frac{4}{3}\pi(\text{R}^{3}-\text{r}^{3})\text{cu units.}$$

where, r = inner radius and R = outer radius.

11. Hemisphere : A plane passing through the centre cuts the sphere in two equal parts called a hemisphere.

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(i) Volume of hemisphere = $$=\frac{2}{3}\pi r^{2}\space\text{cu units}$$ (ii) Curved surface area of hemisphere = 2πr2 sq. units. (iii) Total surface area = 2πr2 + πr2 = 3πr2 sq. units.
  • Sol. Given, height of glass = 10 cm
  • Diameter of glass = Diameter of the hemisphere
  • = 5 cm
  • Thus, radius of glass = Radius of the hemisphere
  • = 2.5 cm
  • Now, apparent capacity of the glass
  • = π(2.5)2 10 cm3
  • = 196.25 cm3
  • And, actual capacity of the glass
  • = Apparent capacity of the glass – Volume of the hemisphere
  • =[196.25-(2/3)π(2.5)3]cm3
  • = 163.54 cm3

Q. A gulabjamun when ready for eating contains sugar syrup of about 30% of its volume. Find approximately how much syrup would be found in 45 such gulabjamuns if each of them is shaped like a cylinder with two hemispherical ends. The complete length of each of them is 5 cm and the diameter is 2.8 cm.[Use π =22/7]

Sol. Given, length of each gulabjamun = 5 cm

  • diameter of each gulabjamun = 2.8 cm
  • Thus, radius of each gulabjamun, r = 1.4 cm
  • Total number of gulabjamuns = 45
  • 3. Oxygen is used for respiration by both autotrophs and heterotrophs for oxidation of glucose to release chemical energy in the form of ATP.P ercentage of syrup in each gulabjamun = 30%
  • and length of the cylindrical part
  • = [5 – (1.4 + 1.4)] cm
  • = (5 – 2.8) cm = 2.2 cm
Q. Two cubes each of volume 64 cm3 are joined end to end to form a solid. Find the surface area and volume of the resulting cuboid.
Ans. 160 cm2, 128 cm3.

Q. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform of 22 m by 14 m. Find the height of the platform.

  • Ans. 2.5 m.

Q. A copper wire 3 mm in diameter is wound around a cylinder whose length is 12 m and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of the copper wire to be 8.88 g/cm.

  • Ans. 12.57 m, 789.41 g.
Chapter 13 Surface Areas and Volumes
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Q. Selvi’s house has an overhead tank in the shape of a cylinder. It is filled up by pumping water from an underground tank that is cuboid in shape. The dimensions of the cuboid are 1.57 m × 1.44 m × 0.95 m. The radius of the overhead tank is 60 cm and its height is 95 cm. Find the height of the water-level in the underground tank after the overhead tank has been filled up completely. Compare the capacities of both the tanks. (Use π = 3.14)

  • Ans. 47.5 cm, 1 : 2.

Q. Water in a canal, 6 m deep and 1.5 m wide is flowing at a speed of 10 km/hr. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed for irrigation ?

  • Ans. 562500 m2.

Q. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 3 km/hr then in hw much time will the tank be completely filled ?

  • Ans. 1 hour and 40 minutes.

Q. A wooden toy rocket is in the shape of a cone mounted on a cylinder. The height of the entire rocket is 26 cm while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm while the diameter of the cylinder is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. [Use π = 3.14]

  • Ans. 195.47 cm2.

Q. A wooden toy was made from the rest of the solid cylinder after scooping out a hemisphere of same radius from each of it. If the height of the cylinder is 10 cm and its base radius is 3.5 cm, find the total surface area. [Use π =22/7]

  • Ans. 374 cm2.

Q. A tent is in the form of a right-circular cylinder of base diameter 4 m and height 2.1 m surmounted by a right circular cone of the same base radius and slant height 2.8 m. Find the area of the canvas used and the cost of canvas at ₹ 500 per square metre.

  • Ans. ₹ 22000.

Q. A solid is in the shape of a cone mounted on a hemisphere, the radius of each of them being 1 cm and the total height of the cone equal to its radius. Find the volume of the solid in terms of π.

  • Ans. π cm3.