Surface Areas and Volumes Revision Notes

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Jasmine Grover

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For 2 dimensional objects like Rectangle Circle, square, they also have surface area but as they are not in 3D space the more appropriate term is Area. Now, a question may come to mind: what is the difference between surface area and volume as well as what is the significance of those. Suppose, you are asked to make a cheese box which is made of a unique number of squares, in that case you need to know the surface area of the given square, or you may be asked to find out the quantity of water that has been supplied in a given time from your water pipeline. In that case you need to know the volume of the water pipe line. There are more examples in our daily life but for that we need to understand the geometrical shape of each object.

Key Terms: Volume and surface area of Cube, Volume and surface area of Cuboid, Volume and surface area of sphere and hemisphere


Explanation: Surface Area

[Click Here for Sample Questions]

Surface area for Rectangle:

If the length of the rectangle is ‘a’, and writhe is ‘b’,

Surface area of Rectangle = ab

Surface area of a Square:

If the length of the rectangle is ‘a’, and writhe is ‘a’,

Surface area of Rectangle = a2

Surface area of a Circle:

If the radius of the circle is ‘r’, then the surface area of the Circle is π×r

Surface area of a Cuboid:

If the length is ‘l’

If the height is ‘h’

If the breadth is ‘b’

Surface area  = 2(lh+hb+bl)

surface area of cuboid

Surface Area for Cube:

If each side of the cube is ‘a’ unit,

Surface area of the Cube = 6a2

Surface Area of a Cylinder:

If the radius is ‘r’ and height is ‘h’,

Surface area of the cylinder = 2πrh

Surface area for a solid cylinder = 2πr (r + h)

Surface Area of a Cone:

If the height is ‘h’

Radius is ‘r’,

Length of slant height is ‘l’,

Curved surface area of a Cone = πrl

Total Surface Area of a solid cone = πr (r + l)

Surface area of a Sphere:

If the radius is ‘r’,Then

Surface area is = 4πr2

Surface area of a hemisphere:

If the radius is ‘r’,

Surface area = 2πr2

Surface area of a solid Hemisphere = 2πr2+πr2=3πr2

surface area of hemi sphere

The video below explains this:

Surface Area and Volume Detailed Video Explanation:


Volume

[Click Here for Sample Questions]

Volume of a Cuboid:

If the length is ‘l’

If the height is ‘h’

If the breadth is ‘b’

Volume of the cuboid is = l×h×b

volume of cuboid

Volume of a Cube:

If each side of the cube is ‘a’ unit, Then

Volume of the cube is = a3

volume of cube

Volume of a Cylinder:

If the radius is ‘r’ and length is ‘l’, Then

Volume of cylinder = πr2h

volume of cylinder

Volume of a Cone:

If the height is ‘h’

Radius is ‘r’,

Length of slant height is ‘l’, then

Volume of the Cone =\(\frac{1}{3}\)πr2h

volume of cone

Volume of a Sphere:

If the radius is ‘r’,then

Surface Area of Sphere =\(\frac{4}{3}\)πr3

volume of Sphere

Volume of a Hemisphere:

If the radius is ‘r’,

Then the surface area is =\(\frac{2}{3}\)πr3 

volume of hemi sphere


Sample Questions

Ques. If volume and surface area of a hemisphere are numerically equal then What is the diameter of the hemisphere? (2 marks)

Ans. Volume of the hemisphere=Surface area of the hemisphere 

2/3×π×r3=2×π×r2

r=3;

The diameter of the hemisphere is 6 units.

Ques. If Volume and surface area of a solid sphere are numerically equal then what is the diameter of the hemisphere? (2 marks)

Ans. Volume of the sphere=Surface area of the sphere

 4/3×π×r3=4×π×r2

r=3;

The diameter of the hemisphere is 6 units.

Ques. Two cubes, each side 2 cm, are joined end to end. Find the surface area of the resulting cuboid? (2 marks)

Ans. Length of resulting cuboid, I = 2(2) = 4 cm

Breadth of resulting cuboid, b = 2 cm

Height of resulting cuboid, h = 2 cm

Total surface area of the resulting cuboid is = 2(4*2+2*2+2*4) = 40 square cm. 

Ques. The sum of the radius of base and height of a solid right circular cylinder is 30 cm. If the total surface area of the solid cylinder is 1600 sq. cm, find the volume of the cylinder.  (3 marks)

Ans. If the radius is ‘r’ and length is ‘l’,

Then surface area of the cylinder is = 2×π×r×l

The surface area for a solid cylinder is = 2πrl+2πr2

 r+l=30; 

2πrl+2πr2=1600 ;

2πrr+l=1600 ;

2πr×30=1800 ;

 r=9.55cm

So, the volume of the cylinder is= πr2l=( 3.14 x 9.55 x 9.55 x 30) – 9.55 = 5858.9 cubic cm.

Ques. The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 40 cm and height 20 m, how much concrete mixture would be required to build 12 such pillars? (3 marks)

Ans. Since the concrete mixture that is to be used to build up the pillars is going to occupy the entire space of the pillar, what we need to find here is the volume of the cylinders.

Radius of base of a cylinder = 40 cm

Height of the cylindrical pillar = 20 m = 2000 cm

So, volume of each cylinder = πrrl=3.14 x 40 x 40 x 2000 = 10,048,000 cm3 = 10.048 m3

Therefore, volume of 12 pillars = volume of each cylinder × 12 = 10.04 × 12 m3 = 120.576 m3

So, 12 pillars would need 120.57 m3 of concrete mixture.

Ques. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas. (3 marks)

Ans. Let, side of small cube is ‘a’;

Then the volume would be a3 unit.

Then, Volume of big solid cube = number of small cube * volume of small cube

12 x 12 x 12=8 x a3;

So, a = 6;

So, the side of each small cube would be 6 units.

Surface area of any cube = 6a2;

Surface area of big cube/ surface area of small cube = 12 x 12/6 x 6= 4:1

Ques. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute? (2 marks)

Ans. River has depth of 3 m;

It has 40 m wide;

Flow rate =2 km/hour = 2000 m/hour = 2000/60 m/min = 33.33 m/min;

So, Volume of the water falling in sea per minute would be = 3 × 40 × 33.33=4000 m3/min;

Ques. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps. (2 marks)

Ans. Joker cap has radius of 7 cm;

It has height of 24 cm;

Then the slant height of the cone would be 242+72=25 cm

Then the surface area of the curved surface would be = =½ × l × 2πr= πrl=3.14 × 7×25 cm2 ; = 549.5 cm2,

Then the surface area of the curved surface is 549.5 cm2.

Ques. The height and the slant height of a cone are 21 cm and 28 cm respectively. Find the volume of the cone. (2 marks)

Ans. We know h2+r2=l2; where r is the radius of the base, h is the height of the cone; l is the slant height;

We have height h = 4 cm, slant height l =5 cm,

Then the radius would be be 52-42 = 3 cm;

Volume of the Cone is =1/3×π×r2×l

= 1/3 x 3.14 x 9 * 5 = 47.099 cm3;

So, the volume of the cone is 47.099 cm3.

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CBSE X Related Questions

1.

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

      2.
      Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

          3.
          An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
          An umbrella has 8 ribs which are equally spaced

              4.

              Form the pair of linear equations for the following problems and find their solution by substitution method.

              (i) The difference between two numbers is 26 and one number is three times the other. Find them.

              (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

              (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

              (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.

              (v) A fraction becomes\(\frac{ 9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.

              (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

                  5.

                  Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

                      6.
                      Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
                      (i) 2, 4, 8, 16, . . . .
                      (ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
                      (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
                      (iv) – 10, – 6, – 2, 2, . . .
                      (v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
                      (vi) 0.2, 0.22, 0.222, 0.2222, . . . .
                      (vii) 0, – 4, – 8, –12, . . . .
                      (viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
                      (ix) 1, 3, 9, 27, . . . .
                      (x) a, 2a, 3a, 4a, . . . .
                      (xi) a, \(a^2, a^3, a^4,\)  . . . .
                      (xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
                      (xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
                      (xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
                      (xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

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