Important Questions Class 10 Maths Chapter 13 Surface Area and Volumes
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Important Questions Class 10 Maths Chapter 13 Surface Area and Volumes

The surface area of any solid object is the sum of areas occupied by all of its faces. The volume of any object can be understood as the amount of space displaced by the object. Every basic solid shape like a cuboid, cone, cylinder, sphere, etc will have a surface area and volume. If we cut a cone from its center into two parts we obtain a smaller cone and another part is known as the frustum of the cone. For calculating the surface area and volume of solid objects there are many formulas and methods available in mathematics.

The important question for class 10 maths Chapter 13 Surface Area and Volumes can be related to these entities

  • Calculating the surface area of an object formed by other two basic solid shapes.
  • Calculating the volume of an object formed by two basic solid shapes.
  • Description of frustum cone and formula associated with it.

Important Questions Chapter 13 – Surface Areas and Volume

Q1. A canal is 300 cm wide and 120 cm deep. The water in the canal is flowing at a speed of 20 km/h. How much area will it irrigate in 20 minutes if 8 cm of standing water is desired?

Solution:

The volume of water flows in the canal in one hour = width of the canal × depth

of the canal × speed of the canal water = 3 × 1.2 × 20 × 1000 m^3

= 72000 m3

In 20 minutes the volume of water = (72000 × 20)/ 60 = 24000 m^3

Area irrigated in 20 minutes, if 8 cm, i.e., 0.08 m standing water is required

=24000/0.08

= 300000 m^2

= 30 hectares

Q2. A cubical ice-cream brick of edge 22 cm is to be distributed among some children by filling ice-cream cones of radius 2 cm and height 7 cm up to its brim. How many children will get the ice cream cones? 

Solution:

Let n be the number of ice-cream cones.

Volume of cubical ice-cream brick = 22 cm × 22 cm × 22 cm

Radius of cone = r = 2 cm

Height of cone = h = 7 cm

Volume of cone = (1/3)πr^2h = (1/3) × (22/7) × 2 × 2 × 7

So, 

n × Volume of one cone = Volume of cubical ice-cream brick

n × (1/3) × (22/7) × 2 × 2 × 7 = 22 × 22 × 22

n × (1/3) × 4 = 22 × 22

n = (22 × 22 × 3)/4

n = 363

Therefore, 363 children will get the ice cream cones.

Q3. Find the number of solid spheres each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm.

Solution:

Given,

Diameter of solid sphere = 6 cm

Diameter of cylinder = 4 cm

Height of cylinder = h = 45 cm

Radius of sphere = r1 = 6/2 = 3 cm

Radius of cylinder = r2 = 4/2 = 2 cm

Let n be the number of spheres.

n × Volume of one sphere = Volume of cylinder

n × (4/3)πr1^3 = πr2^2h

n × (4/3) × (22/7) × 3 × 3 × 3 = (22/7) × 2 × 2 × 45

n × 9 = 45

n = 45/9

n = 5

Q4. Two cubes, each of side 4 cm are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Length of resulting cuboid, I = 2(4) = 8 cm

Breadth of resulting cuboid, b = 4 cm

Height of resulting cuboid, h = 4 cm

Surface area of resulting cuboid

= 2(lb + bh +hl) = 2 [8(4) + 4(4) + 4(8)]

= 2 (32 + 16 + 32) = 2 (80) = 160 cm^2

Q5. Two cubes each of volume 27 cm3 are joined end to end to form a solid. Find the surface area of the resulting cuboid.

Solution:

Volume of a cube = 27 cm3

⇒ (Side)^3 = (3)^3 ∴ Side = 3 cm

Length of resulting cuboid, l = 2 × 3 = 6 cm

Breadth of resulting cuboid, b = 3 cm

Height of resulting cuboid, h = 3 cm

Surface area of resulting cuboid = 2(lb + bh + hl)

= 2(6 × 3 + 3 × 3 + 3 × 6)

= 2(18 + 9 + 18) = 2(45) = 90 cm^2

Q6. A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the total (inner) suface area of the vessel. (Use π = 22/7

Solution:

r = 14/2 = 7 cm

Inner surface area of the vessel = C.S. area of Hemi-sphere + C.S. area of Cylinder

= 2πr^2 + 2πrh = 2πr(r + h) …C.S. area = curved surface area

= 2 × 22/7 × (77 + 6) = 44 × 13 = 572 cm^2

Q7. Two identical solid hemispheres of equal base radius r сm are struck together along their bases. What will be the total surface area of the combination?

Solution:

The resultant solid will be a sphere of radius r whose total surface area is 4πr^2.

Q8. If two cubes of edge 5 cm each are joined end to end, find the surface area of the resulting cuboid.

Solution:

Total length (l) = 5 + 5 = 10 cm

Breadth (b) = 5 cm, Height (h) = 5 cm

Surface Area = 2 (lb + bh + lh)

= 2(10 × 5 + 5 × 5 + 5 × 10) = 2 × 125 = 250 cm^2

Q9. The slant height of the frustum of a cone is 5 cm. If the difference between the radii of its two circular ends is 4 cm, find the height of the frustum.

Solution:

Let r and R be radii of the circular ends of the frustum of the cone.

Then, R – r = 4, l = 5

We know, l^2 = (R – r)^2 + h^2

⇒ 5^2 = 4^2 + h^2 or h^2 = 25 – 16 = 9

⇒ h = 3 cm

Q10. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Solution:

We have, slant height, l = 4 cm

Let R and r be the radii of two circular ends respectively. Therefore, we have

⇒ 2πR = 18 = πR = 9

⇒ 2πr = 6 = πr = 3

∴ Curved surface area of the frustum = (πR + πr)l

= (9 + 3) × 4 = 12 × 4 = 48 cm^2