Points to Remember- Mensuration Class 8 Notes | EduRev

Class 8 Mathematics by Full Circle

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Class 8 : Points to Remember- Mensuration Class 8 Notes | EduRev

The document Points to Remember- Mensuration Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by Full Circle.
All you need of Class 8 at this link: Class 8

Points to Remember 

  • Area of a parallelogram = Base * Height
  • Area of a triangle = 1/2 * Base * Height
  • Area of a trapezium = 1/2 * [Sum of parallel sides] * Height
  • Area of a rhombus = 1/2 * Product of diagonals
  • Surface area of
    (i) a cuboid = 2[lb + bh + hl]
    (ii) a cube = 6l2
    (iii) a cylinder = 2πr(r + h)
  • Volume of
    (i) cuboid = l * b * h
    (ii) cube = l3
    (iii) cylinder = πr2h
  • 1 m3 = 1000 litres

We Know That
A square, a rectangle, a trapezium, a rhombus, a parallelogram, a triangle, a circle, etc., are plane figures and the surfaces enclosed by their boundaries are called areas. We have formulae to find their areas. The perimeter is the distance around a figure. The plane figures are also called 2-D shapes. The solids such as cubes, cuboids, cylinders are called 3-D shapes. A 3-D shape is bounded by faces. These faces can be rectilinear or curved or both. 

We also know that:
(i) The area of a square = Side * Side
(ii) The area of a rectangle = Length * Breadth
(iii) The area of a circle = πr2   [where r is the radius]
(iv) The area of triangle = 1/2 * Base * Altitude
(v) The area of a parallelogram = Base * Height

Note: 
I. All angles of a regular polygon have equal degree measures.
II. All sides of a regular polygon are equal in length.

Solved Examples:
Q1. What length of a solid cylinder which is 2 cm in diameter must be taken to recast into a hollow cylinder of external diameter 20 cm, 0.25 cm thick and 15 cm long?
a. 54.06 cm
b. 74.06 cm
c. 34.06 cm
d. 64.06 cm
Solution: 
B. The diameter of the solid cylinder = 2 cm or the radius = 1 cm; height h =?
V1 = πr²h = π(1)²h = πh
For the hollow cylinder, H = 15 cm; external diameter = 20 cm or external radius = 10 cm. Hence, internal diameter = 10-0.25 (thickness+ = 9.75 cm. Therefore,
V2 = π [ 10² – (9.75²) ] × 15 = 15π × 19.75 × 0.25
Also, V= V2, which gives
h = 74.06 cm

Q2. If the lateral surface of the cylinder is 500 cm² and its height is 10 cm, then find the radius of its base.
a. 7.96 m
b. or 7.96 cm
c. 7.96 cm²
d. 9.61 cm²
Solution: 
B. The lateral surface area is A =2πrh. The curved surface area is A =  500 cm² and its height is 10 cm, hence
A =2πrh
500 = 2 × 3.14 × r × 10
500 = 62.8r
r = 500/62.8
= 7.96
Therefore the radius of the cylinder is 7.96 cm

Q3. A horse is tethered by a rope 10 m long at a point. Find the area of the region where it can graze (π = 3.14)
Solution:
The area of the region the horse can graze is circular with a radius equal to the length of the rope.
Area of the circle is πr²
= 3.14 × 10²
= 3.14 × 100
=314
Hence the area of the region the horse can graze is 314cm²

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