Chapter 16: Mensuration

Introduction
    In this chapter we learn about the perimeter and area of a rectangle and a square and circumference (perimeter) of a circle and area of a triangle, a parallelogram and a circle etc.

PERIMETER and area
1.    Perimeter
    The perimeter of a plane figure is the length of its boundary. The units of perimeter are the same as the units of length, i.e., cm, m, etc.
    
2.    Area
    The measurement of the region enclosed by a plane figure is called its area.
    
3.    STANDARD UNITS OF AREA
    The units of area are sq cm (written as cm²), sq metres (written as m²), etc. 
    Conversion of units
    Length units            Area units
    1 cm = 10 mm         1 cm² = (10 × 10) mm² = 100 mm²    
    1 dm = 10 cm         1 dm² = (10 × 10) cm² = 100 cm²    
    1 m = 10 dm          1 m² = (10 × 10) dm² = 100 dm²    
    1 m = 100 cm        1 m² = (100 × 100) cm² = 10000 cm²    
                                  1 hectare = 10000 m²    

4.    AREA  AND PERIMETER OF A RECTANGLE
    Consider a rectangle with length = l units and breadth = b units. Then, we have :
    (i)     Area of the rectangle = (l × b) sq units

Illustration 1     
        A door of length 1 m and breadth 0.5 m is on a wall. The length of the wall is 4.5 m and the breadth is 3.6 m as shown in Fig. Find the cost of white washing the wall, if the rate of white washing the wall is Rs 20 per m². 

Solution
        We have,                                             
        l = Length of the wall = 4.5 m, 
        b = Breadth of the wall = 3.6 m
    ∴    Area of the wall = l × b = 4.5 m × 3.6 m = 16.2 m² 
          Area of the door = 1 m × 0.5 m = 0.5 m² 
    ∴   Area to be white washed = 16.2 m² – 0.5 m² = 15.7 m² 
        Cost of white washing the wall = Rs (15.7 × 20) = Rs 314

Illustration 2    
        The length and breadth of a playground are 75 m 20 cm and 34 m 80 cm, respectively. Find the cost of levelling it at Rs 1.50 per square metre. How long will a boy take to go three times round the field, if he walks at the rate of 1.5 m/sec.

Solution        
         We have,
        Length of the playground  = 75 m 20 cm = 75.20 cm 
        Breadth of the playground  = 34 m 80 cm = 34.80 m .
    ∴    Area of the playground      = 75.20 × 34.80 m² = 2616.96 m² 
   ∴   Cost of levelling = Rs 2616.96 × 1.50 = Rs 3925.44 
        Perimeter of the playground  = 2 (Length + Breadth) 
            = 2 (75.20 + 34.80) m = 2 × 110m = 220m 

Illustration 3    
        A room is 9 m long, 8 m broad and 6.5 m high. It has one door of dimensions 2 m ´ 1.5 m and three windows each of dimensions 1.5 m × 1 m. Find the cost of white washing the walls at Rs 3.80 per square metre.
  Solution             
        The dimensions of the room are:
        l = Length = 9 m, b = Breadth = 8 m and h = Height = 6.5 m
  ∴       Area of 4 walls of the room = 2 {(l + b) × h} = 2 {9 + 8) × 6.5}m² =221m² 
             Area of 1 door = (2 × 1.5) m² =3 m²
             Area of 3 windows = {3 × (1.5 × l)}m² = 4.5 m²
   ∴     Area to be white washed = Area of 4 walls – Area of 1 door –  Area of 3 windows
                       = (221 –3 – 4.5)m² =213.5m² 
    So, cost of white washing = Rs (213.5 × 3.80) = Rs 811.30

AREA OF PATHS AND VERANDAHS
    Area of verandah is nothing but the difference between the areas of two rectangles.
    Illustration 4    
        A rectangular field is 45 m × 30 m. A path 1.5 m wide is to be constructed along the sides inside the field. Find the area of the path and its cost of construction at the rate of Rs 1.50 per square metre. 
    Solution    
        Length of the field = 45 m                                                
        Width of the field = 30 m 
        Area of the field  = 45 × 30 sq m = 1350 sq m                
        Length of the inner rectangle = 45 – 3 or 42 m 
        Width of the inner rectangle = 30 – 3 or 27 m
        Area of the inner rectangle = 42 × 27 sq m = 1134 sq m
        Area of path = 1350 – 1134 or 216 sq m 
        Cost of constructing 1 sq m of path = Rs 1.50 
        Cost of constructing 216 sq m of path = Rs 1.50 × 216 = Rs 324

Area of Cross Roads
    ABCD is a rectangular piece of ground. Two roads one parallel to the length and the other parallel to the breadth together are called cross roads.

Illustration 5     
        A field is 75 m long and 32 m wide. Two 3 m wide roads are constructed in the centre of the field one parallel to its length and the other parallel to its breadth. Find (a) area of cross roads (B) the cost of levelling the roads at Rs 1.80 per square metre (C)        area of the remaining field.

Solution 

(A)     Area of the road parallel to the length = 75 × 3 m² = 225 m² 
        Area of the road parallel to the breadth = 32 × 3 m²  = 96 m² 
        Area of shaded portion     = 3 × 3 or 9 m²
        ∴     Area of the roads = (225 + 96 – 9) m² = 312 m² 
 (B)    Cost of levelling 1 m² road = Rs 1.80
        Cost of levelling 312 m² road = Rs 312 × 1.80 = Rs 561.60
(C)       Area of field = 75 × 32 or 2400 m²
        Area of remaining field = (2400 – 312) m² = 2088 m²

Illustration 8 
        The base of a parallelogram is thrice its height. If the area is 876 cm², find the base and height of the parallelogram.
    Solution
        Let the height of the parallelogram  be x cm. 
        Then, base = 3x cm.
    ∴   Area of the parallelogram= (x × 3x) cm² = 3x² cm²
        But, area of the parallelogram is given as 867 cm².
    ∴    3x² = 867 
    ⇒     x² = 289 
    ⇒     x² = 172 
    ⇒     x = 17 cm.
    Thus, height = 17 cm and base = (3 × 17) cm = 51 cm. 

Area related to circle
    Circle : Circle is a path of a moving point, which moves in such a manner that its distance from a fixed point is always equal. The fixed point is called centre of the circle and the fixed distance is called radius of the circle.