Chapter 12 : Heron's Formula - Ncert Solutions for Class 9 Maths CBSE

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Chapter 12 - Heron's Formula Excercise Ex. 12.1

Solution 1
Side of traffic signal board = a
Perimeter of traffic signal board = 3  a
By Heron's formula
Perimeter of traffic signal board = 180 cm
Side of traffic signal board
Using equation (1), area of traffic of signal board
Solution 2
We may observe that sides of triangle a, b, c are of 122 m, 22 m, and 120 m respectively
    Perimeter of triangle = (122 + 22 + 120) m
                 2s = 264 m
                   s = 132 m
    By Heron's formula
Rent of 1 m2 area per year = Rs.5000
    Rent of 1 m2 area per month = Rs
    Rent of 1320 m2 area for 3 months 
                            = Rs.(5000  330) = Rs.1650000
    So, company had to pay Rs.1650000.

Solution 3
We may observe that the area to be painted in colour is a triangle, having its sides as 11 m, 6 m, and 15 m.
    Perimeter of such triangle = (11 + 6 + 15) m
                                    2 s = 32 m
                                       s = 16 m
By Heron's formula
    So, the area painted in colour is .
Solution 4
Let third side of triangle be x.
  Perimeter of given triangle = 42 cm
        18 cm + 10 cm + x = 42
                                 x = 14 cm
By Heron's formula
Solution 5
Let the common ratio between the sides of given triangle be x.
So, side of triangle will be 12x, 17x, and 25x.
  Perimeter of this triangle = 540 cm
             12x + 17x + 25x = 540 cm
                         54x = 540 cm
                            x = 10 cm
  Sides of triangle will be 120 cm, 170 cm, and 250 cm.
By Heron's formula
    So, area of this triangle will be 9000 cm2.
Solution 6
Let third side of this triangle be x.
Perimeter of triangle = 30 cm
  12 cm + 12 cm + x = 30 cm
                           x = 6 cm
By Heron's formula

Chapter 12 - Heron's Formula Excercise Ex. 12.2

Solution 1
Let us join BD.
In BCD applying Pythagoras theorem
BD2 = BC2 + CD2
       = (12)2 + (5)2
       = 144 + 25
BD2 = 169
  BD = 13 m
Area of BCD
                 
For ABD
                   
By Heron's formula
Area of triangle 
                        
      Area of park = Area of ABD + Area of BCD
                         = 35.496 + 30 m2
                         = 65.496 m2
                         = 65. 5 m2 (approximately)

Solution 2
For ABC
AC2 = AB2 + BC2
(5)2 = (3)2 + (4)2
So, ABC is a right angle triangle, right angled at point B.
Area of ABC
For DAC
Perimeter = 2s = DA + AC + CD = (5 + 5 + 4) cm = 14 cm
            s = 7 cm
By Heron's formula
Area of triangle
Area of ABCD = Area of ABC + Area of ACD
    = (6 + 9.166) cm2 = 15.166 cm2 = 15.2 cm2 (approximately)

Solution 3
                                  
For triangle I

This triangle is a isosceles triangle.
Perimeter = 2s = (5 + 5 + 1) cm = 11cm
          
For quadrilateral II
This quadrilateral is a rectangle.
Area = l  b = (6.5  1) cm2 = 6.5 cm2

For quadrilateral III
This quadrilateral is a trapezium.
Perpendicular height of parallelogram
                                                     
Area = Area of parallelogram + Area of equilateral triangle
  = 0.866 + 0.433 = 1.299 cm2
Area of triangle (iv) = Area of triangle in (v)
Total area of the paper used = 2.488 + 6.5 + 1.299 + 4.5  2
                                         = 19.287 cm2

Solution 4
For triangle
Perimeter of triangle = (26 + 28 + 30) cm = 84 cm
2s = 84 cm
s = 42 cm
By Heron's formula
Area of triangle
Area of triangle 
                     
Let height of parallelogram be h
Area of parallelogram = Area of triangle
                h  28 cm = 336 cm2
                            h = 12 cm
So, height of the parallelogram is 12 cm.

Solution 5
 
Let ABCD be a rhombus shaped field.

For BCD
Semi perimeter,
Area of triangle
Therefore area of BCD
                                
Area of field = 2  Area of BCD
 = (2  432) m2 = 864 m2
    Area for grazing for 1 cow= = 48 m2
    Each cow will be getting 48 m2 area of grass field.

Solution 6
For each triangular piece
 Semi perimeter  
 
By Heron's formula
Area of triangle
 

Since, there are 5 triangular pieces made of two different colours cloth.

So, area of each cloth required

                                           
Solution 7
We know that
Area of square  (diagonal)2
Area of given kite 
 
Area of 1st shade = Area of 2nd shade
 

So, area of paper required in each shape = 256 cm2.

For IIIrd triangle
Semi perimeter  
 
By Heron's formula
 
Area of triangle

 
Area of IIIrd triangle 
 
                              
 
Area of paper required for IIIrd shade  = 17.92 cm2

Solution 8
We may observe that
Semi perimeter of each triangular shaped tile
         
By Heron's formula
 
Area of triangle
 
 
Area of each tile
                       
 
                        = (36  2.45) cm2
                        =88.2 cm2
Area of 16 tiles = (16  88.2) cm2= 1411.2 cm2

Cost of polishing per cm2 area = 50 p
Cost of polishing 1411.2 cm2 area = Rs. (1411.2  0.50) = Rs.705.60
So, it will cost Rs.705.60 while polishing all the tiles.
 

Solution 9
Draw a line BE parallel to AD and draw a perpendicular BF on CD.

Now we may observe that ABED is a parallelogram.
BE = AD = 13 m
ED = AB = 10 m
EC = 25 - ED = 15 m
For BEC

Semi perimeter  
 
By Heron's formula
 
Area of triangle
 
Area of BEC
 
                    m2 = 84 m2
 
 
 
        Area of  BEC
 
           
 
Area of ABED = BF  DE = 11.2  10
                   = 112 m2
Area of field = 84 + 112
                   = 196 m2

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