NCERT Solution for Class 9 Mathematics Chapter 12 - Heron's Formula RR ACADEMY MEERUT

NCERT Solution for Class 9 Mathematics Chapter 12 - Heron's Formula Page/Excercise 12.1

Question 1

A traffic signal board, indicating 'SCHOOL AHEAD', is an equilateral triangle with side 'a'. Find the area of the signal board, using Heron's formula. If its perimeter is 180 cm, what will be the area of the signal board?

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  

Question 2

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122m, 22m, and 120m (see the given figure). The advertisements yield an earning of Rs.5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?   

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.

Question 3

There is a slide in the park. One of its side walls has been painted in the same colour with a message "KEEP THE PARK GREEN AND CLEAN" (see the given figure). If the sides of the wall are 15m, 11m, and 6m, find the area painted in colour.   

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 .

Question 4

Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

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 

Question 5

Sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540 cm. Find its area.

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.

Question 6

An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

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 

NCERT Solution for Class 9 Mathematics Chapter 12 - Heron's Formula Page/Excercise 12.2

Question 1

A park, in the shape of a quadrilateral ABCD, has  = 90�, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

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)

Question 2

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

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)

Question 3

Radha made a picture of an aeroplane with coloured papers as shown in the given figure. Find the total area of the paper used.

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

Question 4

A triangle and a parallelogram have the same base and the same area. If the sides of triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

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.

Question 5

A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

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.

Question 6

An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see the given figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?                                            

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  
                                            

Question 7

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangles of base 8 cm and sides 6 cm each is to be made of three different shades as shown in the given figure. How much paper of each shade has been used in it?   

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 

Question 8

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see the given figure). Find the cost of polishing the tiles at the rate of 50p per 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.   

Question 9

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

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