NCERT Class 9 Maths Solutions Chapter - 12 Heron's Formula, Ex - 12.2

Ex - 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

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

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

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

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

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

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

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 III^rd triangle

Semi perimeter  

By Heron's formula

Area of triangle 

Area of IIIrd triangle  

                               

Area of paper required for III^rd 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

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

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

^Ex-12.1

Ex-13.1