TOPICS
Exercise - 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 :-
  Side of traffic signal board = a 
  Perimeter of traffic signal board = 3 × a
  2s = 3a
   s = 3a/2
  (s - a) = (s - b) = (s - c) = 3a/2 - a = a/2
  By Heron's formula,
  area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √3a/2 x a/2 x a/2 x a/2
= a²√3/4 square unit  

  Perimeter of traffic signal board = 180 cm    
  Side of traffic signal board (a) = 180/3 cm = 60 cm
  Using equation (1), area of traffic signal board = √3/4 x (60)² cm²
= √3/4 x 3600 cm²
= √3 x 900 cm²
= 900√3 cm²
    

Question-2 :-  The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisements yield an earning of ` 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay? flyover

Solution :-
  The sides of the triangle are of 122 m, 22 m, and 120 m respectively.   
  Perimeter of triangle = (122 + 22 + 120) m   
  2s = 264 m 
   s = 132 m   
  s - a = 132 - 122 = 10m
  s - b = 132 - 22 = 110m
  s - c = 132 - 120 = 12m
  area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √132 x 10 x 110 x 12
= 1320 m² 

  Rent of 1 m² area per year = Rs 5000 
  Rent of 1 m² area per month = Rs 5000/12
  Rent of 1320 m² area for 3 months = 5000/12 x 3 x 1320
= Rs (5000 × 330) = Rs 1650000
  Therefore, the company had to pay Rs 1650000.
    

Question-3 :-  There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour. park

Solution :-
  Let length, breadth, and height of the rectangular hall be l m, b m, and h m respectively.   
  Area of four walls = 2lh + 2bh = 2(l + b) h   
  Perimeter of the floor of hall = 2(l + b) = 250 m   
  Area of four walls = 2(l + b) h = 250h m²  

  Cost of painting per m² area = Rs 10   
  Cost of painting 250h m² area = Rs (250h × 10) = Rs 2500h  
         
  However, it is given that the cost of paining the walls is Rs 15000.   
  15000 = 2500h 
  h = 6 m   
  Therefore, the height of the hall is 6 m. 
    

Question-4 :-  Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Solution :-
  Let the third side of the triangle be x.   
  Perimeter of the given triangle = 42 cm   
  18 cm + 10 cm + x = 42
  x + 28 = 42
  x = 42 - 28
  x = 14 cm  
  s = (a + b + c)/2
  s = (18 + 10 + 14)/2
  s = 42/2
  s = 21cm
  s - a = 21 - 18 = 3cm
  s - b = 21 - 10 = 11cm
  s - c = 21 - 14 = 7cm
  area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √21 x 3 x 11 x 7
= 21√11 cm² 
    

Question-5 :-  Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.

Solution :-
  Let the common ratio between the sides of the given triangle be x.   
  Therefore, the side of the triangle will be 12x, 17x, and 25x.   
  Perimeter of this triangle = 540 cm   
  12x + 17x + 25x = 540 cm   
  54x = 540 cm 
    x = 10 cm 
  a = 12x = 12 x 10 = 120 cm
  b = 17x = 17 x 10 = 170 cm
  c = 25x = 25 x 10 = 250 cm
  s = (a + b + c)/2
  s = (120 + 170 + 250)/2
  s = 540/2
  s = 270cm      
  s - a = 270 - 120 = 150cm
  s - b = 270 - 170 = 100cm
  s - c = 270 - 250 = 20cm
  area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √270 x 150 x 100 x 20
= 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 :-
  Let the third side of this triangle be x.   
  Perimeter of triangle = 30 cm
  12 cm + 12 cm +  x = 30 cm    
  x = 6 cm 
  s = (a + b + c)/2
  s = (12 + 12 + 6)/2
  s = 30/2
  s = 15cm      
  s - a = 15 - 12 = 3cm
  s - b = 15 - 12 = 3cm
  s - c = 15 - 6 = 9cm
  area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √15 x 3 x 3 x 9
= 9√15 cm²
    
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