TOPICS
Unit-12(Examples)

Example-1 :-  Find the area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32 cm. triangle

Solution :-
  Given That : 
  Perimeter of the triangle = 32 cm, 
  a = 8 cm and b = 11 cm. 
  Third side c = 32 cm – (8 + 11) cm = 13 cm 
  So, 2s = 32,  
  s = 16 cm, 
  s – a = (16 – 8) cm = 8 cm, 
  s – b = (16 – 11) cm = 5 cm, 
  s – c = (16 – 13) cm = 3 cm.

  Therefore, area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √16 x 8 x 5 x 3
= 8√30 cm² 
    

Example-2 :-  A triangular park ABC has sides 120m, 80m and 50m. A gardener Dhania has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of Rs 20 per metre leaving a space 3m wide for a gate on one side. triangle

Solution :-
  Given that :
  a = 120m, b = 80m, c = 50m 
  2s = 50 m + 80 m + 120 m = 250 m. 
  s = 125 m 
  Now, s – a = (125 – 120) m = 5 m, 
  s – b = (125 – 80) m = 45 m, 
  s – c = (125 – 50) m = 75 m.

  Therefore, area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √125 x 5 x 45 x 75
= 375√15 m² 

  Also, perimeter of the park =  AB + BC + CA = 250 m 
  Therefore, length of the wire needed for fencing = 250 m – 3 m (to be left for gate) = 247 m 
  And so the cost of fencing = Rs 20 × 247 = Rs 4940
    

Example-3 :-  The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeter is 300 m. Find its area.

Solution :-
  Suppose that the sides, in metres, are 3x, 5x and 7x. 
  Then, we know that 3x + 5x + 7x = 300 (perimeter of the triangle) 
  Therefore, 15x = 300, which gives x = 20. 
  So the sides of the triangle are 3 × 20 m, 5 × 20 m and 7 × 20 m i.e., 60 m, 100 m and 140 m
 triangle
  We have s = (a + b + c)/2 = (60 + 100 + 140)/2 = 300/2 = 150 m
  Now, s – a = (150 – 60) m = 90 m, 
  s – b = (150 – 100) m = 50 m, 
  s – c = (150 – 140) m = 10 m.

  Therefore, area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √150 x 90 x 50 x 10
= 1500√3 m² 
    

Example-4 :-  Kamla has a triangular field with sides 240 m, 200 m, 360 m, where she grew wheat. In another triangular field with sides 240 m, 320 m, 400 m adjacent to the previous field, she wanted to grow potatoes and onions. She divided the field in two parts by joining the mid-point of the longest side to the opposite vertex and grew patatoes in one part and onions in the other part. How much area (in hectares) has been used for wheat, potatoes and onions? (1 hectare = 10000 m2). triangle

Solution :-
  Let ABC be the field where wheat is grown. 
  Also let ACD be the field which has been divided in two parts by joining C to the mid-point E of AD. 
  Given that : 
  a = 200 m, b = 240 m, c = 360 m
  s = (a + b + c)/2 = (200 + 240 + 360)/2 = 800/2 = 400 m
  Now, s - a = 400 - 200 = 200 m
  s - b = 400 - 240 = 160 m
  s - c = 400 - 360 = 40 m

  Therefore, So, area for growing wheat =  √s x (s-a) x (s-b) x (s-c)
= √400 x 200 x 160 x 40
= 16000√2 m² 
= 1.6√2 hectares
=  2.26 hectares

  Let us now calculate the area of triangle ACD.
  Here, we have s = (a + b + c)/2 = (240 + 320 + 400)/2 = 960/2 = 480 m
  s - a = 480 - 240 = 240 m
  s - b = 480 - 320 = 160 m
  s - c = 480 - 400 = 80 m

  So, area of Δ ACD =  √s x (s-a) x (s-b) x (s-c)
= √480 x 240 x 160 x 80
= 38400 m² 
= 3.84 hectares

  Therefore, area for growing potatoes = area for growing onions = (3.84 ÷ 2) hectares = 1.92 hectares. 
    

Example-5 :-  Students of a school staged a rally for cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA; while the other through AC, CD and DA. Then they cleaned the area enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m and ∠ B = 90º, which group cleaned more area and by how much? Find the total area cleaned by the students (neglecting the width of the lanes). triangle

Solution :-
  Given that : 
  AB = 9 m and BC = 40 m, ∠ B = 90°,
  So, AC = √AB² + BC²
  AC = √9² + 40²
  AC = √81 + 1600
  AC = √1681
  AC = 41 m
  Therefore, the first group has to clean the area of triangle ABC, which is right angled.
  Area of Δ ABC = 1/2 x base x height = 1/2 x 40 x 9 = 180 m²

  The second group has to clean the area of triangle ACD, which is scalene having sides 41 m, 15 m and 28 m.
  s = (a + b + c)/2 = (41 + 15 + 28)/2 = 84/2 = 42 m
  s - a = 42 - 41 = 1m
  s - b = 42 - 15 = 27m
  s - c = 42 - 28 = 14m
  Therefore, area of Δ ACD = √s x (s-a) x (s-b) x (s-c)
= √42 x 1 x 27 x 14
= 126 m² 

  So first group cleaned 180 m² which is (180 – 126) =  54 m² more than the area cleaned by the second group. 
  Total area cleaned by all the students = (180 + 126) m² = 306 m².
    

Example-6 :-  Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and produce different crops. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get for their crops? triangle

Solution :-
  Let ABCD be the field. 
  Perimeter = 400 m 
  So, each side = 400 m ÷ 4 = 100 m. 
  i.e. AB = AD = 100 m. 
  Let diagonal BD = 160 m. 
  Then semi-perimeter s of Δ ABD is given by
  s = (a + b + c)/2 = (100 + 100 + 160)/2 = 360/2 = 180 m 
  s - a = 180 - 100 = 80m
  s - b = 180 - 100 = 80m
  s - c = 180 - 160 = 20m
  Therefore, area of Δ ABD = √s x (s-a) x (s-b) x (s-c)
= √180 x 80 x 80 x 20
= 4800 m²
  Therefore, each of them will get an area of 4800 m². 
    
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