TOPICS

Unit-12(Examples)

Heron's Formula

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

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.

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.

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 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).

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).

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?

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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