Welcome dear students! Today we are going to learn about Heron's Formula from Class 9 Maths. We know that the area of a triangle when its height is given, is 1/2 × base × height. Now suppose that we know the lengths of the sides of a scalene triangle and not the height. Can you still find its area? For instance, you have a triangular park whose sides are 40 m, 32 m, and 24 m. How will you calculate its area? Definitely if you want to apply the formula, you will have to calculate its height. But we do not have a clue to calculate the height. Try doing so. If you are not able to get it, then go to the next section. Heron was born in about 10 C.E. possibly in Alexandria in Egypt. He worked in applied mathematics. His works on mathematical and physical subjects are so numerous and varied that he is considered to be an encyclopedic writer in these fields. His geometrical works deal largely with problems on mensuration written in three books. Book I deals with the area of squares, rectangles, triangles, trapezoids (trapezia), various other specialised quadrilaterals, the regular polygons, circles, surfaces of cylinders, cones, spheres etc. In this book, Heron has derived the famous formula for the area of a triangle in terms of its three sides. The formula given by Heron about the area of a triangle, is also known as Hero's formula. It is stated as: Area of a triangle = √(s(s-a)(s-b)(s-c)). [CHECKPOINT]
where a, b and c are the sides of the triangle, and s = semi-perimeter, i.e., half the perimeter of the triangle = (a+b+c)/2. This formula is helpful where it is not possible to find the height of the triangle easily. Let us apply it to calculate the area of the triangular park ABC, mentioned above (see Fig. 10.2). In the figure, we see a triangle with vertices A, B, and C. The side lengths are given as 40 m, 24 m, and 32 m. Let us take a = 40 m, b = 24 m, c = 32 m, so that we have s = (40+24+32)/2 m = 48 m. s – a = (48 – 40) m = 8 m, s – b = (48 – 24) m = 24 m, s – c = (48 – 32) m = 16 m. Therefore, area of the park ABC = √(s(s-a)(s-b)(s-c)) = √(48×8×24×16) m² = 384 m². We see that 32² + 24² = 1024 + 576 = 1600 = 40². Therefore, the sides of the park make a right triangle. The largest side, i.e., BC which is 40 m will be the hypotenuse and the angle between the sides AB and AC will be 90°. [CHECKPOINT]
We can check that the area of the park is 1/2 × 32 × 24 m² = 384 m². We find that the area we have got is the same as we found by using Heron’s formula. Now using Heron’s formula, you verify this fact by finding the areas of other triangles discussed earlier viz., (i) equilateral triangle with side 10 cm. (ii) isosceles triangle with unequal side as 8 cm and each equal side as 5 cm. You will see that For (i), we have s = (10+10+10)/2 cm = 15 cm. Area of triangle = √(15(15-10)(15-10)(15-10)) cm² = √(15×5×5×5) cm² = 25√3 cm². For (ii), we have s = (8+5+5)/2 cm = 9 cm. Area of triangle = √(9(9-8)(9-5)(9-5)) cm² = √(9×1×4×4) cm² = 12 cm². Let us now solve some more 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 (see Fig. 10.3). In the accompanying figure, we see a triangle with sides labelled 8 cm, 11 cm, and an unknown third side. [CHECKPOINT]
Solution : Here we have 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, i.e., 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(s-a)(s-b)(s-c)) = √(16×8×5×3) cm² = 8√30 cm². Example 2 : A triangular park ABC has sides 120m, 80m and 50m (see Fig. 10.4). 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 ₹20 per metre leaving a space 3m wide for a gate on one side. In the figure, we see triangle ABC with sides 120 m, 80 m, and 50 m. [CHECKPOINT]
Solution : For finding area of the park, we have 2s = 50 m + 80 m + 120 m = 250 m. i.e., 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 park = √(125×5×45×75) m² = 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 = ₹20 × 247 = ₹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. In the figure, we see a triangle with sides proportional to 3, 5, and 7. [CHECKPOINT]
Solution : Suppose that the sides, in metres, are 3x, 5x and 7x (see Fig. 10.5). 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. Can you now find the area [Using Heron’s formula]? We have s = (60+100+140)/2 m = 150 m. and area will be √(150(150-60)(150-100)(150-140)) m² = √(150×90×50×10) m² = 1500√3 m². Now let us move on to the exercises. Exercise 10.1, 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? [CHECKPOINT]
Solution: For an equilateral triangle with side a, the semi-perimeter s = (a+a+a)/2 = 3a/2. Using Heron’s formula, area = √(s(s-a)(s-a)(s-a)). Substituting s, we get √((3a/2)(3a/2-a)(3a/2-a)(3a/2-a)). This simplifies to √((3a/2)(a/2)(a/2)(a/2)) = √(3a⁴/16) = (a²√3)/4. Now, if the perimeter is 180 cm, then 3a = 180, so a = 60 cm. Substituting a = 60 into the area formula, we get (60²√3)/4 = 3600√3/4 = 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 (see Fig. 10.6). 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? In the figure, we see a triangle with sides 122, 22, and 120 m. [CHECKPOINT]
Solution: First, find the area. Let a = 122 m, b = 22 m, c = 120 m. Perimeter = 122 + 22 + 120 = 264 m. So s = 264/2 = 132 m. Then s – a = 132 – 122 = 10 m. s – b = 132 – 22 = 110 m. s – c = 132 – 120 = 12 m. Area = √(132×10×110×12) m² = 1320 m². The earning is ₹5000 per m² per year. For 3 months, which is 1/4 of a year, the rent per m² is 5000/4 = ₹1250. Total rent = 1320 m² × ₹1250 = ₹16,50,000. 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” (see Fig. 10.7). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour. [CHECKPOINT]
In the figure, we see a triangle with sides 15, 11, and 6 m. Solution: Let a = 15 m, b = 11 m, c = 6 m. Perimeter = 15 + 11 + 6 = 32 m. So s = 16 m. s – a = 16 – 15 = 1 m. s – b = 16 – 11 = 5 m. s – c = 16 – 6 = 10 m. Area = √(16×1×5×10) m² = √800 m² = 20√2 m². Question 4: Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm. Solution: Let a = 18 cm, b = 10 cm. Perimeter is 42 cm, so the third side c = 42 – (18+10) = 14 cm. Semi-perimeter s = 42/2 = 21 cm. s – a = 21 – 18 = 3 cm. s – b = 21 – 10 = 11 cm. s – c = 21 – 14 = 7 cm. Area = √(21×3×11×7) cm² = √4851 cm². Factoring it gives 3²×7²×11. So the square root is 21√11 cm². [CHECKPOINT]
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 sides be 12x, 17x, and 25x. The perimeter is 12x + 17x + 25x = 54x. Given perimeter is 540 cm, so 54x = 540, which gives x = 10. Therefore, the sides are 120 cm, 170 cm, and 250 cm. Semi-perimeter s = 540/2 = 270 cm. s – a = 270 – 120 = 150 cm. s – b = 270 – 170 = 100 cm. s – c = 270 – 250 = 20 cm. Area = √(270×150×100×20) cm² = 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 equal sides be a = 12 cm and b = 12 cm. The perimeter is 30 cm, so the third side c = 30 – (12+12) = 6 cm. Semi-perimeter s = 30/2 = 15 cm. s – a = 15 – 12 = 3 cm. s – b = 15 – 12 = 3 cm. s – c = 15 – 6 = 9 cm. Area = √(15×3×3×9) cm² = √1215 cm². Factoring gives 3⁵×5. The square root is 9√15 cm². [CHECKPOINT]
Let us now review the summary of this chapter. In this chapter, you have studied the following point. The area of a triangle with its sides as a, b and c is calculated by using Heron’s formula, stated as Area of triangle = √(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2. This formula is extremely useful when the height of the triangle is not readily available. Remember to always calculate the semi-perimeter first, then find the differences s – a, s – b, and s – c before taking the square root. Practice these steps regularly to master the calculations for your examinations. Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]