1. Day 116 – Perimeter and area of a triangle on x-y plane

2. Introduction
We have different ways in which we can use to find the area of a triangle. Some of these ways include using the formula 1 2 𝑏ℎ, 1 2 𝑎𝑏 sin 𝜃 and using Heron’s formula. However, some of these methods may not help us to find the area of a triangle when we are only given the coordinates of its vertices. We have to perform a number of computations to use them. In this lesson, we will find the area and the perimeter of triangle on xy- plane.

3. Vocabulary
Perimeter
This is the distance around a shape on a plane.

4. Finding perimeter of a triangle on xy- plane From the definition of the perimeter, we find the perimeter of a triangle by adding the lengths of all the three sides. The length from one vertex to another is found using the distance formula. The distance formula If 𝐴( 𝑥 1 , 𝑦 1 ) and B( 𝑥 2 , 𝑦 2 ) are two points on xy- plane, the distance from A to B is given by the formula, ( 𝑥 2 − 𝑥 1 ) 2 + ( 𝑦 2 − 𝑦 1 ) 2

5. Example 1 Find the length of a triangle with vertices and hence, the perimeter at 𝐴 −4,−4 , 𝐵(8,−4) and 𝐶(8,1). Solution Distance from A to B is 8−−4 2 + (−4−−4) 2 =12 Distance from B to C is −4−1 2 + (8−8) 2 =5 Distance from B to C is 8−−4 2 + (1−−4) 2 =13 Perimeter of ∆𝐴𝐵𝐶=𝐴𝐵+𝐵𝐶+𝐴𝐶 = = 30 𝑢𝑛𝑖𝑡𝑠

6. Finding the area of a triangle on xy- plane We can use Heron’s formula to find the area of a triangle after getting the perimeter. According to Heron’s formula, Area of Triangle = 𝑠(𝑠−𝑎)(𝑠−𝑏)(𝑠−𝑐) where s = 1 2 𝑡ℎ𝑒 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 and a,b,c are lengths of each side. If we use this formula to find area of the ∆𝐴𝐵𝐶 in example 1 above, 𝑠= 30 2 , 𝑎=12, 𝑏=5 and 𝑐=13. 𝐴= 15(15−12)(15−5)(15−13) 𝐴= 15×3×10×2 𝐴= 900 =30 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠

7. However, Heron’s formula can be tedious especially when the lengths of the sides are irrational numbers. Given a triangle with vertices at 𝑥 1 , 𝑦 1 , 𝑥 2 , 𝑦 2 , ( 𝑥 3 , 𝑦 3 ) a simpler way to find the area of triangle is to use the formula, 𝐴= 1 2 ( 𝑥 1 𝑦 2 − 𝑦 3 + 𝑥 2 𝑦 3 − 𝑦 1 + 𝑥 3 𝑦 1 − 𝑦 2 ) This formula can easily be remembered by finding half the value of the determinant of the matrix, 𝑥 1 𝑥 2 𝑥 3 𝑦 1 𝑦 2 𝑦 3

8. Example 2 Find the area of a triangle with vertices at 1,1 , 4,2 and (1,4). Solution Let 1,1 , 4,2 and (1,4) be 𝑥 1 , 𝑦 1 , 𝑥 2 , 𝑦 2 and ( 𝑥 3 , 𝑦 3 ) 𝐴= 1 2 ( 𝑥 1 𝑦 2 − 𝑦 3 + 𝑥 2 𝑦 3 − 𝑦 1 + 𝑥 3 𝑦 1 − 𝑦 2 ) 𝐴= 1 2 (1 2−4 +4 4−1 +1(1−2) 𝐴= 1 2 −2+4 3 −1 𝐴=4.5 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠

9. homework
Find the perimeter of a triangle with vertices at 4,2 , 12,2 𝑎𝑛𝑑 2,8 .

10. Answers to homework
24 𝑢𝑛𝑖𝑡𝑠

11. THE END