Faculty of Mathematics Centre for Education in Waterloo, Ontario N2L 3G1 Mathematics and Computing Grade 6 Math Circles Fall 2018 - November 13/14 Area of Triangles Today we will learn how to find the area of a triangle in several different ways. Before we begin calculating areas, let's review some knowledge about triangles. The Basics A triangle, denoted 4ABC, has the following properties: •4 ABC has 3 vertices, 3 sides, and 3 interior angles • The sum of the interior angles of 4ABC is 180◦ We can label a triangle, 4ABC, as follows: A x c b y z B a C ◦ Sum of interior angles = 180 = \BAC + \ABC + \BCA = x + y + z Pythagorean Theorem Pythagorean Theorem is a very popular mathematical the- orem about the three sides of a right triangle. It states that the square of the hypotenuse is equal to the sum of b c the squares of the other two sides. If we write this using mathematical notation, we get: a a2 + b2 = c2 3 Types of Triangles Equilateral Isosceles Scalene j j j j j All side lengths are equal Two side lengths are equal All side lengths are different All angles are equal Two angles are equal All angles are different Find the Area of a Triangle Method 1: Basic Formula A Basic Formula for Area of a Triangle Let b be the base of the triangle and h be the height of the h triangle. The area of the triangle, A4ABC , is... b×h 1 A4ABC = 2 = 2 (b × h) B C b Example 1: Find the area of 4DEF . Using Pythagorean Theorem, D 13 m 52 + x2 = 132 25 + x2 = 169 x x2 = 144 x = 12 m j j E 5 m F 5×12 2 So, A4DEF = 2 = 30 m Page 2 Method 2: Heron's Formula A Heron's Formula Let a, b, and c be the side lengths of 4ABC. Then, c A = ps(s − a)(s − b)(s − c) b 4ABC 1 B where s = (a + b + c) is the semi-perimeter of 4ABC. a 2 C Example 2: Find the area of 4P QR: 1 P s = (10 + 21 + 17) = 24 2 10 cm R p A4P QR = 24(24 − 10)(24 − 21)(24 − 17) = p24(14)(3)(7) 17 cm p = 7056 21 cm 2 A4P QR = 84 cm Q Example 3: Suppose you have 4ABC. What is the area of 4XYZ? (Image is not to scale.) Notice that 4AXY is isosceles ) XY = 16. Then, A we see that AX = XC = ZX = YZ = 10. Now we can use Heron's formula as follows: 16 10 j jj 1 1 s = (XY + YZ + ZX) = (16 + 10 + 10) = 18 X Y jj 3 2 2 B A = p18(18 − 16)(18 − 10)(18 − 10) j j 4XYZ j 12 = p18(2)(8)(8) p Z = 2304 11 2 C = 48 units The area of 4XYZ is 48 units2. Page 3 Method 3: Complete the Rectangle Suppose we have the coordinates of 4ABC. We do not know what the side lengths are but we are given the coordinates, kind of like the game Battleship! Draw the triangle on the graph below. A(2,3) B(0,0) C(4,1) 2 A 2 3 1 2 2 2 Row 3 1 C 3 1 B 0 0 1 24 3 4 Column Let's find the area of 4ABC. Since we do not know the side lengths, we will enclose 4ABC inside a rectangle. Notice the rectangle is made up of four triangles: 4ABC, 1 , 2 , and 3 . How can we calculate the area of 4ABC? Subtract the areas of 1 , 2 , and 3 from the area of the rectangle! Arectangle = length × width = 4 × 3 = 12 b × h 2 × 3 A = = = 3 1 2 2 2 × 2 A = = 2 2 2 4 × 1 A = = 2 3 2 A4ABC = Arectangle− A 1 − A 2 − A 3 A4ABC = 12 − 3 − 2 − 2 2 A4ABC = 5 units Page 4 Complete the Rectangle If we have a triangle, 4ABC, and we enclose it in a rectangle, then A4ABC = Arectangle− A 1 − A 2 − A 3 where 1 , 2 , and 3 are triangles that form a rectangle with 4ABC. Example 4: Alain has a 7 m × 9 m backyard and decides to construct a triangular-shaped patio. He plots his backyard plan on a graph as follows: 7 m 4 m 4 m 6 5 4 4 m Row 5 m 3 2 1 m 1 8 m 0 0 1 2 3 4 5 6 7 8 9 m Column What is the area of his patio? How much space does he have left in his backyard? Apatio = Arectangle− A 1 − A 2 − A 3 5 × 4 4 × 4 8 × 1 = (5 × 8) − − − 2 2 2 = 40 − 10 − 8 − 4 = 18 m2 The area of Alain's patio is 18 m2. After building his patio, he will have (7 × 9) − 18 = 63 − 18 = 45 m2 of his backyard left. Page 5 Method 4: Shoelace Theorem Also known as \Shoelace Formula," or \Gauss' Area Formula" Shoelace Theorem (for a Triangle) Suppose a triangle has the following coordinates: (a1; b1), (a2; b2), (a3; b3) where a1; a2; a3; b1; b2; and b3 can be any positive number. Then, a b 1 1 1 a2 b2 1 A3 = = j(a2b1 + a3b2 + a1b3) − (a1b2 + a2b3 + a3b1)j 2 a b 2 3 3 a1 b1 where jaj is called the absolute value of a. (i.e. |−3j = 3, j4j = 4) Example 5: Given the coordinates below, what is the area of 4ABC? A(3,4) B(7,9) C(11,4) 3 4 1 7 9 A3 = 2 11 4 3 4 1 = j(7 × 4 + 11 × 9 + 3 × 4) − (3 × 9 + 7 × 4 + 11 × 4)j 2 1 = j(28 + 99 + 12) − (27 + 28 + 44)j 2 1 = j139 − 99j 2 1 = j40j 2 1 = × 40 2 2 A3 = 20 m The area of 4ABC is 20 units2. Page 6 We can actually generalize the Shoelace Theorem and use it to find the area of a shape with any number of sides! Shoelace Theorem (for a shape with n sides) Suppose a shape has n sides, so there are n sets of coordinates: (a1; b1); (a2; b2); :::; (an; bn). Then, a b 1 1 a2 b2 1 1 . An = . = j(a2b1 + a3b2 + ::: + a1bn) − (a1b2 + a2b3 + ::: + anb1)j 2 2 a b n n a1 b1 Example 6: Calculate the area of a hexagon with the following coordinates: (3,9), (9,9), (11,5), (9,1), (3,1), (1,5) 3 9 9 9 11 5 1 A6 = 9 1 2 3 1 1 5 3 9 1 = j(9 × 9 + 11 × 9 + 9 × 5 + 3 × 1 + 1 × 1 + 3 × 5) 2 −(3 × 9 + 9 × 5 + 11 × 1 + 9 × 1 + 3 × 5 + 1 × 9)j 1 = j(81 + 99 + 45 + 3 + 1 + 15) − (27 + 45 + 11 + 9 + 15 + 9)j 2 1 = j244 − 116j 2 1 = j128j 2 1 = × 128 2 2 A6 = 64 units The area of the hexagon is 64 units2. Page 7 Problem Set 1. Find the area of the following triangles: (a) A 3 cm B 2 cm C 2 × 3 A = = 3 cm2 4ABC 2 (b) E 4 mm j D j j F Use Heron's Formula: 1 s = (4 + 4 + 4) = 6 2 p p 2 A4DEF = 6(2)(2)(2) = 48 ≈ 6:928 mm (c) 4DEF where D(2; 8);E(8; 0);F (4; 4) 2 8 1 8 0 1 A4DEF = = j(8 × 8 + 4 × 0 + 2 × 4) − (2 × 0 + 8 × 4 + 4 × 8)j 2 4 4 2 2 8 1 = j72 − 64j 2 1 = j8j 2 2 A4DEF = 4 units 2. Given the perimeter, P4ABC = 42 cm, what is the area of 4ABC? Page 8 B 15 cm 14 cm A 13 cm C First, we need to find x using the perimeter. P4ABC = 42 = x + 15 + 13 ) x = 14 cm Now, we can use Heron's formula to find the area. 1 s = (14 + 15 + 13) = 21 cm 2 p p p 2 A4ABC = 21(21 − 14)(21 − 15)(21 − 13) = 21(7)(6)(8) = 7056 = 84 cm 3. Given DE = 12, EF = 16, and FD = 20, what is the area of 4DEF ? Using Heron's Formula, we get: 1 s = (12 + 16 + 20) = 24 2 p p p 2 A4DEF = 24(24 − 12)(24 − 16)(24 − 20) = 24(12)(8)(4) = 9216 = 96 units Page 9 4. The perimeter of 4ST U is 32 cm. If \ST U = \SUV and TU = 12 cm, what is the area of 4ST U? Drawing 4ST U gives us: S T U 12 cm Since \ST U = \SUV , 4ST U is an isosceles triangle which means that ST = US. Use the perimeter to solve for ST and US. 32 − 12 P = 32 = 12 + ST + US = 12 + 2ST ) ST = = 10 cm 4ST U 2 Using Heron's Formula, we can find the area of 4ST U: 1 s = (12 + 10 + 10) = 16 2 p p p 2 A4ST U = 16(16 − 12)(16 − 10)(16 − 10) = 16(4)(6)(6) = 2304 = 48 cm 5. Find the area of a triangle with the following coordinates: (2,3), (6,3), (8,7) Using Shoelace Theorem, we get that... 2 3 1 6 3 A3 = 2 8 7 2 3 1 = j(6 × 3 + 8 × 3 + 2 × 7) − (2 × 3 + 6 × 7 + 8 × 3)j 2 1 = j56 − 72j 2 1 = |−16j 2 1 = (16) 2 2 A3 = 8 units Page 10 6.