Area of a Triangle (Using Sine)

Area of a Triangle (Using Sine)

Starter Use the Sine rule and Cosine rule (one for each triangle) to calculate the missing lengths… 13cm ? 32º 84º ? 5.1m 56º 18cm

Area of a Triangle (Using Sine) • We have so far looked at the Sine rule • And the Cosine rule a SinA b SinB c SinC SinA a SinBb SinC c OR = = = = B c a2 = b2 + c2 - 2bc CosA a A b C

Area of a Triangle (Using Sine) a2 = b2 + c2 - 2bc CosA • Reminder of the Cosine Rule C a2 = b2 + c2 - 2bc CosA b 13cm a2 = 132 + 182 – (2x13x18) Cos84 A a2 = 493 – 468Cos84 84º ? a a2 = 444.080… 18cm a = 21.07cm (2dp) c B

Area of a Triangle (Using Sine) a SinA b SinB c SinC = = • Reminder of the Sine Rule b SinB c SinC C = b ? 32º c SinC b = x Sin B A a 5.1 Sin32 b = x Sin 56 5.1m 56º c b = 7.98m (2dp) B

Area of a Triangle (Using Sine) • The formula for the area of a triangle is; 1/2 base x height OR (base x height) ÷ 2 This relies on us knowing the perpendicular height though Another formula you can use is; 1/2abSinC (1/2 x a x b x SinC)

Area of a Triangle (Using Sine) Learning Objectives All will be able to recognise appropriate situations to use the formula for the area of a triangle using Sine (Grade A) Most will be able to then correctly use the formula to calculate areas (Grade A) Some will then be able to use this in the context of a harder question, such as finding the area of a segment (Grade A*)

Area of a Triangle (Using Sine) 1/2 ab SinC • Work out the area of the triangle opposite… 1/2 ab SinC 1/2 x 6 x 4 x Sin46 1/2 x 17.264…  8.63cm2 (2dp) C 46° 6cm a 4cm b B c A

Area of a Triangle (Using Sine) 1/2 ab SinC • Work out the area of the triangle opposite… 1/2 ab SinC 1/2 x 12.5 x 9.3 x Sin83 1/2 x 115.383…  57.69m2 (2dp) B 12.5m c a 83° C 9.3m A b

Plenary B c a 60º C A 5cm b Area = 1/2 ab sin C Area = 1/2 (5 x 5) sin 60 Area = 1/2 25 sin 60 Area = 1/2 21.65… Area = 10.825… 32.48

Area of a Triangle (Using Sine) Learning Objectives All will be able to recognise appropriate situations to use the formula for the area of a triangle using Sine (Grade A) Most will be able to then correctly use the formula to calculate areas (Grade A) Some will then be able to use this in the context of a harder question, such as finding the area of a segment (Grade A*)

Plenary The Area of the square is a (poor) approximation for π A formula for π 1cm 1cm 90˚ 1cm 1cm 1cm A = πr2 Area = 1/2 abSinC A = π x 12 Area = 1/2 x 1 x 1 x Sin90 A = π Area = 0.5 The Area of a circle with a radius of 1cm will be exactly π… x 4 Area of the Square = 2

Plenary The Area of the Hexagon is a better approximation for π A formula for π 1cm 1cm 60˚ 1cm 1cm 1cm A = πr2 Area = 1/2 abSinC A = π x 12 Area = 1/2 x 1 x 1 x Sin60 A = π Area = 0.433…. The Area of a circle with a radius of 1cm will be exactly π… x 6 Area of the Hexagon = 2.598

Plenary The Area of an Octagon is an even better approximation for π A formula for π 1cm 1cm 1cm 45˚ 1cm 1cm A = πr2 Area = 1/2 abSinC A = π x 12 Area = 1/2 x 1 x 1 x Sin45 A = π Area = 0.353…. The Area of a circle with a radius of 1cm will be exactly π… x 8 Area of the Octagon = 2.828

Plenary The more sides the polygon has, the closer the area is to π A formula for π The angle in the triangle is 360 ÷ n where n is the number of sides 1cm 1cm 360 n 1cm We can just ignore the 1 x 1 part of the area formula 1cm Area = 1/2 abSinC We must remember to multiply by n at the end 360 n ) ( Area = 1/2 x 1 x 1 x Sin 360 n ) ( Area = 1/2 Sin x n The larger the value of n, the more sides the polygon has, and the closer you get to π! 360 n ) ( Area = 1/2nSin

Plenary π = 3.141592654… n 1/2nSin(360/n) 10 2.94 20 3.09 50 3.13330839 100 3.139525976 1000 3.141571983

Summary • We have recapped using the Sine and Cosine Rules • We have seen how to find the area of a triangle without the perpendicular height • Some questions involved using several rules one at a time

Area of a Triangle (Using Sine)