1. Page 288 – Area of Triangles Surveyors calculate measures of distances and angles so that they can represent boundary lines of parcels of land. The diagram at the right is a plot of John and Renee Walter’s land. What is the area of the region to the nearest square foot? A B C D E 202 ft 180.25 ft 158 ft 125 ft 201.5 ft 97º 82.5º 124.5º 75º 161º

2. Area Formula K will represent the area. K = ½bh Notice in the triangle that sin A = h/c So, h = c sin A Substitute into the area formula and you get K = ½bc sin A You can also use: K = ½ab Sin C & K = ½ac Sin B Use these when you know 2 sides and the included angle AB C a b c h

3. Find the area of triangle ABC if a = 7.5, b = 9 and C = 100º. Round to the nearest tenth. K = ½ab Sin C = ½(7.5)(9)sin 100º = (33.75)(0.9848) = 33.237 The area is about 33.2 sq units. A B C9 7.5 100º

4. If you know the measure of one side and two angles. Remember, K = ½bc sinA The Law of Sines states: b. sin B = c.s sin C So, b = c sin B sin C K = ½c² sinA sin B sinC K = ½a² sinB sinC sin A K = ½b² sinA sinC sinB

5. Find the area of triangle ABC if a = 18.6, A = 19º20’, and B = 63º50’. Round to the nearest tenth. Find C: 180 – (19º20’ + 63º50’) = 96º50’ The find the area of the triangle: A B C 18.6 19º20’ 63º50’ K = ½a² sinB sinC sinA = ½(18.6)² sin63º50’ sin96º50’ sin19º20’ = ½(345.96) (0.8975)(0.9929) (0.3311) = 465.6 So the are a of the triangle is about 465.6 sq units

6. If you know 3 sides of the triangle you can use the Law of cosines and the formula. Find the area of ABC if a = √2, b = 2, and c = 3. First, solve for A using the Law of Cosines a² = b² + c² - 2bc Cos A (√2)² = 2² +3² - 2(2)(3) Cos A 2 = 4 + 9 – 12 Cos A 2 = 13 – 12 Cos A 0.9167 = Cos A 23º33’ = A √22 3AB C Now find the area: K = ½(2)(3) sin 23º33’ = 3(0.3995) = 1.199 So the area is about 1.2 sq units

7. Hero’s Formula If you know the measures of all three sides of the triangle you can also use Hero’s Formula K = √ s(s - a)(s - b)(s - c) s = ½(a + b + c)

8. Use Hero’s Formula to find the area of ABC if a = 20, b = 30 and c = 40. First find s: s = ½(20 + 30 + 40) = ½(90) = 45 Then use Hero’s Formula: K = √45(45 – 20)(45 – 30)(45 – 40) = √45(25)(15)(5) = √84,375 = 290.5 So the area of the triangle is about 290.5 sq units.

9. To solve some applications, you may have to use more than formula. Separate the region into triangles. A B C D E 202 ft 180.25 ft 158 ft 125 ft 201.5 ft 97º 82.5º 124.5º 75º 161º Now the area of the region is the sum of the areas of the triangles. Let a be the side with measure 202, b is 158, c is 201.5, and e is 180.25. a b c e

10. Area of each triangle = ½ac sin B = ½(202)(201.5)sin 82.5º = (20,351.5)(0.9914) = 20,176.5 A B C D E 202 ft 180.25 ft 158 ft 125 ft 201.5 ft 97º 82.5º 124.5º 75º 161º a b c e = ½eb sin D = ½(180.25)(158) sin 75º = 14,239.75)(0.9659) = 13,754.2 This gives you the areas of ABC and CDE

11. Area of ACE Use the Law of Cosines to find AC and CE AC = √202² + 201.5² - 2(202)(201.5)cos82.5º A B C D E 202 ft 180.25 ft 158 ft 125 ft 201.5 ft 97º 82.5º 124.5º 75º 161º a b c e 20.176.5 13,754.2 = 266.0 CE = √180.25² + 158² - 2(180.25)(158) cos 75º = 206.7 The use Hero’s Formula: s = ½(125 + 266 + 206.7) = 298.85 K = √298.85(298.85 – 125)(298.85 – 266)(298.85 – 206.7) = √298.85(173.85)(32.85)(92.15) = 12,540.9 So the area of the land is 20,176.5 + 13,754.2 + 12,540.9or 46,472 sq ft

12. Assignment Page 293 – 294 #12 –20, 27 - 30