1. Honors Geometry Unit 6 Lesson 5 Finding Areas with Trigonometry

2. Objectives I can use trigonometry to find the surface area of a regular polygon I can use trigonometry to find the area of a triangle.

3. Application We can use right triangles to find the surface areas of two figures – Oblique triangles – triangles with no right angle – Regular polygons – n-gons with all sides and angles congruent

4. Vocabulary PQRST is a regular pentagon Center of a regular polygon – the center of a circle circumscribed around the polygon (X) Radius of a regular polygon – radius of the circumscribed circle ( ) Apothem – segment from the polygon center to a side, perpendicular to that side( ) Central angle of a regular polygon – angle with vertex @ polygon center, joining two polygon vertices ( )

5. Find the measure of a central angle of PQRST A pentagon is a regular polygon with 5 sides. Thus, the measure of each central angle of pentagon PQRST is or 72.

6. A.A B.B C.C D.D A.m  DGH = 45° B.m  DGC = 60° C.m  CGD = 72° D.m  GHD = 90° In the figure, regular hexagon ABCDEF is inscribed in Find the measure of a central angle.

7. Formula

8. Use the Formula for the Area of a Regular Polygon Find the area of the regular hexagon. Round to the nearest tenth. Step 1 Find the measure of a central angle. A regular hexagon has 6 congruent central angles, so Step 2 Find the apothem. Apothem PS is the height of isosceles ΔQPR. It bisects  QPR, so m  SPR = 30. It also bisects QR, so SR = 2.5 meters. ΔPSR is a 30°-60°-90° triangle with a shorter leg that measures 2.5 meters, so ≈ 65.0 m 2 30° 2.5

9. A.A B.B C.C D.D A.48 cm 2 B.144 cm 2 C.166.3 cm 2 D.182.4 cm 2 What is the area of a regular hexagon with side length of 8 centimeters? Round to the nearest tenth if necessary. Practice

10. A.A B.B C.C D.D A.784 in 2 B.676 in 2 C.400 in 2 D.196 in 2 What is the area of a square with an apothem length of 14 inches? Round to the nearest tenth if necessary. Practice

11. A.A B.B C.C D.D A.346 m 2 B.299.6 m 2 C.173 m 2 D.149.8 m 2 Find the area of a regular triangle with a side length of 18.6 meters. Practice

12. Next Application… Area of an oblique triangle – Given two sides of any triangle and the measure of an angle between them – Use trigonometry to find its surface area Recall previous formula for the area of a triangle: A = ½ bh

13. We will use an obtuse triangle Label sides a, b, and c, opposite their corresponding angles Draw a height, h, inside

14. Next… In order to use A = ½ bh, we need b and h, but all we know are a, b, and the measure of angle C (for example) we need “h”! Look at triangle BDC inside: – How can we write a trig ratio using sides h and a? – We can use this to solve for “h”!

15. So Far we have… Solve this for “h”: h = a sin C Now we have the info we need to use A = 1/2bh! A = ½ bh substitute “a sin C” for “h” A = ½ a b sin C

16. IN CONCLUSION The area of an oblique triangle is one-half the product of the lengths of two sides, times the sine of their included angle! For any triangle, ABC Area = ½ bc sinA = ½ ab sinC = ½ ac sinB

17. Practice Find the area of a triangular lot having two sides of lengths 90m and 52m and an included angle of 102°. Draw it: Area = ½ (90)(52) sin 102 ≈ 2288.87 m 2

18. Practice Find the area of a triangle with sides 6 and 10 and an included angle of 110° Round to the nearest hundredth. Area = 28.19

19. Practice Find the area of a triangle with side lengths 92 and 30 with an included angle 130°. Area = 1057.14