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Honors Geometry Unit 6 Lesson 5 Finding Areas with Trigonometry 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. 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 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 ( ) XQ • Apothem – segment from the polygon center to a side, perpendicular to that side(XN ) • Central angle of a regular polygon – angle with vertex @ polygon center, joining two polygon vertices (QXR ) 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 360 pentagon PQRST is or 72. 5 In the figure, regular hexagon ABCDEF is inscribed in Find the measure of a central angle. A. mDGH = 45° B. mDGC = 60° C. mCGD = 72° D. mGHD = 90° A. B. C. D. A B C D Formula 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 mSPR = 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 30° 2.5 ≈ 65.0 m2 Practice What is the area of a regular hexagon with side length of 8 centimeters? Round to the nearest tenth if necessary. A. 48 cm2 B. 144 cm2 C. 166.3 cm2 D. 182.4 cm2 A. B. C. D. A B C D Practice What is the area of a square with an apothem length of 14 inches? Round to the nearest tenth if necessary. A. 784 in2 B. 676 in2 C. 400 in2 D. 196 in2 A. B. C. D. A B C D Practice Find the area of a regular triangle with a side length of 18.6 meters. A. 346 m2 B. 299.6 m2 C. 173 m2 D. 149.8 m2 A. B. C. D. A B C D 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 We will use an obtuse triangle B c A h a D b • Label sides a, b, and c, opposite their corresponding angles • Draw a height, h, inside C B Next… c h A a C D b • 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”! B • Look at triangle BDC inside: – How can we write a trig ratio using sides h and a? – We can use this to sin C solve for “h”! h h a D a C So Far we have… h sin C a • Solve this for “h”: h = a sin C • Now we have the info we need to use A = 1/2bh! • A = ½ bh • A = ½ a b sin C substitute “a sin C” for “h” 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 Practice • Find the area of a triangular lot having two sides of lengths 90m and 52m and an included angle of 102°. • Draw it: 52 102 90 • Area = ½ (90)(52) sin 102 ≈ 2288.87 m2 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 Practice • Find the area of a triangle with side lengths 92 and 30 with an included angle 130°. • Area = 1057.14