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

VocabularyPQRST 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 ( )

XQ

XN

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

pentagon PQRST is or 72.5

360

A. AB. BC. CD. D

A. mDGH = 45°

B. mDGC = 60°

C. mCGD = 72°

D. mGHD = 90°

In the figure, regular hexagon ABCDEF is inscribed in Find the measure of a central angle.

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 isoscelesΔQPR. It bisects QPR, so mSPR = 30. It also bisects QR, so SR = 2.5 meters. ΔPSR is a 30°-60°-90° triangle with ashorter leg that measures 2.5 meters, so ≈ 65.0 m2

30°

2.5

A. AB. BC. CD. D

A. 48 cm2

B. 144 cm2

C. 166.3 cm2

D. 182.4 cm2

What is the area of a regular hexagon with side length of 8 centimeters? Round to the nearest tenth if necessary.

Practice

A. AB. BC. CD. D

A. 784 in2

B. 676 in2

C. 400 in2

D. 196 in2

What is the area of a square with an apothem length of 14 inches? Round to the nearest tenth if necessary.

Practice

A. AB. BC. CD. D

A. 346 m2

B. 299.6 m2

C. 173 m2

D. 149.8 m2

Find the area of a regular triangle with a side length of 18.6 meters.

Practice

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

• Label sides a, b, and c, opposite their corresponding angles

• Draw a height, h, inside

C

B

A

h

b

ac

D C

B

A

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”!

h

b

ac

D C

B

A

ah

C

B

D

a

hC sin

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

a

hC sin

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, ABCArea = 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:

• Area = ½ (90)(52) sin 102 ≈ 2288.87 m2

90

52

102

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