Page 137 - Math Course 1 (Book 2)

Geometric Area: Polygons and Circles

Write a trigonometric ratio to f nd the length of GF.
A = πr 2 Area of a circle

DF length of opposite side
tan ∠DGF = tan� = = π(44) 2 Substitution
GF length of adjacent side
9 ≈ 6082.1 Use a calculator
tan 36° = m∠DGF = 36, DF = 9
GF The area of the cover is 6082.1 square inches.
To convert to square yards, divide by 1296.

(GF) tan 36° = 9 Multiply each side by GF.
The area of the cover is 4.7
Answer square yards to the nearest
9 tenth.
GF = Divide each side by tan 36°
tan 36°

GF ≈ 12.4 Use a calculator.
Area of an Inscribed Polygon

Area: 1 Area of a regular Example
A = Pa
2 polygon
1
≈ (90)(12.4) P ≈ 90, a ≈ 12.4 Find the area of the shaded region. Assume that
2 the triangle is equilateral. Round to the nearest
tenth.
≈ 557 Simplify.
The area of the shaded
region is the difference
The area of the pentagon is between the area of the
Answer
about 557 square meters. circle and the area of the
triangle. First, f nd the
area of the circle.

Real World Example

A = πr 2 Area of a circle
An outdoor accessories company manufactures
circular covers for outdoor umbrellas. If the cover
is 8 inches longer than the umbrella on each side, = π(7) 2 Substitution
f nd the area of the cover in square yards.
≈ 153.9 Use a calculator
The diameter of the umbrella is 72 inches, and
the cover must extend 8 inches in each direction. To f nd the area of the triangle, use properties
So the diameter of the cover is 8 + 72 + 8 or 88
inches. Divide by 2 to f nd that the radius is 44 of 300–600–900 triangles. First, f nd the length
inches. of the base. The hypotenuse of △RSZ is 7 so RS
3
is 3.5 and SZ is 3.5 . Since YZ = 2(SZ),
YZ = 7 3
X
8 in.

Y Z

129