Question

Adam is building a deck in his backyard around a tree. The deck is in the shape of a regular octagon. In the center of the deck, there is a hole for the tree. If a = 10.9 ft, s = 9 ft, and the diameter of the hole is equal to 7 ft, find the area of the deck to the nearest tenth of a square foot. (Use 3.14 as an approximation of pi.)

Answer 1

I believe the answer is 334.31.

Explanation:

a= the height of a triangle. s= the base of the triangle. b x h / 2 = area of the triangle. 9*10.9=98.1. 98.1/2= 49.05. There are 8 triangles in an octagon, so 8 x 49.05 = 392.4

Since the circle is a hole, we have to subtract that from the area. Half of the diameter is the radius, which is 3.5. Formula for a circle is pi x r^2

Pi = 3.14, so 3.14 x 3.5^2 = 38.465

372.775-38.465= 334.31

Answer 2

Area of deck = 535.20 square feet

Explanation:

Given : Side of regular octagon, a = 10.9 feet

`\text{Area of regular octagon = }2(1+√(2))\cdot a^2\\\\\implies Area = 2(1+√(2))10.9^2\\\\\implies Area = 573.67\text{ feet square}`

Diameter of circular hole = 7 feet

So, radius of the circular hole = 3.5 feet

Area of hole = π·radius²

= 3.14 × (3.5)²

= 38.47 square feet

Therefore, Area of deck = Area of octagon - Area of hole

= 573.67 - 38.47

= 535.20 square feet