Question

practice problems 6 from unit 1, lesson 3 here is an octagon. (note: the diagonal sides of the octagon are not 4 inches long.) a. while estimating the area of the octagon, lin reasoned that it must be less than 100 square inches. do you agree? explain your reasoning. b. find the exact area of the octagon. show your reasoning.

Explanation:

Step1: Find the area of the outer - square

The outer - square has side length \(s = 10\) inches. The area of a square is \(A_{square}=s^2\). So, \(A_{square}=10\times10 = 100\) square inches.

Step2: Find the area of the four corner right - triangles

Each of the four corner right - triangles has base \(b = 3\) inches and height \(h = 3\) inches. The area of a right - triangle is \(A_{triangle}=\frac{1}{2}bh\). For each triangle, \(A_{triangle}=\frac{1}{2}\times3\times3=\frac{9}{2}\) square inches. The total area of the four corner right - triangles is \(4\times\frac{9}{2}=18\) square inches.

Step3: Calculate the area of the octagon

The area of the octagon \(A_{octagon}\) is the area of the outer - square minus the area of the four corner right - triangles. So, \(A_{octagon}=100 - 18=82\) square inches.

Answer:

a. Yes. The octagon is inscribed in a square of area 100 square inches. Since the octagon is inside the square, its area must be less than the area of the square.
b. 82 square inches. The area of the enclosing square is \(10\times10 = 100\) square inches. The four corner right - triangles each have an area of \(\frac{1}{2}\times3\times3=\frac{9}{2}\) square inches, and the total area of the four corner right - triangles is \(4\times\frac{9}{2}=18\) square inches. Subtracting the area of the four corner right - triangles from the area of the square gives \(100 - 18 = 82\) square inches for the area of the octagon.