Question

from a square with sides of length 5‚triangular pieces from the four corners are removed to form a regular octagon . find the area removed to the nearest integer?​

Answer 1

Answer:

4

Step-by-step explanation:

5 – 2x = $\sqrt{2}$ x

​

=> x = $\frac{5}{2 +\sqrt{2} }$

Area removed = 2 $x^{2}$    

 

                        = $\frac{2 * 25}{2(\sqrt{2}+1) }$

                        = 25 $(\sqrt{2} - 1)^{2}$ ≈  4.3

So area removed to nearest integer is 4



Answer 2

Answer:

Four 4

Step-by-step explanation: