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

Information given here:

Length of sides of triangle = 5

To form a regular octagon the sides of square are removed in triangular pieces

To find:

Area removed from square to the nearest integer  = ?

Let’s assume that the side removed from the one corner of square = x

In that case, side of octagon =  5 – 2x

                                                = $\sqrt{2}$ x

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

Now, Area removed from the square = 2$x^{2}$

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

After solving this, we get

                        = 25 ($\sqrt{2}$ - 1)^{2}[\tex]

                        ≈  4.3

Since 4 is the nearest integer from 4.3

Hence the area removed to the nearest integer is 4.

Answer 2

Answer:

4

Step-by-step explanation:

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?

Let say Equal Sides of Triangle cut = b

Then side of Hexagon = 5 - 2b

Side of Hexagon = √b² + b² = b√2

5 - 2b = b√2

=> b(2 + √2) = 5

=> b = 5/(2 + √2)

Area of Triangle = (1/2)b * b = (1/2)  25/(4 + 2 + 4√2)   = 25/(12 + 8√2)

Area of 4 Triangles = 25/(3 + 2√2)

= 25(3 - 2√2)/(9 - 8)

= 25 ( 3 - 2√2)

= 4.3

= 4  (to the nearest integer)