The endpoints of the side of a square lie on the circle and the opposite side of the square tangent to the circle at its midpoint. Find the ratio of the side and the radius.

Respuesta :

Answer:

[tex]\displaystyle \frac{L}{r}=\frac{8}{5}[/tex]

Step-by-step explanation:

Circle and Square

We have a geometric construction as shown in the image below. We can see that

[tex]\displaystyle r+h=L[/tex]

Or, equivalently

[tex]\displaystyle h=L-r[/tex]

The triangle formed by r,h and L/2 is right, because the opposite side of the square is tangent to the circle at its midpoint. This means we can use Pythagoras's theorem:

[tex]\displaystyle r^2=h^2+\left(\frac{L}{2}\right)^2[/tex]

Replacing h

[tex]\displaystyle r^2=(L-r)^2+\left(\frac{L}{2}\right)^2[/tex]

Expanding squares

[tex]\displaystyle r^2=L^2-2Lr+r^2+\frac{L^2}{4}[/tex]

Simplifying

[tex]\displaystyle 2Lr=L^2+\frac{L^2}{4}[/tex]

Multiplying by 4

[tex]\displaystyle 8r=4L+L[/tex]

Joining terms

[tex]\displaystyle 8r=5L[/tex]

Solving for the ratio L/R as required

[tex]\displaystyle \frac{L}{r}=\frac{8}{5}[/tex]

Ver imagen elcharly64

The  ratio of the side and the radius is 8:5

calculation;

Since

r + h = L

So we can write h = L - r

Now we applied the Pythagoras theorem

[tex]r^2 = h^2 +( \frac{L}{2}) ^2\\\\r^2 = (L-r)^2 + ( \frac{L^}{2}) ^2\\\\r^2 = L^2- 2Lr+ r^2 + \frac{L^2}{4}\\\\ 2Lr = L^2 \frac{L^2}{4}\\\\[/tex]

Now

8r = 4L + L

8r = 5L

Learn more about the ratio here: https://brainly.com/question/13176103?referrer=searchResults