On the figure given show that #bar(OC)# is #sqrt(2)#?

Question

Maverick Smith · Accepted Answer

WOW...I finally got it...although it seems too easy...and probably it is not the way you wanted it!

I considered the two small circles as equal and having radius #1#, each of them (or #u# as unity in distance #bar(PO)#...I think). So the entire base of the triangle (diameter of big circle) should be #3#. According to this, the distance #bar(OM)# should be #0.5# and the distance #bar(MC)# should be one big cirlce radius or #3/2=1.5#. Now, I applied Pythagoras to the triangle #OMC# with: #bar(OC)=x# #bar(OM)=0.5# #bar(MC)=1.5# and I got: #1.5^2=x^2+0.5^2# or: #x^2=1.5^2-0.5^2=(3/2)^2-(1/2)^2=8/4=2# so: #x=sqrt(2)# Does that make sense?

Ella Armstrong · Answer

undefined To show that the length of bar OC is √2, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

In the given figure, OC represents the hypotenuse of a right triangle OAC. Let's denote OA as 'a' and AC as 'a'. Then, using the Pythagorean theorem:

OC² = OA² + AC²

Substituting the values:

OC² = a² + a²
OC² = 2a²

Now, taking the square root of both sides to find the length of OC:

OC = √(2a²)

Since 'a' represents the length of a side of the square, we have:

a = 1 (assuming the side length of the square is 1)

OC = √(2 * 1²)
OC = √2

Thus, the length of bar OC is √2.